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+# 387. First Unique Character in a String
+
+## 1st
+
+### ①
+
+素直に文字ごとの出現頻度を辞書にして、2パス目で左から1回しか出ていない文字を確認するようにした。
+
+defaultdictの代わりにCounter()を使っても良い。
+
+所要時間: 4:00
+
+n: len(s)
+- 時間計算量: O(n)
+- 空間計算量: O(n)
+
+```py
+class Solution:
+    def firstUniqChar(self, s: str) -> int:
+        char_to_frequency = defaultdict(int)
+        for c in s:
+            char_to_frequency[c] += 1
+        for i, c in enumerate(s):
+            if char_to_frequency[c] == 1:
+                return i
+        return -1
+```
+
+### ②
+
+Counter版
+
+所要時間: 0:41
+
+n: len(s)
+- 時間計算量: O(n)
+- 空間計算量: O(n)
+
+```py
+class Solution:
+    def firstUniqChar(self, s: str) -> int:
+        char_to_frequency = Counter(s)
+        for i, c in enumerate(s):
+            if char_to_frequency[c] == 1:
+                return i
+        return -1
+```
+
+### ③
+
+辞書を作らないのであれば、左右から検索して最初に見つかった位置が一致していればunique, という探し方もできる。
+
+所要時間: 1:33
+
+n: len(s)
+- 時間計算量: O(n^2)
+- 空間計算量: O(1)
+
+```py
+class Solution:
+    def firstUniqChar(self, s: str) -> int:
+        for c in s:
+            left_index = s.find(c)
+            if left_index == s.rfind(c):
+                return left_index
+        return -1
+```
+
+## 2nd
+
+### 参考
+
+- https://discord.com/channels/1084280443945353267/1198621745565937764/1244101017507991613
+- https://discord.com/channels/1084280443945353267/1227073733844406343/1233831347051565137
+
+議論にあったdefaultdictを書いてみる
+
+継承
+
+```py
+class DefaultDict(dict):
+  def __init__(self, default_factory=None):
+    self.default_factory = default_factory
+
+  def __missing__(self, key):
+    if self.default_factory is None:
+      raise KeyError
+    self[key] = self.default_factory()
+    return self[key]
+```
+
+委譲。こんな感じか？ `del d['foo']` とかはできるようにしていない
+
+```py
+class DefaultDict:
+    def __init__(self, default_factory=None):
+        self.d = {}
+        self.default_factory = default_factory
+
+    def __missing__(self, key):
+        if self.default_factory is None:
+            raise KeyError
+        self.d[key] = self.default_factory()
+        return self.d[key]
+
+    def __getitem__(self, key):
+        if not key in self:
+            return self.__missing__(key)
+        return self.d[key]
+
+    def __setitem__(self, key, value):
+        self.d[key] = value
+
+    def __contains__(self, key):
+        return key in self.d
+```
+
+- https://discord.com/channels/1084280443945353267/1200089668901937312/1209879956436291645
+  - https://github.com/hayashi-ay/leetcode/pull/28/files
+
+1-passの解法が面白かった。自分も参考に書いてみる。
+
+```py
+class Solution:
+    def firstUniqChar(self, s: str) -> int:
+        unique_occurrence_indexes = {} # 2回以上出現した場合はlen(s)が入る
+        for i, c in enumerate(s):
+            if c in unique_occurrence_indexes:
+                unique_occurrence_indexes[c] = len(s)
+                continue
+            unique_occurrence_indexes[c] = i
+        candidate = min(iter(unique_occurrence_indexes.values()))
+        if candidate == len(s):
+            return -1
+        return candidate
+```
+
+## 3rd
+
+```py
+class Solution:
+    def firstUniqChar(self, s: str) -> int:
+        char_to_frequency = defaultdict(int)
+        for c in s:
+            char_to_frequency[c] += 1
+        for i, c in enumerate(s):
+            if char_to_frequency[c] == 1:
+                return i
+        return -1
+```
+
+## 4th
+
+微妙な感じ。
+
+```py
+class Solution:
+    def firstUniqChar(self, s: str) -> int:
+        char_to_first_occurence_index = {}
+        seen = set()
+        for i, c in enumerate(s):
+            if c in seen:
+                if c in char_to_first_occurence_index:
+                    del char_to_first_occurence_index[c]
+                continue
+            char_to_first_occurence_index[c] = i
+            seen.add(c)
+        if len(char_to_first_occurence_index) == 0:
+            return -1
+        return min(char_to_first_occurence_index.values())
+```
+
+[こちら](https://github.com/hayashi-ay/leetcode/pull/28/files#diff-5ec7c3c87171edf4d61e9eb79fd926cafa27caf068da7474222897c8e9e7ab96R92)を参考に、書き直し。
+
+```py
+class Solution:
+    def firstUniqChar(self, s: str) -> int:
+        char_to_first_occurence_index = {}
+        duplicated_chars = set()
+        for i, c in enumerate(s):
+            if c in duplicated_chars:
+                continue
+            if c in char_to_first_occurence_index:
+                del char_to_first_occurence_index[c]
+                duplicated_chars.add(c)
+                continue
+            char_to_first_occurence_index[c] = i
+        if not char_to_first_occurence_index:
+            return -1
+        return min(char_to_first_occurence_index.values())
+```
+
+
+上で参考にしたコードはdictが挿入順に順序付けられることを利用しているが、あえてそれを保証しない3.7以前のPythonを気にするならこう書くだろうか (気にする場面は少なくなっているだろうが)。
+
+```py
+class Solution:
+    def firstUniqChar(self, s: str) -> int:
+        char_to_first_occurence_index = OrderedDict()
+        duplicated_chars = set()
+        for i, c in enumerate(s):
+            if c in duplicated_chars:
+                continue
+            if c in char_to_first_occurence_index:
+                del char_to_first_occurence_index[c]
+                duplicated_chars.add(c)
+                continue
+            char_to_first_occurence_index[c] = i
+        if not char_to_first_occurence_index:
+            return -1
+        return char_to_first_occurence_index.popitem(last=False)[1]
+```