LeetCode 76: Minimum Window Substring — Step-by-Step Visual Trace

#python #leetcode #algorithms

Hard — Sliding Window | Two Pointers | Hash Table | String

The Problem

Find the minimum window substring in string s that contains all characters of string t. If no such window exists, return an empty string.

Approach

Use a sliding window technique with two pointers to expand the window until all required characters are found, then contract from the left to find the minimum valid window. Track character frequencies to determine when a valid window is formed.

Time: O(|s| + |t|) · Space: O(|s| + |t|)

Code

class Solution:
    def minWindow(self, s: str, t: str) -> str:
        if not s or not t:
            return ""

        char_freq_t = {}
        for char in t:
            char_freq_t[char] = char_freq_t.get(char, 0) + 1

        left, right = 0, 0
        char_freq_temp = {}
        required_chars = len(char_freq_t)
        formed_chars = 0
        min_length = float("inf")
        min_window = ""

        while right < len(s):
            char_freq_temp[s[right]] = char_freq_temp.get(s[right], 0) + 1

            if (
                s[right] in char_freq_t
                and char_freq_temp[s[right]] == char_freq_t[s[right]]
            ):
                formed_chars += 1

            while left <= right and formed_chars == required_chars:
                if right - left + 1 < min_length:
                    min_length = right - left + 1
                    min_window = s[left : right + 1]

                char_freq_temp[s[left]] -= 1
                if (
                    s[left] in char_freq_t
                    and char_freq_temp[s[left]] < char_freq_t[s[left]]
                ):
                    formed_chars -= 1

                left += 1

            right += 1

        return min_window