README_EN.md

76. Minimum Window Substring

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Description

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.

A substring is a contiguous sequence of characters within the string.

 

Example 1:

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

Example 2:

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

Example 3:

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?

Solutions

Python3

Java

TypeScript

function minWindow(s: string, t: string): string {
    let n1 = s.length,
        n2 = t.length;
    if (n1 < n2) return '';
    let need = new Array(128).fill(0);
    let window = new Array(128).fill(0);
    for (let i = 0; i < n2; ++i) {
        ++need[t.charCodeAt(i)];
    }

    let left = 0,
        right = 0;
    let res = '';
    let count = 0;
    let min = n1 + 1;
    while (right < n1) {
        let cur = s.charCodeAt(right);
        ++window[cur];
        if (need[cur] > 0 && need[cur] >= window[cur]) {
            ++count;
        }
        while (count == n2) {
            cur = s.charCodeAt(left);
            if (need[cur] > 0 && need[cur] >= window[cur]) {
                --count;
            }
            if (right - left + 1 < min) {
                min = right - left + 1;
                res = s.slice(left, right + 1);
            }
            --window[cur];
            ++left;
        }
        ++right;
    }
    return res;
}

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