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      <div class="breadcrumb"><a href="index.html">Home</a> / Advanced DP</div>
      <h1>Advanced Dynamic Programming</h1>
      <p>Beyond the basics: tree DP, bitmask DP, digit DP, interval DP, knapsack variants, convex hull trick, divide &amp; conquer optimization, Knuth's optimization, SOS DP, and profile DP. Every technique you need to solve the hardest DP problems in competitive programming -- with full code in Python and C++.</p>
    </div>

    <div class="toc">
      <h4>Table of Contents</h4>
      <a href="#tree-dp">1. DP on Trees</a>
      <a href="#bitmask-dp">2. Bitmask DP</a>
      <a href="#digit-dp">3. Digit DP</a>
      <a href="#interval-dp">4. Interval DP</a>
      <a href="#knapsack-variants">5. Knapsack Variants</a>
      <a href="#cht">6. Convex Hull Trick</a>
      <a href="#dnc-opt">7. Divide &amp; Conquer Optimization</a>
      <a href="#knuth-opt">8. Knuth's Optimization</a>
      <a href="#sos-dp">9. SOS DP (Sum over Subsets)</a>
      <a href="#profile-dp">10. Profile DP</a>
      <a href="#practice">11. Practice Problems</a>
    </div>

    <!-- ============================================================ -->
    <!-- SECTION 1: DP ON TREES -->
    <!-- ============================================================ -->
    <section id="tree-dp" class="section">
      <h2>1. DP on Trees</h2>
      <p>Tree DP exploits the recursive structure of trees. We root the tree and compute values bottom-up (subtree DP), then sometimes push values top-down (rerooting). These techniques solve problems about paths, subtrees, diameters, and optimal placement on trees.</p>

      <h3>1.1 Subtree DP -- Tree Diameter</h3>
      <p>The <strong>diameter</strong> of a tree is the longest path between any two nodes. We compute it in a single DFS: at each node, track the two longest paths descending into children. The diameter through a node is the sum of its two longest child-paths.</p>

      <div class="formula-box">
        <div class="label">Formula</div>
        <p>Let depth[v] = longest descending path from v. At node v with children c1, c2, ..., ck:</p>
        <p><strong>depth[v] = 1 + max(depth[ci])</strong></p>
        <p><strong>diameter = max over all v of (top two depth[ci] values summed)</strong></p>
      </div>

      <pre><code><span class="lang-label">Python</span>
<span class="keyword">import</span> sys
<span class="keyword">from</span> collections <span class="keyword">import</span> defaultdict

<span class="keyword">def</span> <span class="function">tree_diameter</span>(n, edges):
    <span class="comment"># Build adjacency list</span>
    adj = defaultdict(<span class="builtin">list</span>)
    <span class="keyword">for</span> u, v <span class="keyword">in</span> edges:
        adj[u].append(v)
        adj[v].append(u)

    diameter = <span class="number">0</span>

    <span class="keyword">def</span> <span class="function">dfs</span>(node, parent):
        <span class="keyword">nonlocal</span> diameter
        max1 = max2 = <span class="number">0</span>  <span class="comment"># two longest descending paths</span>
        <span class="keyword">for</span> child <span class="keyword">in</span> adj[node]:
            <span class="keyword">if</span> child == parent:
                <span class="keyword">continue</span>
            depth = dfs(child, node) + <span class="number">1</span>
            <span class="keyword">if</span> depth >= max1:
                max2 = max1
                max1 = depth
            <span class="keyword">elif</span> depth > max2:
                max2 = depth
        diameter = <span class="builtin">max</span>(diameter, max1 + max2)
        <span class="keyword">return</span> max1

    dfs(<span class="number">0</span>, -<span class="number">1</span>)
    <span class="keyword">return</span> diameter
</code></pre>

      <pre><code><span class="lang-label">C++</span>
<span class="keyword">#include</span> &lt;bits/stdc++.h&gt;
<span class="keyword">using namespace</span> std;

<span class="keyword">int</span> diameter;
vector&lt;vector&lt;<span class="keyword">int</span>&gt;&gt; adj;

<span class="keyword">int</span> <span class="function">dfs</span>(<span class="keyword">int</span> node, <span class="keyword">int</span> parent) {
    <span class="keyword">int</span> max1 = <span class="number">0</span>, max2 = <span class="number">0</span>;
    <span class="keyword">for</span> (<span class="keyword">int</span> child : adj[node]) {
        <span class="keyword">if</span> (child == parent) <span class="keyword">continue</span>;
        <span class="keyword">int</span> depth = <span class="function">dfs</span>(child, node) + <span class="number">1</span>;
        <span class="keyword">if</span> (depth >= max1) { max2 = max1; max1 = depth; }
        <span class="keyword">else if</span> (depth > max2) { max2 = depth; }
    }
    diameter = <span class="function">max</span>(diameter, max1 + max2);
    <span class="keyword">return</span> max1;
}
</code></pre>

      <div class="tip-box">
        <div class="label">Complexity</div>
        <p>O(n) time and space. A single DFS visits each node exactly once.</p>
      </div>

      <h3>1.2 Rerooting Technique</h3>
      <p>Many tree DP problems ask "compute f(v) for every node v as if v were the root." Naive approach: run DFS from each node = O(n^2). Rerooting does it in O(n) with two passes.</p>
      <p><strong>Pass 1 (bottom-up):</strong> Root at node 0, compute dp_down[v] = answer when v is root of its subtree.</p>
      <p><strong>Pass 2 (top-down):</strong> For each node, "reroot" by removing the child's contribution from the parent's answer and combining with the rest of the tree.</p>

      <div class="example-box">
        <div class="label">Example: Sum of distances to all other nodes</div>
        <p>Given a tree, compute for each node v the sum of distances from v to every other node.</p>
        <p>Let subtree_size[v] = number of nodes in subtree of v. Let dp_down[v] = sum of distances from v to all nodes in its subtree.</p>
        <p>Then dp_down[v] = sum over children c of (dp_down[c] + subtree_size[c]).</p>
        <p>For rerooting: ans[child] = ans[parent] + (n - subtree_size[child]) - subtree_size[child] = ans[parent] + n - 2*subtree_size[child].</p>
      </div>

      <pre><code><span class="lang-label">Python</span>
<span class="keyword">def</span> <span class="function">sum_of_distances</span>(n, edges):
    adj = [[] <span class="keyword">for</span> _ <span class="keyword">in</span> <span class="builtin">range</span>(n)]
    <span class="keyword">for</span> u, v <span class="keyword">in</span> edges:
        adj[u].append(v)
        adj[v].append(u)

    sz = [<span class="number">1</span>] * n
    dp_down = [<span class="number">0</span>] * n

    <span class="comment"># Pass 1: bottom-up from root 0</span>
    order = []
    parent = [-<span class="number">1</span>] * n
    visited = [<span class="keyword">False</span>] * n
    stack = [<span class="number">0</span>]
    visited[<span class="number">0</span>] = <span class="keyword">True</span>
    <span class="keyword">while</span> stack:
        node = stack.pop()
        order.append(node)
        <span class="keyword">for</span> nb <span class="keyword">in</span> adj[node]:
            <span class="keyword">if</span> <span class="keyword">not</span> visited[nb]:
                visited[nb] = <span class="keyword">True</span>
                parent[nb] = node
                stack.append(nb)

    <span class="keyword">for</span> node <span class="keyword">in</span> <span class="builtin">reversed</span>(order):
        <span class="keyword">for</span> nb <span class="keyword">in</span> adj[node]:
            <span class="keyword">if</span> nb == parent[node]:
                <span class="keyword">continue</span>
            sz[node] += sz[nb]
            dp_down[node] += dp_down[nb] + sz[nb]

    <span class="comment"># Pass 2: top-down rerooting</span>
    ans = [<span class="number">0</span>] * n
    ans[<span class="number">0</span>] = dp_down[<span class="number">0</span>]
    <span class="keyword">for</span> node <span class="keyword">in</span> order:
        <span class="keyword">for</span> nb <span class="keyword">in</span> adj[node]:
            <span class="keyword">if</span> nb == parent[node]:
                <span class="keyword">continue</span>
            ans[nb] = ans[node] + n - <span class="number">2</span> * sz[nb]

    <span class="keyword">return</span> ans
</code></pre>

      <div class="tip-box">
        <div class="label">When to Use Rerooting</div>
        <p>Rerooting works when the DP combines children's answers using an <strong>invertible operation</strong> (addition, multiplication, XOR, max with counts). If you can "remove" a child's contribution, you can reroot.</p>
      </div>

      <h3>1.3 Subtree DP -- Maximum Path Sum</h3>
      <p>Given a tree with weighted nodes, find the path (between any two nodes) with maximum sum. This is a generalization of the diameter problem.</p>

      <pre><code><span class="lang-label">C++</span>
<span class="keyword">#include</span> &lt;bits/stdc++.h&gt;
<span class="keyword">using namespace</span> std;

<span class="keyword">long long</span> ans;
vector&lt;vector&lt;<span class="keyword">int</span>&gt;&gt; adj;
vector&lt;<span class="keyword">int</span>&gt; val;

<span class="keyword">long long</span> <span class="function">dfs</span>(<span class="keyword">int</span> u, <span class="keyword">int</span> par) {
    <span class="keyword">long long</span> best = <span class="number">0</span>;  <span class="comment">// best descending path sum (can be 0 = take nothing)</span>
    <span class="keyword">for</span> (<span class="keyword">int</span> v : adj[u]) {
        <span class="keyword">if</span> (v == par) <span class="keyword">continue</span>;
        <span class="keyword">long long</span> child_best = <span class="function">dfs</span>(v, u);
        ans = <span class="function">max</span>(ans, best + child_best + val[u]);
        best = <span class="function">max</span>(best, child_best);
    }
    <span class="keyword">return</span> best + val[u];
}

<span class="keyword">int</span> <span class="function">main</span>() {
    <span class="keyword">int</span> n;
    cin &gt;&gt; n;
    adj.resize(n);
    val.resize(n);
    <span class="keyword">for</span> (<span class="keyword">int</span> i = <span class="number">0</span>; i &lt; n; i++) cin &gt;&gt; val[i];
    <span class="keyword">for</span> (<span class="keyword">int</span> i = <span class="number">0</span>; i &lt; n - <span class="number">1</span>; i++) {
        <span class="keyword">int</span> u, v; cin &gt;&gt; u &gt;&gt; v; u--; v--;
        adj[u].push_back(v);
        adj[v].push_back(u);
    }
    ans = *<span class="function">max_element</span>(val.begin(), val.end());
    <span class="function">dfs</span>(<span class="number">0</span>, -<span class="number">1</span>);
    cout &lt;&lt; ans &lt;&lt; <span class="string">"\n"</span>;
}
</code></pre>
    </section>

    <!-- ============================================================ -->
    <!-- SECTION 2: BITMASK DP -->
    <!-- ============================================================ -->
    <section id="bitmask-dp" class="section">
      <h2>2. Bitmask DP</h2>
      <p>When the state involves a <strong>subset</strong> of a small set (n &le; 20-23), we represent each subset as a bitmask integer. This gives us 2^n states, each encodable in a single int. Classic applications: TSP, assignment problem, Hamiltonian path, and subset enumeration.</p>

      <div class="warning-box">
        <div class="label">Size Constraint</div>
        <p>Bitmask DP is feasible only for small n. Memory = O(2^n * n), time = O(2^n * n^2) for TSP. For n=20, 2^20 = ~1M which is fine. For n=25, 2^25 = ~33M which is borderline. Beyond n=25, consider meet-in-middle or other approaches.</p>
      </div>

      <h3>2.1 Travelling Salesman Problem (TSP)</h3>
      <p>Given n cities and distances between every pair, find the shortest route visiting every city exactly once and returning to the start. The classic Held-Karp algorithm solves this in O(2^n * n^2).</p>

      <div class="formula-box">
        <div class="label">Recurrence</div>
        <p><strong>dp[mask][i]</strong> = minimum cost to visit exactly the cities in <em>mask</em>, ending at city <em>i</em>.</p>
        <p>Base: dp[1 &lt;&lt; start][start] = 0</p>
        <p>Transition: dp[mask | (1 &lt;&lt; j)][j] = min(dp[mask][i] + dist[i][j]) for all i in mask where j is not in mask.</p>
        <p>Answer: min over all i of dp[(1 &lt;&lt; n) - 1][i] + dist[i][start]</p>
      </div>

      <pre><code><span class="lang-label">Python</span>
<span class="keyword">def</span> <span class="function">tsp</span>(dist):
    <span class="string">"""Held-Karp TSP. dist is n x n matrix. Returns min tour cost."""</span>
    n = <span class="builtin">len</span>(dist)
    INF = <span class="builtin">float</span>(<span class="string">'inf'</span>)
    dp = [[INF] * n <span class="keyword">for</span> _ <span class="keyword">in</span> <span class="builtin">range</span>(<span class="number">1</span> &lt;&lt; n)]
    dp[<span class="number">1</span>][<span class="number">0</span>] = <span class="number">0</span>  <span class="comment"># start at city 0</span>

    <span class="keyword">for</span> mask <span class="keyword">in</span> <span class="builtin">range</span>(<span class="number">1</span>, <span class="number">1</span> &lt;&lt; n):
        <span class="keyword">for</span> u <span class="keyword">in</span> <span class="builtin">range</span>(n):
            <span class="keyword">if</span> dp[mask][u] == INF:
                <span class="keyword">continue</span>
            <span class="keyword">if</span> <span class="keyword">not</span> (mask &amp; (<span class="number">1</span> &lt;&lt; u)):
                <span class="keyword">continue</span>
            <span class="keyword">for</span> v <span class="keyword">in</span> <span class="builtin">range</span>(n):
                <span class="keyword">if</span> mask &amp; (<span class="number">1</span> &lt;&lt; v):
                    <span class="keyword">continue</span>
                new_mask = mask | (<span class="number">1</span> &lt;&lt; v)
                dp[new_mask][v] = <span class="builtin">min</span>(dp[new_mask][v], dp[mask][u] + dist[u][v])

    full = (<span class="number">1</span> &lt;&lt; n) - <span class="number">1</span>
    <span class="keyword">return</span> <span class="builtin">min</span>(dp[full][i] + dist[i][<span class="number">0</span>] <span class="keyword">for</span> i <span class="keyword">in</span> <span class="builtin">range</span>(n))
</code></pre>

      <pre><code><span class="lang-label">C++</span>
<span class="keyword">int</span> <span class="function">tsp</span>(vector&lt;vector&lt;<span class="keyword">int</span>&gt;&gt;&amp; dist) {
    <span class="keyword">int</span> n = dist.size();
    <span class="keyword">const int</span> INF = <span class="number">1e9</span>;
    vector&lt;vector&lt;<span class="keyword">int</span>&gt;&gt; dp(<span class="number">1</span> &lt;&lt; n, vector&lt;<span class="keyword">int</span>&gt;(n, INF));
    dp[<span class="number">1</span>][<span class="number">0</span>] = <span class="number">0</span>;

    <span class="keyword">for</span> (<span class="keyword">int</span> mask = <span class="number">1</span>; mask &lt; (<span class="number">1</span> &lt;&lt; n); mask++) {
        <span class="keyword">for</span> (<span class="keyword">int</span> u = <span class="number">0</span>; u &lt; n; u++) {
            <span class="keyword">if</span> (dp[mask][u] == INF || !(mask &amp; (<span class="number">1</span> &lt;&lt; u))) <span class="keyword">continue</span>;
            <span class="keyword">for</span> (<span class="keyword">int</span> v = <span class="number">0</span>; v &lt; n; v++) {
                <span class="keyword">if</span> (mask &amp; (<span class="number">1</span> &lt;&lt; v)) <span class="keyword">continue</span>;
                <span class="keyword">int</span> nm = mask | (<span class="number">1</span> &lt;&lt; v);
                dp[nm][v] = <span class="function">min</span>(dp[nm][v], dp[mask][u] + dist[u][v]);
            }
        }
    }
    <span class="keyword">int</span> ans = INF;
    <span class="keyword">int</span> full = (<span class="number">1</span> &lt;&lt; n) - <span class="number">1</span>;
    <span class="keyword">for</span> (<span class="keyword">int</span> i = <span class="number">0</span>; i &lt; n; i++)
        ans = <span class="function">min</span>(ans, dp[full][i] + dist[i][<span class="number">0</span>]);
    <span class="keyword">return</span> ans;
}
</code></pre>

      <h3>2.2 Assignment Problem</h3>
      <p>Given n workers and n tasks with cost[i][j], assign each worker to exactly one task (and each task to exactly one worker) minimizing total cost. This is equivalent to minimum weight perfect matching in a bipartite graph.</p>

      <pre><code><span class="lang-label">Python</span>
<span class="keyword">def</span> <span class="function">assignment</span>(cost):
    <span class="string">"""Bitmask DP for assignment problem. O(2^n * n)."""</span>
    n = <span class="builtin">len</span>(cost)
    INF = <span class="builtin">float</span>(<span class="string">'inf'</span>)
    dp = [INF] * (<span class="number">1</span> &lt;&lt; n)
    dp[<span class="number">0</span>] = <span class="number">0</span>

    <span class="keyword">for</span> mask <span class="keyword">in</span> <span class="builtin">range</span>(<span class="number">1</span> &lt;&lt; n):
        <span class="keyword">if</span> dp[mask] == INF:
            <span class="keyword">continue</span>
        worker = <span class="builtin">bin</span>(mask).count(<span class="string">'1'</span>)  <span class="comment"># which worker to assign next</span>
        <span class="keyword">if</span> worker >= n:
            <span class="keyword">continue</span>
        <span class="keyword">for</span> task <span class="keyword">in</span> <span class="builtin">range</span>(n):
            <span class="keyword">if</span> mask &amp; (<span class="number">1</span> &lt;&lt; task):
                <span class="keyword">continue</span>
            new_mask = mask | (<span class="number">1</span> &lt;&lt; task)
            dp[new_mask] = <span class="builtin">min</span>(dp[new_mask], dp[mask] + cost[worker][task])

    <span class="keyword">return</span> dp[(<span class="number">1</span> &lt;&lt; n) - <span class="number">1</span>]
</code></pre>

      <h3>2.3 Subset Enumeration</h3>
      <p>A critical bitmask trick: iterating over all submasks of a given mask in O(3^n) total across all masks.</p>

      <pre><code><span class="lang-label">C++</span>
<span class="comment">// Enumerate all submasks of mask (including mask itself, excluding 0)</span>
<span class="keyword">for</span> (<span class="keyword">int</span> sub = mask; sub &gt; <span class="number">0</span>; sub = (sub - <span class="number">1</span>) &amp; mask) {
    <span class="comment">// process submask 'sub'</span>
}
<span class="comment">// Don't forget to handle sub = 0 separately if needed</span>

<span class="comment">// Total work across all masks: sum over mask of 2^popcount(mask) = 3^n</span>
</code></pre>

      <div class="tip-box">
        <div class="label">Bit Manipulation Cheatsheet</div>
        <p><strong>Check bit i:</strong> mask &amp; (1 &lt;&lt; i)</p>
        <p><strong>Set bit i:</strong> mask | (1 &lt;&lt; i)</p>
        <p><strong>Clear bit i:</strong> mask &amp; ~(1 &lt;&lt; i)</p>
        <p><strong>Toggle bit i:</strong> mask ^ (1 &lt;&lt; i)</p>
        <p><strong>Lowest set bit:</strong> mask &amp; (-mask)</p>
        <p><strong>Clear lowest set bit:</strong> mask &amp; (mask - 1)</p>
        <p><strong>Count set bits:</strong> __builtin_popcount(mask) in C++, bin(mask).count('1') in Python</p>
      </div>
    </section>

    <!-- ============================================================ -->
    <!-- SECTION 3: DIGIT DP -->
    <!-- ============================================================ -->
    <section id="digit-dp" class="section">
      <h2>3. Digit DP</h2>
      <p>Digit DP counts numbers in a range [L, R] satisfying some property by building the number digit by digit. The key idea is a <strong>tight</strong> flag: whether the digits chosen so far exactly match the upper bound's prefix (constraining future digits) or are already smaller (free to choose any digit).</p>

      <div class="formula-box">
        <div class="label">General Template</div>
        <p>count(R) - count(L-1) gives the answer for [L, R].</p>
        <p>State: (position, tight, ...extra state). Position goes from MSB to LSB.</p>
        <p>If tight=True, current digit can be 0..digits[pos]. If tight=False, digit can be 0..9.</p>
      </div>

      <h3>3.1 Template: Count numbers in [1, N] with digit sum = S</h3>

      <pre><code><span class="lang-label">Python</span>
<span class="keyword">from</span> functools <span class="keyword">import</span> lru_cache

<span class="keyword">def</span> <span class="function">count_digit_sum</span>(N, S):
    <span class="string">"""Count integers in [1, N] whose digit sum equals S."""</span>
    digits = [<span class="builtin">int</span>(d) <span class="keyword">for</span> d <span class="keyword">in</span> <span class="builtin">str</span>(N)]
    n = <span class="builtin">len</span>(digits)

    <span class="builtin">@lru_cache</span>(maxsize=<span class="keyword">None</span>)
    <span class="keyword">def</span> <span class="function">solve</span>(pos, tight, rem, started):
        <span class="string">"""
        pos: current digit position (0-indexed from left)
        tight: whether we're still bounded by N
        rem: remaining digit sum needed
        started: whether we've placed a nonzero digit yet
        """</span>
        <span class="keyword">if</span> rem &lt; <span class="number">0</span>:
            <span class="keyword">return</span> <span class="number">0</span>
        <span class="keyword">if</span> pos == n:
            <span class="keyword">return</span> <span class="number">1</span> <span class="keyword">if</span> rem == <span class="number">0</span> <span class="keyword">and</span> started <span class="keyword">else</span> <span class="number">0</span>

        limit = digits[pos] <span class="keyword">if</span> tight <span class="keyword">else</span> <span class="number">9</span>
        result = <span class="number">0</span>
        <span class="keyword">for</span> d <span class="keyword">in</span> <span class="builtin">range</span>(<span class="number">0</span>, limit + <span class="number">1</span>):
            result += solve(
                pos + <span class="number">1</span>,
                tight <span class="keyword">and</span> d == limit,
                rem - d,
                started <span class="keyword">or</span> d &gt; <span class="number">0</span>
            )
        <span class="keyword">return</span> result

    <span class="keyword">return</span> solve(<span class="number">0</span>, <span class="keyword">True</span>, S, <span class="keyword">False</span>)
</code></pre>

      <pre><code><span class="lang-label">C++</span>
<span class="keyword">#include</span> &lt;bits/stdc++.h&gt;
<span class="keyword">using namespace</span> std;

<span class="keyword">string</span> num;
<span class="keyword">int</span> target;
<span class="keyword">int</span> memo[<span class="number">20</span>][<span class="number">2</span>][<span class="number">200</span>][<span class="number">2</span>];

<span class="keyword">int</span> <span class="function">solve</span>(<span class="keyword">int</span> pos, <span class="keyword">bool</span> tight, <span class="keyword">int</span> rem, <span class="keyword">bool</span> started) {
    <span class="keyword">if</span> (rem &lt; <span class="number">0</span>) <span class="keyword">return</span> <span class="number">0</span>;
    <span class="keyword">if</span> (pos == (<span class="keyword">int</span>)num.size())
        <span class="keyword">return</span> (rem == <span class="number">0</span> &amp;&amp; started) ? <span class="number">1</span> : <span class="number">0</span>;

    <span class="keyword">int</span>&amp; res = memo[pos][tight][rem][started];
    <span class="keyword">if</span> (res != -<span class="number">1</span>) <span class="keyword">return</span> res;
    res = <span class="number">0</span>;

    <span class="keyword">int</span> limit = tight ? (num[pos] - <span class="string">'0'</span>) : <span class="number">9</span>;
    <span class="keyword">for</span> (<span class="keyword">int</span> d = <span class="number">0</span>; d &lt;= limit; d++) {
        res += <span class="function">solve</span>(pos + <span class="number">1</span>, tight &amp;&amp; (d == limit),
                     rem - d, started || (d &gt; <span class="number">0</span>));
    }
    <span class="keyword">return</span> res;
}

<span class="keyword">int</span> <span class="function">count_digit_sum</span>(<span class="keyword">long long</span> N, <span class="keyword">int</span> S) {
    num = to_string(N);
    target = S;
    <span class="function">memset</span>(memo, -<span class="number">1</span>, <span class="keyword">sizeof</span> memo);
    <span class="keyword">return</span> <span class="function">solve</span>(<span class="number">0</span>, <span class="keyword">true</span>, S, <span class="keyword">false</span>);
}
</code></pre>

      <div class="tip-box">
        <div class="label">Digit DP Pattern</div>
        <p>Almost every digit DP problem follows this skeleton. The only thing that changes is the "extra state" (digit sum, count of certain digits, modular remainder, etc.) and the base case condition. Master this template and you can solve 90% of digit DP problems.</p>
      </div>

      <div class="example-box">
        <div class="label">Walkthrough: count_digit_sum(123, 5)</div>
        <p>We want numbers in [1, 123] with digit sum = 5. The digits array is [1, 2, 3].</p>
        <p>At pos=0, tight=True: we can pick d=0 (then free for next digits) or d=1 (stay tight).</p>
        <p>If d=0: need remaining sum 5 from 2 digits (00-99). Numbers: 05, 14, 23, 32, 41, 50 = 6 numbers.</p>
        <p>If d=1, tight: pos=1, tight=True, rem=4. Can pick d=0..2.</p>
        <p>If d=1,0: need 4 from 1 digit (0-9). Only 4. Number: 104.</p>
        <p>If d=1,1: need 3 from 1 digit (0-9). Only 3. Number: 113.</p>
        <p>If d=1,2, tight: need 2 from 1 digit (0-3). Only 2. Number: 122.</p>
        <p>Total: 6 + 3 = 9 numbers.</p>
      </div>
    </section>

    <!-- ============================================================ -->
    <!-- SECTION 4: INTERVAL DP -->
    <!-- ============================================================ -->
    <section id="interval-dp" class="section">
      <h2>4. Interval DP</h2>
      <p>Interval DP operates on contiguous subarrays/substrings. The state is dp[i][j] = optimal answer for the interval [i, j]. We try all ways to split the interval and combine results. Typical time complexity: O(n^3).</p>

      <h3>4.1 Matrix Chain Multiplication</h3>
      <p>Given matrices A1(p0 x p1), A2(p1 x p2), ..., An(p_{n-1} x p_n), find the minimum number of scalar multiplications to compute their product.</p>

      <div class="formula-box">
        <div class="label">Recurrence</div>
        <p><strong>dp[i][j]</strong> = minimum cost to multiply matrices i through j.</p>
        <p>dp[i][j] = min over k in [i, j-1] of: dp[i][k] + dp[k+1][j] + p[i-1] * p[k] * p[j]</p>
        <p>Base: dp[i][i] = 0 (single matrix costs nothing).</p>
      </div>

      <pre><code><span class="lang-label">Python</span>
<span class="keyword">def</span> <span class="function">matrix_chain</span>(p):
    <span class="string">"""p = list of dimensions. n matrices: p[0]xp[1], p[1]xp[2], ..., p[n-1]xp[n]."""</span>
    n = <span class="builtin">len</span>(p) - <span class="number">1</span>  <span class="comment"># number of matrices</span>
    INF = <span class="builtin">float</span>(<span class="string">'inf'</span>)
    dp = [[<span class="number">0</span>] * n <span class="keyword">for</span> _ <span class="keyword">in</span> <span class="builtin">range</span>(n)]

    <span class="keyword">for</span> length <span class="keyword">in</span> <span class="builtin">range</span>(<span class="number">2</span>, n + <span class="number">1</span>):  <span class="comment"># chain length</span>
        <span class="keyword">for</span> i <span class="keyword">in</span> <span class="builtin">range</span>(n - length + <span class="number">1</span>):
            j = i + length - <span class="number">1</span>
            dp[i][j] = INF
            <span class="keyword">for</span> k <span class="keyword">in</span> <span class="builtin">range</span>(i, j):
                cost = dp[i][k] + dp[k + <span class="number">1</span>][j] + p[i] * p[k + <span class="number">1</span>] * p[j + <span class="number">1</span>]
                dp[i][j] = <span class="builtin">min</span>(dp[i][j], cost)

    <span class="keyword">return</span> dp[<span class="number">0</span>][n - <span class="number">1</span>]
</code></pre>

      <h3>4.2 Burst Balloons (LeetCode 312)</h3>
      <p>Given n balloons with values, bursting balloon i gives nums[left]*nums[i]*nums[right]. Find max coins by bursting all balloons. The trick: think of which balloon is burst <strong>last</strong> in each interval.</p>

      <pre><code><span class="lang-label">Python</span>
<span class="keyword">def</span> <span class="function">max_coins</span>(nums):
    nums = [<span class="number">1</span>] + nums + [<span class="number">1</span>]  <span class="comment"># pad with 1s</span>
    n = <span class="builtin">len</span>(nums)
    dp = [[<span class="number">0</span>] * n <span class="keyword">for</span> _ <span class="keyword">in</span> <span class="builtin">range</span>(n)]

    <span class="keyword">for</span> length <span class="keyword">in</span> <span class="builtin">range</span>(<span class="number">2</span>, n):  <span class="comment"># gap between i and j</span>
        <span class="keyword">for</span> i <span class="keyword">in</span> <span class="builtin">range</span>(n - length):
            j = i + length
            <span class="keyword">for</span> k <span class="keyword">in</span> <span class="builtin">range</span>(i + <span class="number">1</span>, j):  <span class="comment"># k is the LAST balloon burst in (i,j)</span>
                dp[i][j] = <span class="builtin">max</span>(
                    dp[i][j],
                    dp[i][k] + dp[k][j] + nums[i] * nums[k] * nums[j]
                )

    <span class="keyword">return</span> dp[<span class="number">0</span>][n - <span class="number">1</span>]
</code></pre>

      <h3>4.3 Merge Stones</h3>
      <p>Given n piles of stones, merge exactly k consecutive piles into one each step (cost = sum of merged piles). Find minimum total cost to merge into one pile, or -1 if impossible.</p>

      <pre><code><span class="lang-label">C++</span>
<span class="keyword">int</span> <span class="function">mergeStones</span>(vector&lt;<span class="keyword">int</span>&gt;&amp; stones, <span class="keyword">int</span> K) {
    <span class="keyword">int</span> n = stones.size();
    <span class="keyword">if</span> ((n - <span class="number">1</span>) % (K - <span class="number">1</span>) != <span class="number">0</span>) <span class="keyword">return</span> -<span class="number">1</span>;

    vector&lt;<span class="keyword">int</span>&gt; prefix(n + <span class="number">1</span>, <span class="number">0</span>);
    <span class="keyword">for</span> (<span class="keyword">int</span> i = <span class="number">0</span>; i &lt; n; i++) prefix[i + <span class="number">1</span>] = prefix[i] + stones[i];

    <span class="comment">// dp[i][j] = min cost to merge stones[i..j] as far as possible</span>
    vector&lt;vector&lt;<span class="keyword">int</span>&gt;&gt; dp(n, vector&lt;<span class="keyword">int</span>&gt;(n, <span class="number">0</span>));

    <span class="keyword">for</span> (<span class="keyword">int</span> len = K; len &lt;= n; len++) {
        <span class="keyword">for</span> (<span class="keyword">int</span> i = <span class="number">0</span>; i + len &lt;= n; i++) {
            <span class="keyword">int</span> j = i + len - <span class="number">1</span>;
            dp[i][j] = <span class="number">1e9</span>;
            <span class="keyword">for</span> (<span class="keyword">int</span> mid = i; mid &lt; j; mid += K - <span class="number">1</span>) {
                dp[i][j] = <span class="function">min</span>(dp[i][j], dp[i][mid] + dp[mid + <span class="number">1</span>][j]);
            }
            <span class="keyword">if</span> ((len - <span class="number">1</span>) % (K - <span class="number">1</span>) == <span class="number">0</span>)
                dp[i][j] += prefix[j + <span class="number">1</span>] - prefix[i];
        }
    }
    <span class="keyword">return</span> dp[<span class="number">0</span>][n - <span class="number">1</span>];
}
</code></pre>

      <div class="tip-box">
        <div class="label">Complexity</div>
        <p>Matrix chain and burst balloons: O(n^3) time, O(n^2) space. Merge stones with k groups: O(n^3 / k) time.</p>
      </div>
    </section>

    <!-- ============================================================ -->
    <!-- SECTION 5: KNAPSACK VARIANTS -->
    <!-- ============================================================ -->
    <section id="knapsack-variants" class="section">
      <h2>5. Knapsack Variants</h2>
      <p>The 0/1 knapsack is foundational DP, but real problems demand variants: unbounded (infinite copies), bounded (limited copies), multi-dimensional (multiple constraints), and meet-in-middle (when n is too large for standard DP but small enough for splitting).</p>

      <h3>5.1 Unbounded Knapsack</h3>
      <p>Each item can be used unlimited times. The only change: iterate items in the inner loop or process weights going forward (not backward).</p>

      <pre><code><span class="lang-label">Python</span>
<span class="keyword">def</span> <span class="function">unbounded_knapsack</span>(W, weights, values):
    <span class="string">"""Max value with capacity W, unlimited copies of each item."""</span>
    dp = [<span class="number">0</span>] * (W + <span class="number">1</span>)
    <span class="keyword">for</span> w <span class="keyword">in</span> <span class="builtin">range</span>(<span class="number">1</span>, W + <span class="number">1</span>):
        <span class="keyword">for</span> i <span class="keyword">in</span> <span class="builtin">range</span>(<span class="builtin">len</span>(weights)):
            <span class="keyword">if</span> weights[i] &lt;= w:
                dp[w] = <span class="builtin">max</span>(dp[w], dp[w - weights[i]] + values[i])
    <span class="keyword">return</span> dp[W]
</code></pre>

      <div class="tip-box">
        <div class="label">0/1 vs Unbounded</div>
        <p>In 0/1 knapsack, iterate capacity <strong>right to left</strong> (each item used at most once). In unbounded, iterate <strong>left to right</strong> (reuse allowed). This single direction change is the entire difference.</p>
      </div>

      <h3>5.2 Bounded Knapsack (Binary Lifting Trick)</h3>
      <p>Each item i has a limit c[i]. Naive: treat as c[i] separate items = O(W * sum(c[i])). Binary lifting: decompose c[i] into powers of 2 (1, 2, 4, ..., remainder) to get O(W * n * log(max_c)) items.</p>

      <pre><code><span class="lang-label">C++</span>
<span class="keyword">int</span> <span class="function">bounded_knapsack</span>(<span class="keyword">int</span> W, vector&lt;<span class="keyword">int</span>&gt;&amp; wt, vector&lt;<span class="keyword">int</span>&gt;&amp; val, vector&lt;<span class="keyword">int</span>&gt;&amp; cnt) {
    <span class="keyword">int</span> n = wt.size();
    vector&lt;<span class="keyword">int</span>&gt; dp(W + <span class="number">1</span>, <span class="number">0</span>);

    <span class="keyword">for</span> (<span class="keyword">int</span> i = <span class="number">0</span>; i &lt; n; i++) {
        <span class="comment">// Binary decomposition of cnt[i]</span>
        <span class="keyword">int</span> remaining = cnt[i];
        <span class="keyword">for</span> (<span class="keyword">int</span> k = <span class="number">1</span>; remaining &gt; <span class="number">0</span>; k *= <span class="number">2</span>) {
            <span class="keyword">int</span> batch = <span class="function">min</span>(k, remaining);
            remaining -= batch;
            <span class="keyword">int</span> bw = wt[i] * batch, bv = val[i] * batch;
            <span class="comment">// 0/1 knapsack for this batch (right to left)</span>
            <span class="keyword">for</span> (<span class="keyword">int</span> w = W; w &gt;= bw; w--)
                dp[w] = <span class="function">max</span>(dp[w], dp[w - bw] + bv);
        }
    }
    <span class="keyword">return</span> dp[W];
}
</code></pre>

      <h3>5.3 Multi-Dimensional Knapsack</h3>
      <p>When there are multiple constraints (weight AND volume, for example), add an extra dimension to the DP table.</p>

      <pre><code><span class="lang-label">Python</span>
<span class="keyword">def</span> <span class="function">knapsack_2d</span>(W, V, items):
    <span class="string">"""items = list of (weight, volume, value). Max value with capacity W and volume V."""</span>
    dp = [[<span class="number">0</span>] * (V + <span class="number">1</span>) <span class="keyword">for</span> _ <span class="keyword">in</span> <span class="builtin">range</span>(W + <span class="number">1</span>)]

    <span class="keyword">for</span> w, v, val <span class="keyword">in</span> items:
        <span class="keyword">for</span> cw <span class="keyword">in</span> <span class="builtin">range</span>(W, w - <span class="number">1</span>, -<span class="number">1</span>):
            <span class="keyword">for</span> cv <span class="keyword">in</span> <span class="builtin">range</span>(V, v - <span class="number">1</span>, -<span class="number">1</span>):
                dp[cw][cv] = <span class="builtin">max</span>(dp[cw][cv], dp[cw - w][cv - v] + val)

    <span class="keyword">return</span> dp[W][V]
</code></pre>

      <h3>5.4 Meet-in-the-Middle Knapsack</h3>
      <p>When n is up to ~40 and W is very large, standard DP is impossible. Split items into two halves, enumerate all 2^(n/2) subsets for each, then combine with binary search. Time: O(2^(n/2) * log(2^(n/2))).</p>

      <pre><code><span class="lang-label">Python</span>
<span class="keyword">from</span> bisect <span class="keyword">import</span> bisect_right

<span class="keyword">def</span> <span class="function">meet_in_middle_knapsack</span>(W, weights, values):
    n = <span class="builtin">len</span>(weights)
    half = n // <span class="number">2</span>

    <span class="keyword">def</span> <span class="function">enumerate_subsets</span>(wts, vals):
        <span class="string">"""Return list of (weight, value) for all subsets."""</span>
        m = <span class="builtin">len</span>(wts)
        result = []
        <span class="keyword">for</span> mask <span class="keyword">in</span> <span class="builtin">range</span>(<span class="number">1</span> &lt;&lt; m):
            tw = tv = <span class="number">0</span>
            <span class="keyword">for</span> i <span class="keyword">in</span> <span class="builtin">range</span>(m):
                <span class="keyword">if</span> mask &amp; (<span class="number">1</span> &lt;&lt; i):
                    tw += wts[i]
                    tv += vals[i]
            <span class="keyword">if</span> tw &lt;= W:
                result.append((tw, tv))
        <span class="keyword">return</span> result

    left = enumerate_subsets(weights[:half], values[:half])
    right = enumerate_subsets(weights[half:], values[half:])

    <span class="comment"># Sort right by weight, build prefix-max of values</span>
    right.sort()
    rw = [r[<span class="number">0</span>] <span class="keyword">for</span> r <span class="keyword">in</span> right]
    rv = [r[<span class="number">1</span>] <span class="keyword">for</span> r <span class="keyword">in</span> right]
    <span class="comment"># Make rv prefix-max: rv[i] = max value among right subsets with weight <= rw[i]</span>
    <span class="keyword">for</span> i <span class="keyword">in</span> <span class="builtin">range</span>(<span class="number">1</span>, <span class="builtin">len</span>(rv)):
        rv[i] = <span class="builtin">max</span>(rv[i], rv[i - <span class="number">1</span>])

    ans = <span class="number">0</span>
    <span class="keyword">for</span> lw, lv <span class="keyword">in</span> left:
        remaining = W - lw
        idx = bisect_right(rw, remaining) - <span class="number">1</span>
        <span class="keyword">if</span> idx &gt;= <span class="number">0</span>:
            ans = <span class="builtin">max</span>(ans, lv + rv[idx])
        <span class="keyword">else</span>:
            ans = <span class="builtin">max</span>(ans, lv)

    <span class="keyword">return</span> ans
</code></pre>

      <div class="warning-box">
        <div class="label">Common Mistakes</div>
        <p>1. Forgetting to iterate right-to-left in 0/1 knapsack (accidentally allowing reuse).</p>
        <p>2. In meet-in-middle, forgetting to build the prefix-max on the sorted right half.</p>
        <p>3. Off-by-one in bounded knapsack binary decomposition (the remainder group).</p>
      </div>
    </section>

    <!-- ============================================================ -->
    <!-- SECTION 6: CONVEX HULL TRICK -->
    <!-- ============================================================ -->
    <section id="cht" class="section">
      <h2>6. Convex Hull Trick</h2>
      <p>The Convex Hull Trick (CHT) optimizes DP recurrences of the form: <strong>dp[i] = min/max over j &lt; i of (a[j] * x[i] + b[j])</strong>, where each previous state j defines a linear function f_j(x) = a[j]*x + b[j], and we query at x = x[i]. If we maintain a convex hull of these lines, each query takes O(log n) or amortized O(1).</p>

      <div class="formula-box">
        <div class="label">When to Use CHT</div>
        <p>Your DP recurrence must be expressible as: dp[i] = min/max(m_j * x_i + b_j) + cost(i)</p>
        <p>where m_j and b_j depend on j (a previously computed state) and x_i depends only on i.</p>
        <p>If slopes m_j are monotone AND query points x_i are monotone, use a simple deque (amortized O(1) per operation). Otherwise, use Li Chao tree (O(log n) per operation).</p>
      </div>

      <h3>6.1 Simple CHT with Monotone Slopes and Queries</h3>

      <pre><code><span class="lang-label">Python</span>
<span class="keyword">from</span> collections <span class="keyword">import</span> deque

<span class="keyword">class</span> <span class="function">ConvexHullTrick</span>:
    <span class="string">"""Maintains lower hull for minimum queries.
    Requirements: lines added in decreasing slope order,
    queries in increasing x order."""</span>

    <span class="keyword">def</span> <span class="function">__init__</span>(self):
        self.hull = deque()  <span class="comment"># list of (slope, intercept)</span>

    <span class="keyword">def</span> <span class="function">_bad</span>(self, l1, l2, l3):
        <span class="string">"""Check if l2 is never optimal between l1 and l3."""</span>
        <span class="keyword">return</span> (l3[<span class="number">1</span>] - l1[<span class="number">1</span>]) * (l1[<span class="number">0</span>] - l2[<span class="number">0</span>]) &lt;= \
               (l2[<span class="number">1</span>] - l1[<span class="number">1</span>]) * (l1[<span class="number">0</span>] - l3[<span class="number">0</span>])

    <span class="keyword">def</span> <span class="function">add_line</span>(self, m, b):
        <span class="string">"""Add line y = m*x + b. Slopes must be decreasing."""</span>
        line = (m, b)
        <span class="keyword">while</span> <span class="builtin">len</span>(self.hull) &gt;= <span class="number">2</span> <span class="keyword">and</span> self._bad(self.hull[-<span class="number">2</span>], self.hull[-<span class="number">1</span>], line):
            self.hull.pop()
        self.hull.append(line)

    <span class="keyword">def</span> <span class="function">query</span>(self, x):
        <span class="string">"""Query minimum y at x. x must be increasing."""</span>
        <span class="keyword">while</span> <span class="builtin">len</span>(self.hull) &gt;= <span class="number">2</span>:
            m1, b1 = self.hull[<span class="number">0</span>]
            m2, b2 = self.hull[<span class="number">1</span>]
            <span class="keyword">if</span> m1 * x + b1 &gt; m2 * x + b2:
                self.hull.popleft()
            <span class="keyword">else</span>:
                <span class="keyword">break</span>
        m, b = self.hull[<span class="number">0</span>]
        <span class="keyword">return</span> m * x + b
</code></pre>

      <pre><code><span class="lang-label">C++</span>
<span class="keyword">struct</span> Line {
    <span class="keyword">long long</span> m, b;
    <span class="keyword">long long</span> <span class="function">eval</span>(<span class="keyword">long long</span> x) { <span class="keyword">return</span> m * x + b; }
};

<span class="keyword">struct</span> CHT {
    deque&lt;Line&gt; hull;

    <span class="keyword">bool</span> <span class="function">bad</span>(Line l1, Line l2, Line l3) {
        <span class="keyword">return</span> (__int128)(l3.b - l1.b) * (l1.m - l2.m) &lt;=
               (__int128)(l2.b - l1.b) * (l1.m - l3.m);
    }

    <span class="keyword">void</span> <span class="function">add_line</span>(<span class="keyword">long long</span> m, <span class="keyword">long long</span> b) {
        Line line = {m, b};
        <span class="keyword">while</span> (hull.size() &gt;= <span class="number">2</span> &amp;&amp; <span class="function">bad</span>(hull[hull.size()-<span class="number">2</span>], hull.back(), line))
            hull.pop_back();
        hull.push_back(line);
    }

    <span class="keyword">long long</span> <span class="function">query</span>(<span class="keyword">long long</span> x) {
        <span class="keyword">while</span> (hull.size() &gt;= <span class="number">2</span> &amp;&amp; hull[<span class="number">0</span>].<span class="function">eval</span>(x) &gt;= hull[<span class="number">1</span>].<span class="function">eval</span>(x))
            hull.pop_front();
        <span class="keyword">return</span> hull.front().<span class="function">eval</span>(x);
    }
};
</code></pre>

      <h3>6.2 Example: Minimize dp[i] = min(dp[j] + a[j]*b[i])</h3>

      <div class="example-box">
        <div class="label">Walkthrough: APIO 2014 Sequence</div>
        <p>Suppose dp[i] = min over j &lt; i of (dp[j] + cost(j, i)), where cost(j, i) can be decomposed as m_j * x_i + b_j.</p>
        <p>Rewrite: dp[i] = min(m_j * x_i + b_j) where m_j = some function of j's values, b_j = dp[j] + extra, x_i = some function of i.</p>
        <p>Add line (m_j, b_j) when computing dp[j]. Query at x_i to compute dp[i]. If slopes decrease and queries increase, the deque approach gives O(n) total.</p>
      </div>

      <div class="tip-box">
        <div class="label">Li Chao Tree</div>
        <p>When slopes and queries are NOT monotone, use a Li Chao segment tree. It supports add_line in O(log(MAXX)) and query in O(log(MAXX)). Conceptually simpler than maintaining an explicit hull with binary search.</p>
      </div>
    </section>

    <!-- ============================================================ -->
    <!-- SECTION 7: DIVIDE & CONQUER OPTIMIZATION -->
    <!-- ============================================================ -->
    <section id="dnc-opt" class="section">
      <h2>7. Divide &amp; Conquer Optimization</h2>
      <p>This optimization applies when the DP recurrence is: <strong>dp[i][j] = min over k in [opt[i][j-1], opt[i+1][j]] of (dp[k][j-1] + C(k+1, i))</strong> and the "optimal split point" opt[i][j] is monotone: opt[i][j] &le; opt[i+1][j]. This means as i increases, the optimal k never decreases.</p>

      <div class="formula-box">
        <div class="label">Condition</div>
        <p>Given dp[i][j] = min over k &lt; i of (dp[k][j-1] + C(k, i)), if the optimal k for row i is &le; optimal k for row i+1, we have the monotonicity condition. This reduces O(n^2 * k) to O(n * k * log n).</p>
      </div>

      <pre><code><span class="lang-label">C++</span>
<span class="keyword">#include</span> &lt;bits/stdc++.h&gt;
<span class="keyword">using namespace</span> std;

<span class="keyword">const long long</span> INF = <span class="number">1e18</span>;
<span class="keyword">int</span> N, K;
vector&lt;vector&lt;<span class="keyword">long long</span>&gt;&gt; dp;

<span class="comment">// cost(l, r) = cost of one group covering elements [l, r]</span>
<span class="keyword">long long</span> <span class="function">cost</span>(<span class="keyword">int</span> l, <span class="keyword">int</span> r);

<span class="keyword">void</span> <span class="function">solve</span>(<span class="keyword">int</span> layer, <span class="keyword">int</span> lo, <span class="keyword">int</span> hi, <span class="keyword">int</span> opt_lo, <span class="keyword">int</span> opt_hi) {
    <span class="keyword">if</span> (lo &gt; hi) <span class="keyword">return</span>;
    <span class="keyword">int</span> mid = (lo + hi) / <span class="number">2</span>;
    <span class="keyword">long long</span> best = INF;
    <span class="keyword">int</span> opt = opt_lo;

    <span class="keyword">for</span> (<span class="keyword">int</span> k = opt_lo; k &lt;= <span class="function">min</span>(mid - <span class="number">1</span>, opt_hi); k++) {
        <span class="keyword">long long</span> val = dp[layer - <span class="number">1</span>][k] + <span class="function">cost</span>(k + <span class="number">1</span>, mid);
        <span class="keyword">if</span> (val &lt; best) {
            best = val;
            opt = k;
        }
    }
    dp[layer][mid] = best;

    <span class="function">solve</span>(layer, lo, mid - <span class="number">1</span>, opt_lo, opt);
    <span class="function">solve</span>(layer, mid + <span class="number">1</span>, hi, opt, opt_hi);
}

<span class="keyword">void</span> <span class="function">compute</span>() {
    dp.assign(K + <span class="number">1</span>, vector&lt;<span class="keyword">long long</span>&gt;(N + <span class="number">1</span>, INF));
    dp[<span class="number">0</span>][<span class="number">0</span>] = <span class="number">0</span>;

    <span class="comment">// Base layer: dp[1][i] = cost(1, i)</span>
    <span class="keyword">for</span> (<span class="keyword">int</span> i = <span class="number">1</span>; i &lt;= N; i++)
        dp[<span class="number">1</span>][i] = <span class="function">cost</span>(<span class="number">1</span>, i);

    <span class="keyword">for</span> (<span class="keyword">int</span> j = <span class="number">2</span>; j &lt;= K; j++)
        <span class="function">solve</span>(j, <span class="number">1</span>, N, <span class="number">0</span>, N - <span class="number">1</span>);
}
</code></pre>

      <pre><code><span class="lang-label">Python</span>
<span class="keyword">import</span> sys
sys.setrecursionlimit(<span class="number">300000</span>)

<span class="keyword">def</span> <span class="function">dnc_optimize</span>(n, k, cost_fn):
    <span class="string">"""
    dp[j][i] = min over m < i of (dp[j-1][m] + cost_fn(m+1, i))
    Returns dp[k][n].
    """</span>
    INF = <span class="builtin">float</span>(<span class="string">'inf'</span>)
    prev = [INF] * (n + <span class="number">1</span>)
    prev[<span class="number">0</span>] = <span class="number">0</span>
    <span class="keyword">for</span> i <span class="keyword">in</span> <span class="builtin">range</span>(<span class="number">1</span>, n + <span class="number">1</span>):
        prev[i] = cost_fn(<span class="number">1</span>, i)

    <span class="keyword">for</span> layer <span class="keyword">in</span> <span class="builtin">range</span>(<span class="number">2</span>, k + <span class="number">1</span>):
        cur = [INF] * (n + <span class="number">1</span>)

        <span class="keyword">def</span> <span class="function">rec</span>(lo, hi, opt_lo, opt_hi):
            <span class="keyword">if</span> lo &gt; hi:
                <span class="keyword">return</span>
            mid = (lo + hi) // <span class="number">2</span>
            best = INF
            opt = opt_lo
            <span class="keyword">for</span> m <span class="keyword">in</span> <span class="builtin">range</span>(opt_lo, <span class="builtin">min</span>(mid, opt_hi + <span class="number">1</span>)):
                val = prev[m] + cost_fn(m + <span class="number">1</span>, mid)
                <span class="keyword">if</span> val &lt; best:
                    best = val
                    opt = m
            cur[mid] = best
            rec(lo, mid - <span class="number">1</span>, opt_lo, opt)
            rec(mid + <span class="number">1</span>, hi, opt, opt_hi)

        rec(<span class="number">1</span>, n, <span class="number">0</span>, n - <span class="number">1</span>)
        prev = cur

    <span class="keyword">return</span> prev[n]
</code></pre>

      <div class="tip-box">
        <div class="label">How to Verify Monotonicity</div>
        <p>Compute the cost function C(l, r). If C satisfies the <strong>quadrangle inequality</strong> (C(a,c) + C(b,d) &le; C(a,d) + C(b,c) for a &le; b &le; c &le; d), then the optimal split points are monotone. Common qualifying costs: sum of squares of segment sums, variance-based costs.</p>
      </div>
    </section>

    <!-- ============================================================ -->
    <!-- SECTION 8: KNUTH'S OPTIMIZATION -->
    <!-- ============================================================ -->
    <section id="knuth-opt" class="section">
      <h2>8. Knuth's Optimization</h2>
      <p>Knuth's optimization speeds up interval DP from O(n^3) to O(n^2). It applies when the cost function C(i, j) satisfies the <strong>quadrangle inequality</strong> AND the DP recurrence is: dp[i][j] = min over i &lt; k &lt; j of (dp[i][k] + dp[k][j] + C(i, j)).</p>

      <div class="formula-box">
        <div class="label">Conditions</div>
        <p>1. <strong>Quadrangle Inequality:</strong> C(a,c) + C(b,d) &le; C(a,d) + C(b,c) for all a &le; b &le; c &le; d.</p>
        <p>2. <strong>Monotonicity:</strong> C(a,c) &le; C(b,d) for a &le; b &le; c &le; d.</p>
        <p>Then opt[i][j-1] &le; opt[i][j] &le; opt[i+1][j], limiting the search range for k.</p>
      </div>

      <pre><code><span class="lang-label">C++</span>
<span class="comment">// Knuth's optimization for optimal BST / interval DP</span>
<span class="keyword">void</span> <span class="function">knuth_dp</span>(<span class="keyword">int</span> n, vector&lt;vector&lt;<span class="keyword">long long</span>&gt;&gt;&amp; dp,
              vector&lt;vector&lt;<span class="keyword">int</span>&gt;&gt;&amp; opt,
              vector&lt;vector&lt;<span class="keyword">long long</span>&gt;&gt;&amp; C) {
    <span class="comment">// Base cases</span>
    <span class="keyword">for</span> (<span class="keyword">int</span> i = <span class="number">0</span>; i &lt; n; i++) {
        dp[i][i] = <span class="number">0</span>;
        opt[i][i] = i;
    }

    <span class="keyword">for</span> (<span class="keyword">int</span> len = <span class="number">2</span>; len &lt;= n; len++) {
        <span class="keyword">for</span> (<span class="keyword">int</span> i = <span class="number">0</span>; i + len - <span class="number">1</span> &lt; n; i++) {
            <span class="keyword">int</span> j = i + len - <span class="number">1</span>;
            dp[i][j] = <span class="number">1e18</span>;

            <span class="comment">// Key: search only from opt[i][j-1] to opt[i+1][j]</span>
            <span class="keyword">int</span> lo = opt[i][j - <span class="number">1</span>];
            <span class="keyword">int</span> hi = (j + <span class="number">1</span> &lt; n) ? opt[i + <span class="number">1</span>][j] : j;

            <span class="keyword">for</span> (<span class="keyword">int</span> k = lo; k &lt;= hi; k++) {
                <span class="keyword">long long</span> val = dp[i][k] + dp[k + <span class="number">1</span>][j] + C[i][j];
                <span class="keyword">if</span> (val &lt; dp[i][j]) {
                    dp[i][j] = val;
                    opt[i][j] = k;
                }
            }
        }
    }
}
</code></pre>

      <pre><code><span class="lang-label">Python</span>
<span class="keyword">def</span> <span class="function">knuth_dp</span>(n, C):
    <span class="string">"""
    Interval DP with Knuth's optimization.
    C[i][j] = cost of interval [i, j].
    Returns dp[0][n-1] = optimal cost.
    """</span>
    INF = <span class="builtin">float</span>(<span class="string">'inf'</span>)
    dp = [[<span class="number">0</span>] * n <span class="keyword">for</span> _ <span class="keyword">in</span> <span class="builtin">range</span>(n)]
    opt = [[<span class="number">0</span>] * n <span class="keyword">for</span> _ <span class="keyword">in</span> <span class="builtin">range</span>(n)]

    <span class="keyword">for</span> i <span class="keyword">in</span> <span class="builtin">range</span>(n):
        opt[i][i] = i

    <span class="keyword">for</span> length <span class="keyword">in</span> <span class="builtin">range</span>(<span class="number">2</span>, n + <span class="number">1</span>):
        <span class="keyword">for</span> i <span class="keyword">in</span> <span class="builtin">range</span>(n - length + <span class="number">1</span>):
            j = i + length - <span class="number">1</span>
            dp[i][j] = INF
            lo = opt[i][j - <span class="number">1</span>]
            hi = opt[i + <span class="number">1</span>][j] <span class="keyword">if</span> i + <span class="number">1</span> &lt; n <span class="keyword">else</span> j
            <span class="keyword">for</span> k <span class="keyword">in</span> <span class="builtin">range</span>(lo, hi + <span class="number">1</span>):