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# Minimum Window Substring LeetCode Solution

## Problem – 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.

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?

### Minimum Window Substring LeetCode Solution in Python

``````def minWindow(self, s, t):
need, missing = collections.Counter(t), len(t)
i = I = J = 0
for j, c in enumerate(s, 1):
missing -= need[c] > 0
need[c] -= 1
if not missing:
while i < j and need[s[i]] < 0:
need[s[i]] += 1
i += 1
if not J or j - i <= J - I:
I, J = i, j
return s[I:J]``````

### Minimum Window Substring LeetCode Solution in Java

``````public class Solution {
public String minWindow(String s, String t) {
if(s == null || s.length() < t.length() || s.length() == 0){
return "";
}
HashMap<Character,Integer> map = new HashMap<Character,Integer>();
for(char c : t.toCharArray()){
if(map.containsKey(c)){
map.put(c,map.get(c)+1);
}else{
map.put(c,1);
}
}
int left = 0;
int minLeft = 0;
int minLen = s.length()+1;
int count = 0;
for(int right = 0; right < s.length(); right++){
if(map.containsKey(s.charAt(right))){
map.put(s.charAt(right),map.get(s.charAt(right))-1);
if(map.get(s.charAt(right)) >= 0){
count ++;
}
while(count == t.length()){
if(right-left+1 < minLen){
minLeft = left;
minLen = right-left+1;
}
if(map.containsKey(s.charAt(left))){
map.put(s.charAt(left),map.get(s.charAt(left))+1);
if(map.get(s.charAt(left)) > 0){
count --;
}
}
left ++ ;
}
}
}
if(minLen>s.length())
{
return "";
}

return s.substring(minLeft,minLeft+minLen);
}
``````

### Minimum Window Substring LeetCode Solution in C++

``````class Solution {
public:
string minWindow(string s, string t) {
if (s.size() == 0 || t.size() == 0) return "";
vector<int> remaining(128, 0);
int required = t.size();
for (int i = 0; i < required; i++) remaining[t[i]]++;
// left is the start index of the min-length substring ever found
int min = INT_MAX, start = 0, left = 0, i = 0;
while(i <= s.size() && start < s.size()) {
if(required) {
if (i == s.size()) break;
remaining[s[i]]--;
if (remaining[s[i]] >= 0) required--;
i++;
} else {
if (i - start < min) {
min = i -start;
left = start;
}
remaining[s[start]]++;
if (remaining[s[start]] > 0) required++;
start++;
}
}
return min == INT_MAX? "" : s.substr(left, min);
}
};
``````
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