76. Minimum Window Substring

解題說明

C++ 解法

複雜度分析

虛擬碼

1. Build frequency map for t (need[])
2. left=0, have=0, required=distinct chars in t
3. Expand right pointer:
   a. Add s[right] to window
   b. If s[right] is needed and window count meets need: have++
4. While have == required:
   a. Update min window if smaller
   b. Remove s[left] from window, advance left
   c. If removal breaks a requirement: have--
5. Return min window substring

其他解法

延伸思考