76. Minimum Window Substring

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.

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?

Companies:
Facebook, Amazon, LinkedIn, Google, Adobe, Uber, Apple, Snapchat, VMware, Microsoft, Flipkart, ByteDance

Related Topics:
Hash Table, String, Sliding Window

Similar Questions:

• Substring with Concatenation of All Words (Hard)
• Minimum Size Subarray Sum (Medium)
• Sliding Window Maximum (Hard)
• Permutation in String (Medium)
• 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(N)
// Space: O(C) where C is the range of characters
class Solution {
public:
    string minWindow(string s, string t) {
        unordered_map<char, int> target, cnt;
        int len = INT_MAX, i = 0, N = s.size(), matched = 0, begin = 0;
        for (char c : t) target[c]++;
        for (int j = 0; j < N; ++j) {
            if (++cnt[s[j]] <= target[s[j]]) ++matched;
            while (matched == t.size()) {
                if (j - i + 1 < len) {
                    len = j - i + 1;
                    begin = i;
                }
                if (--cnt[s[i]] < target[s[i]]) --matched;
                ++i;
            }
        }
        return len == INT_MAX ? "" : s.substr(begin, len);
    }
};

Solution 2. Shrinkable Slinding Window

// OJ: https://leetcode.com/problems/minimum-window-substring/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(C) where C is the range of characters
// Ref: https://discuss.leetcode.com/topic/30941/here-is-a-10-line-template-that-can-solve-most-substring-problems
class Solution {
public:
    string minWindow(string s, string t) {
        int cnt[128] = {};
        for (char c : t) cnt[c]++;
        int N = s.size(), i = 0, j = 0, start = -1, minLen = INT_MAX, matched = 0;
        while (j < N) {
            matched += --cnt[s[j++]] >= 0;
            while (matched == t.size()) {
                if (j - i < minLen) minLen = j - i, start = i;
                matched -= ++cnt[s[i++]] > 0;
            }
        }
        return start == -1 ? "" : s.substr(start, minLen);
    }
};