diff --git a/solutions/62. Unique Paths/README.md b/solutions/62. Unique Paths/README.md
index 99aa4ce..2b2222d 100644
--- a/solutions/62. Unique Paths/README.md	
+++ b/solutions/62. Unique Paths/README.md	
@@ -59,3 +59,13 @@ dp[i][j] = dp[i-1][j] + dp[i][j-1]
 * 複雜度：
   * 時間複雜度：O(M * N)
   * 空間複雜度：O(M * N)
+
+仔細觀察，會發現不需要用 M * N 的空間，只需要一個一維的陣列，因為每次只需要用到上一個 row 的資料，所以可以省略掉一個維度。這樣空間複雜度就會變成 O(N)：
+
+```
+dp[j] += dp[j - 1]
+```
+
+* 複雜度：
+  * 時間複雜度：O(M * N)
+  * 空間複雜度：O(N)
\ No newline at end of file
diff --git a/solutions/62. Unique Paths/solution.py b/solutions/62. Unique Paths/solution.py
index 027dabe..f69ff05 100644
--- a/solutions/62. Unique Paths/solution.py	
+++ b/solutions/62. Unique Paths/solution.py	
@@ -1,10 +1,10 @@
 class Solution:
     def uniquePaths(self, m: int, n: int) -> int:
-        dp = [[0] * n for _ in range(m)]
-        for i in range(m):
-            for j in range(n):
-                if i == 0 or j == 0:
-                    dp[i][j] = 1
-                else:
-                    dp[i][j] = dp[i-1][j] + dp[i][j-1]
-        return dp[-1][-1]
+        dp = [1] * n
+    
+        for _ in range(1, m):
+            for j in range(1, n):
+                dp[j] += dp[j - 1]
+        
+        return dp[-1]
+        
\ No newline at end of file