76. Minimum Window Substring.js.md

76. Minimum Window Substring

https://leetcode-cn.com/problems/minimum-window-substring/

https://leetcode.com/problems/minimum-window-substring/

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?

方法：滑动窗口

/**
 * @param {string} s
 * @param {string} t
 * @return {string}
 */
var minWindow = function (s, t) {
    const chars = Object.create(null);
    const lookFor = Object.create(null);
    let subStr = '';
    let subLen = Infinity;
    let count = 0;
    for (let i = 0; i < t.length; i++) {
        lookFor[t[i]] = (lookFor[t[i]] || 0) + 1;
        count++;
    }
    for (let left = 0, right = 0; right < s.length; right++) {
        chars[s[right]] = (chars[s[right]] || 0) + 1;
        if (lookFor[s[right]] !== undefined) {
            if (chars[s[right]] <= lookFor[s[right]]) {
                count -= 1;
            }
        }
        if (count === 0) {
            while (lookFor[s[left]] === undefined || lookFor[s[left]] < chars[s[left]]) {
                chars[s[left]] -= 1;
                left += 1;
            }
            if (right - left + 1 < subLen) {
                subLen = right - left + 1;
                subStr = s.slice(left, right + 1);
            }
            chars[s[left]] -= 1;
            left += 1;
            count += 1;
        }
    }
    return subStr;
};