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+Trapping Rain Water
+
+
+Given n non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it is able to trap after raining.
+
+Input: arr[]   = {3, 0, 2, 0, 4}
+Output: 7
+
+Explanation:
+We can trap "3 units" of water between 3 and 2,
+"1 unit" on top of bar 2 and "3 units" between 2 
+and 4.
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diff --git a/Interview Questions/Trapping Rain Water/Trapping Rain Water.cpp b/Interview Questions/Trapping Rain Water/Trapping Rain Water.cpp
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+// C++ implementation of the approach
+#include<bits/stdc++.h>
+using namespace std;
+
+int maxWater(int arr[], int n)
+{
+
+	// indices to traverse the array
+	int left = 0;
+	int right = n-1;
+
+	// To store Left max and right max
+	// for two pointers left and right
+	int l_max = 0;
+	int r_max = 0;
+
+	// To store the total amount
+	// of rain water trapped
+	int result = 0;
+	while (left <= right)
+	{
+
+	// We need check for minimum of left
+	// and right max for each element
+	if(r_max <= l_max)
+	{
+
+		// Add the difference between
+		// current value and right max at index r
+		result += max(0, r_max-arr[right]);
+
+		// Update right max
+		r_max = max(r_max, arr[right]);
+
+		// Update right pointer
+		right -= 1;
+	}
+	else
+	{
+
+		// Add the difference between
+		// current value and left max at index l
+		result += max(0, l_max-arr[left]);
+
+		// Update left max
+		l_max = max(l_max, arr[left]);
+
+		// Update left pointer
+		left += 1;
+	}
+	}
+	return result;
+}
+
+// Driver code
+int main() {
+	int arr[] = {0, 1, 0, 2, 1, 0, 1, 3, 2, 1, 2, 1};
+	int n = sizeof(arr)/sizeof(arr[0]);
+	cout << maxWater(arr, n) << endl;
+	return 0;
+}
+
+