76. Minimum Window Substring

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76. Minimum Window Substring

Description

Difficulty: Hard

Related Topics: Hash Table, String, Sliding Window

Given two strings s and t of lengths m and n respectively, return the minimum window substring of s such that every character in t (including duplicates) is included in the window. If there is no such substring, return the empty string "".

The testcases will be generated such that the answer is unique.

Example 1:

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Input: s = "ADOBECODEBANC", t = "ABC"
Output: "BANC"
Explanation: The minimum window substring "BANC" includes 'A', 'B', and 'C' from string t.

Example 2:

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Input: s = "a", t = "a"
Output: "a"
Explanation: The entire string s is the minimum window.

Example 3:

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Input: s = "a", t = "aa"
Output: ""
Explanation: Both 'a's from t must be included in the window.
Since the largest window of s only has one 'a', return empty string.

Constraints:

  • m == s.length
  • n == t.length
  • 1 <= m, n <= 105
  • s and t consist of uppercase and lowercase English letters.

Follow up: Could you find an algorithm that runs in O(m + n) time?

Hints/Notes

  • 2023/08/09
  • Sliding window, keep taking new elements while trying to shrink the left boundry
  • Use map to record if all required letters are included
  • 0x3F’s solution(checked)

Solution

Language: C++

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class Solution {
public:
    string minWindow(string s, string t) {
        unordered_map<char, int> need, window;
        for (char c: t) {
            need[c]++;
        }
        int valid = 0, left = 0, right = 0, start = 0, len = INT_MAX;
        while (right < s.size()) {
            char c = s[right++];
            window[c]++;
            if (need.count(c)) {
                if (need[c] == window[c]) {
                    valid++;
                }
            }
            while (valid == need.size()) {
                if (right - left < len) {
                    start = left;
                    len = right - left;
                }
                c = s[left];
                left++;
                if (need.count(c)) {
                    if (need[c] == window[c]) {
                        valid--;
                    }
                    window[c]--;
                }
            }
        }
        return len == INT_MAX ? "": s.substr(start, len);
    }
};