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# 76. Minimum Window Substring
## Problem Statement
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.
## Intuition
```
Approach:
Basically, we need to maintain count of the t string in a map and
maintain the same count for the s string in another map
We move the low pointer when, the count of all alphabets in window >= count of t
Shift low pointer till the condition is unmet again,
Repeat the steps
```
### Links
### Video Links
### Approach 1:
```
Sliding Window
```
{% code title="C++" lineNumbers="true" %}
```cpp
class Solution {
public:
string minWindow(string S, string T) {
string result;
if(S.empty() || T.empty()){
return result;
}
unordered_map map;
unordered_map window;
for(int i = 0; i < T.length(); i++){
map[T[i]]++;
}
int minLength = INT_MAX;
int letterCounter = 0;
for(int slow = 0, fast = 0; fast < S.length(); fast++){
char c = S[fast];
if(map.find(c) != map.end()){
window[c]++;
if(window[c] <= map[c]){
letterCounter++;
}
}
if(letterCounter >= T.length()){
while(map.find(S[slow]) == map.end() || window[S[slow]] > map[S[slow]]){
window[S[slow]]--;
slow++;
}
if(fast - slow + 1 < minLength){
minLength = fast - slow + 1;
result = S.substr(slow, minLength);
}
}
}
return result;
}
};
```
{% endcode %}
### Approach 2:
```
```
{% code title="C++" lineNumbers="true" %}
```cpp
```
{% endcode %}
### Approach 3:
```
```
{% code title="C++" lineNumbers="true" %}
```cpp
```
{% endcode %}
### Approach 4:
```
```
{% code title="C++" lineNumbers="true" %}
```cpp
```
{% endcode %}
### Similar Problems
###
---
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