# Minimum Window Substring Problem

## Description

LeetCode Problem 76.

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:

``````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:

``````1
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Input: s = "a", t = "a"
Output: "a"
Explanation: The entire string s is the minimum window.
``````

Example 3:

``````1
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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 <= 10^5
• s and t consist of uppercase and lowercase English letters.

## Sample C++ Code

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class Solution {
public:
string minWindow(string s, string t) {
int len_s = s.size();
int len_t = t.size();

unordered_map<char, int> ht;
for (int i = 0; i < len_t; i ++) {
ht[t[i]] ++;
}

int i = 0, j = 0;
int cnt = ht.size();
int min_win = INT_MAX;
string ans;
for (i = 0; i < len_s; i ++) {
while ((cnt != 0) && (j < len_s)) {
if (ht.find(s[j]) != ht.end()) {
ht[s[j]] --;
if (ht[s[j]] == 0)
cnt --;
}
j ++;
}
if ((cnt == 0) && (min_win > j - i)) {
min_win = j - i;
ans = s.substr(i, j - i);
}
if (t.find(s[i]) != string::npos) {
ht[s[i]] ++;
if (ht[s[i]] > 0)
cnt ++;
}

}
return ans;
}
};
``````