76

76. Minimum Window Substring (Hard)

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:

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?

Related Topics:
Hash Table, String, Sliding Window

Similar Questions:

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• Minimum Size Subarray Sum (Hard)
• Sliding Window Maximum (Hard)
• Permutation in String (Hard)
• Smallest Range Covering Elements from K Lists (Hard)
• Minimum Window Subsequence (Hard)

Solution 1. Minimum Sliding Window

// OJ: https://leetcode.com/problems/minimum-window-substring/
// Author: github.com/lzl124631x
// Time: O(M+N)
// Space: O(C) where C is the range of characters
class Solution {
public:
    string minWindow(string s, string t) {
        int M = s.size(), N = t.size(), cnt[128] = {}, matched = 0, minLen = 
INT_MAX, start = -1;
        for (char c : t) cnt[c]++;
        for (int i = 0, j = 0; j < M; ++j) {
            matched += cnt[s[j]]-- > 0;
            while (matched >= N) {
                if (j - i + 1 < minLen) {
                    minLen = j - i + 1;
                    start = i;
                }
                matched -= ++cnt[s[i++]] > 0;
            }
        }
        return start == -1 ? "" : s.substr(start, minLen);
    }
};