diff --git a/62_unique_paths_medium/1dp.py b/62_unique_paths_medium/1dp.py
new file mode 100644
index 0000000..03f549c
--- /dev/null
+++ b/62_unique_paths_medium/1dp.py
@@ -0,0 +1,18 @@
+#
+# @lc app=leetcode id=62 lang=python3
+#
+# [62] Unique Paths
+#
+
+# @lc code=start
+class Solution:
+    def uniquePaths(self, m: int, n: int) -> int:
+        num_paths_in_row = [0] * n
+        num_paths_in_row[0] = 1
+        for _ in range(m):
+            for column in range(1, n):
+                num_paths_in_row[column] += num_paths_in_row[column - 1]
+        return num_paths_in_row[-1]
+
+
+# @lc code=end
diff --git a/62_unique_paths_medium/2dp.py b/62_unique_paths_medium/2dp.py
new file mode 100644
index 0000000..d32b1d0
--- /dev/null
+++ b/62_unique_paths_medium/2dp.py
@@ -0,0 +1,21 @@
+#
+# @lc app=leetcode id=62 lang=python3
+#
+# [62] Unique Paths
+#
+
+# @lc code=start
+class Solution:
+    def uniquePaths(self, m: int, n: int) -> int:
+        num_paths = [[0] * n for _ in range(m)]
+        for row in range(m):
+            for column in range(n):
+                if row == 0 or column == 0:
+                    num_paths[row][column] = 1
+                    continue
+                num_paths[row][column] = num_paths[row - 1][column] + num_paths[row][column - 1]
+
+        return num_paths[-1][-1]
+
+
+# @lc code=end
diff --git a/62_unique_paths_medium/README.md b/62_unique_paths_medium/README.md
new file mode 100644
index 0000000..fb0a16f
--- /dev/null
+++ b/62_unique_paths_medium/README.md
@@ -0,0 +1,117 @@
+# 問題へのリンク
+[Unique Paths - LeetCode](https://leetcode.com/problems/unique-paths/description/)
+
+# 言語
+Python
+
+# 問題の概要
+m x n のグリッドが与えられる。左上のマスから右下のマスまで、右または下にのみ移動する場合のユニークな経路の総数を求める問題である。
+
+# 自分の解法
+
+この問題は、最終的に `(m-1) + (n-1)` 回の移動のうち、どの `m-1` 回を「下」への移動にするか（残りは「右」になる）を選択する組み合わせの問題として解釈できる。したがって、組み合わせの公式 `C(m+n-2, m-1)` を用いて解くことができる。
+
+## step1
+`math.comb` を利用して、組み合わせ計算を直接実装した。これは最も簡潔で効率的な解法である。
+
+```python
+import math
+
+
+class Solution:
+    def uniquePaths(self, m: int, n: int) -> int:
+        if m == 1 and n == 1:
+            return 1
+        return math.comb(m + n - 2, m - 1)
+```
+
+- 時間計算量：`O(min(m, n))`。`math.comb`の実装に依存するが、効率的に計算される。
+- 空間計算量：`O(1)`。
+
+## step2
+`math.comb` に頼らず、組み合わせを計算する関数を自前で実装した。これにより、ライブラリの内部実装への理解を深め、特定のバージョンに依存しないコードとなる。
+
+```python
+class Solution:
+    def uniquePaths(self, m: int, n: int) -> int:
+        def combination(n: int, k: int) -> int:
+            assert 0 <= k <= n
+            if n == k == 0:
+                return 1
+            if n - k < k:
+                k = n - k
+            numerator = 1
+            denominator = 1
+            for num in range(n - k + 1, n + 1):
+                numerator *= num
+            for num in range(1, k + 1):
+                denominator *= num
+            return numerator // denominator
+
+        moves_to_right = n - 1
+        moves_to_bottom = m - 1
+        total_moves = moves_to_right + moves_to_bottom
+        return combination(total_moves, moves_to_bottom)
+```
+
+- 時間計算量：`O(min(m, n))`。組み合わせの計算 `C(n, k)` のループ回数は `k` に比例するため。
+- 空間計算量：`O(1)`。
+
+# 別解・模範解答
+
+### 1. 動的計画法 (2D DP)
+`dp[i][j]` を `(i, j)` に到達する経路の数と定義する。`dp[i][j] = dp[i-1][j] + dp[i][j-1]` という漸化式で解くことができる。
+
+```python
+class Solution:
+    def uniquePaths(self, m: int, n: int) -> int:
+        num_paths = [[0] * n for _ in range(m)]
+        for row in range(m):
+            for column in range(n):
+                if row == 0 or column == 0:
+                    num_paths[row][column] = 1
+                    continue
+                num_paths[row][column] = num_paths[row - 1][column] + num_paths[row][column - 1]
+
+        return num_paths[-1][-1]
+```
+
+- 時間計算量：`O(m*n)`。グリッドの全マスを一度ずつ計算するため。
+- 空間計算量：`O(m*n)`。`m x n` のDPテーブルを保持するため。
+
+### 2. 動的計画法 (1D DP)
+2D DPの空間計算量を最適化したアプローチである。`i`行目の計算は`i-1`行目の情報のみに依存するため、1次元配列を使い回すことで空間を削減できる。
+
+```python
+class Solution:
+    def uniquePaths(self, m: int, n: int) -> int:
+        num_paths_in_row = [1] * n
+        for _ in range(1, m):
+            for column in range(1, n):
+                num_paths_in_row[column] += num_paths_in_row[column - 1]
+        return num_paths_in_row[-1]
+```
+
+- 時間計算量：`O(m*n)`。
+- 空間計算量：`O(n)`。1行分の情報を保持する配列のみ使用するため。
+
+### 3. 再帰 (メモ化)
+トップダウンの動的計画法アプローチである。`@cache`デコレータ（メモ化）を使い、重複する計算を避ける。
+
+```python
+from functools import cache
+
+
+class Solution:
+    @cache
+    def uniquePaths(self, m: int, n: int) -> int:
+        if m == 1 or n == 1:
+            return 1
+        return self.uniquePaths(m - 1, n) + self.uniquePaths(m, n - 1)
+```
+
+- 時間計算量：`O(m*n)`。メモ化により、各 `(m, n)` の組み合わせは一度しか計算されないため。
+- 空間計算量：`O(m*n)`。再帰のスタックとキャッシュのための空間が必要である。
+
+# 次に解く問題
+- [Binary Tree Level Order Traversal - LeetCode](https://leetcode.com/problems/binary-tree-level-order-traversal/)
diff --git a/62_unique_paths_medium/bfs.py b/62_unique_paths_medium/bfs.py
new file mode 100644
index 0000000..3620cfc
--- /dev/null
+++ b/62_unique_paths_medium/bfs.py
@@ -0,0 +1,38 @@
+#
+# @lc app=leetcode id=62 lang=python3
+#
+# [62] Unique Paths
+#
+
+# @lc code=start
+from collections import deque
+
+
+class Solution:
+    def uniquePaths(self, m: int, n: int) -> int:
+        top_left = (0, 0)
+        num_paths_grid = [[0 for _ in range(n)] for _ in range(m)]
+        num_paths_grid[top_left[0]][top_left[1]] = 1
+        queue: deque[tuple[int, int]] = deque([top_left])
+        directions = ((0, 1), (1, 0))
+        visited: set[tuple[int, int]] = set([top_left])
+
+        def is_valid(row, column) -> bool:
+            return 0 <= row < m and 0 <= column < n
+
+        while queue:
+            row, column = queue.popleft()
+            num_paths = num_paths_grid[row][column]
+            neighbors = [(row + drow, column + dcol) for drow, dcol in directions]
+            for neighbor_row, neighbor_col in neighbors:
+                if not is_valid(neighbor_row, neighbor_col):
+                    continue
+                num_paths_grid[neighbor_row][neighbor_col] += num_paths
+                if (neighbor_row, neighbor_col) in visited:
+                    continue
+                visited.add((neighbor_row, neighbor_col))
+                queue.append((neighbor_row, neighbor_col))
+        return num_paths_grid[m - 1][n - 1]
+
+
+# @lc code=end
diff --git a/62_unique_paths_medium/recursive.py b/62_unique_paths_medium/recursive.py
new file mode 100644
index 0000000..798266a
--- /dev/null
+++ b/62_unique_paths_medium/recursive.py
@@ -0,0 +1,19 @@
+#
+# @lc app=leetcode id=62 lang=python3
+#
+# [62] Unique Paths
+#
+
+# @lc code=start
+from functools import cache
+
+
+class Solution:
+    @cache
+    def uniquePaths(self, m: int, n: int) -> int:
+        if m == 1 or n == 1:
+            return 1
+        return self.uniquePaths(m - 1, n) + self.uniquePaths(m, n - 1)
+
+
+# @lc code=end
diff --git a/62_unique_paths_medium/step1.py b/62_unique_paths_medium/step1.py
new file mode 100644
index 0000000..3154ad8
--- /dev/null
+++ b/62_unique_paths_medium/step1.py
@@ -0,0 +1,20 @@
+#
+# @lc app=leetcode id=62 lang=python3
+#
+# [62] Unique Paths
+#
+
+# @lc code=start
+
+
+import math
+
+
+class Solution:
+    def uniquePaths(self, m: int, n: int) -> int:
+        if m == 1 and n == 1:
+            return 1
+        return math.comb(m + n - 2, m - 1)
+
+
+# @lc code=end
diff --git a/62_unique_paths_medium/step1_comb_from_scratch.py b/62_unique_paths_medium/step1_comb_from_scratch.py
new file mode 100644
index 0000000..3f668e2
--- /dev/null
+++ b/62_unique_paths_medium/step1_comb_from_scratch.py
@@ -0,0 +1,25 @@
+#
+# @lc app=leetcode id=62 lang=python3
+#
+# [62] Unique Paths
+#
+
+# @lc code=start
+class Solution:
+    def uniquePaths(self, m: int, n: int) -> int:
+        def combination(n: int, k: int) -> int:
+            assert 0 <= k <= n
+            if n == k == 0:
+                return 1
+            numerator = 1
+            denominator = 1
+            for num in range(n - k + 1, n + 1):
+                numerator *= num
+            for num in range(1, k + 1):
+                denominator *= num
+            return numerator // denominator
+
+        return combination(n + m - 2, m - 1)
+
+
+# @lc code=end
diff --git a/62_unique_paths_medium/step2.py b/62_unique_paths_medium/step2.py
new file mode 100644
index 0000000..cfa84fd
--- /dev/null
+++ b/62_unique_paths_medium/step2.py
@@ -0,0 +1,28 @@
+#
+# @lc app=leetcode id=62 lang=python3
+#
+# [62] Unique Paths
+#
+
+# @lc code=start
+class Solution:
+    def uniquePaths(self, m: int, n: int) -> int:
+        def combination(n: int, k: int) -> int:
+            assert 0 <= k <= n
+            if n == k == 0:
+                return 1
+            numerator = 1
+            denominator = 1
+            for num in range(n - k + 1, n + 1):
+                numerator *= num
+            for num in range(1, k + 1):
+                denominator *= num
+            return numerator // denominator
+
+        vertical_moves = m - 1
+        horizontal_moves = n - 1
+        total_moves = vertical_moves + horizontal_moves
+        return combination(total_moves, vertical_moves)
+
+
+# @lc code=end
diff --git a/62_unique_paths_medium/step2_comb_from_scratch.py b/62_unique_paths_medium/step2_comb_from_scratch.py
new file mode 100644
index 0000000..9cc4cda
--- /dev/null
+++ b/62_unique_paths_medium/step2_comb_from_scratch.py
@@ -0,0 +1,30 @@
+#
+# @lc app=leetcode id=62 lang=python3
+#
+# [62] Unique Paths
+#
+
+# @lc code=start
+class Solution:
+    def uniquePaths(self, m: int, n: int) -> int:
+        def combination(n: int, k: int) -> int:
+            assert 0 <= k <= n
+            if n == k == 0:
+                return 1
+            if n - k < k:
+                k = n - k
+            numerator = 1
+            denominator = 1
+            for num in range(n - k + 1, n + 1):
+                numerator *= num
+            for num in range(1, k + 1):
+                denominator *= num
+            return numerator // denominator
+
+        moves_to_right = n - 1
+        moves_to_bottom = m - 1
+        total_moves = moves_to_right + moves_to_bottom
+        return combination(total_moves, moves_to_bottom)
+
+
+# @lc code=end
diff --git a/62_unique_paths_medium/step3_1dp.py b/62_unique_paths_medium/step3_1dp.py
new file mode 100644
index 0000000..03f549c
--- /dev/null
+++ b/62_unique_paths_medium/step3_1dp.py
@@ -0,0 +1,18 @@
+#
+# @lc app=leetcode id=62 lang=python3
+#
+# [62] Unique Paths
+#
+
+# @lc code=start
+class Solution:
+    def uniquePaths(self, m: int, n: int) -> int:
+        num_paths_in_row = [0] * n
+        num_paths_in_row[0] = 1
+        for _ in range(m):
+            for column in range(1, n):
+                num_paths_in_row[column] += num_paths_in_row[column - 1]
+        return num_paths_in_row[-1]
+
+
+# @lc code=end
diff --git a/62_unique_paths_medium/step3_math.py b/62_unique_paths_medium/step3_math.py
new file mode 100644
index 0000000..9281d24
--- /dev/null
+++ b/62_unique_paths_medium/step3_math.py
@@ -0,0 +1,19 @@
+#
+# @lc app=leetcode id=62 lang=python3
+#
+# [62] Unique Paths
+#
+
+# @lc code=start
+import math
+
+
+class Solution:
+    def uniquePaths(self, m: int, n: int) -> int:
+        moves_to_bottom = m - 1
+        moves_to_right = n - 1
+        total_moves = moves_to_bottom + moves_to_right
+        return math.comb(total_moves, moves_to_bottom)
+
+
+# @lc code=end