# 18. Minimum Window Substring

Given two strings `s` and `t`, return *the minimum window in `s` which will contain all the characters in `t`*. If there is no such window in `s` that covers all characters in `t`, return *the empty string `""`*.

**Note** that If there is such a window, it is guaranteed that there will always be only one unique minimum window in `s`.

```
Example 1:
Input: s = "ADOBECODEBANC", t = "ABC"
Output: "BANC"

Example 2:
Input: s = "a", t = "a"
Output: "a"
```

## Solution: (Sliding Window + Hash Maps)

**Algorithm**

1. We start with two pointers, left and right initially pointing to the first element of the string S.
2. We use the right pointer to expand the window until we get a desirable window i.e. a window that contains all of the characters of T.
3. Once we have a window with all the characters, we can move the left pointer ahead one by one. If the window is still a desirable one we keep on updating the minimum window size.
4. If the window is not desirable any more, we repeat step 2 onwards.

![](https://leetcode.com/problems/minimum-window-substring/Figures/76/76_Minimum_Window_Substring_2.png)

The above steps are repeated until we have looked at all the windows. The smallest window is returned

.![](https://leetcode.com/problems/minimum-window-substring/Figures/76/76_Minimum_Window_Substring_3.png)

```cpp
class Solution
{
public:
    string minWindow(string s, string t)
    {

        int minLen = INT_MAX;

        unordered_map<char, int> tc;
        unordered_map<char, int> sc;

        string res = "";

        for (int i = 0; i < t.length(); i++)
        {
            tc[t[i]]++;
        }

        int start = 0;
        int i = 0;
        int count = 0;

        while (i < s.length())
        {
            if (tc.find(s[i]) != tc.end())
            {
                int val = tc[s[i]];

                int sameVal = sc[s[i]];
                if (sameVal < val)
                {
                    count++;
                }
                sc[s[i]]++;
            }

            while (count >= t.length())
            {
                if (i - start + 1 < minLen)
                {
                    res = s.substr(start, i - start + 1);
                    minLen = i - start + 1;
                }

                if (tc.find(s[start]) != tc.end())
                {
                    int val = tc[s[start]];

                    int sameVal = sc[s[start]];
                    if (sameVal <= val)
                    {
                        count--;
                    }
                    sc[s[start]]--;
                }
                start++;
            }

            i++;
        }
        return res;
    }
};
```

**Time Complexity: O(S + T)**\
**Space Complexity: O(S + T)**


---

# Agent Instructions: Querying This Documentation

If you need additional information that is not directly available in this page, you can query the documentation dynamically by asking a question.

Perform an HTTP GET request on the current page URL with the `ask` query parameter:

```
GET https://soumyajit4419.gitbook.io/ds-algo/strings/17.-minimum-window-substring.md?ask=<question>
```

The question should be specific, self-contained, and written in natural language.
The response will contain a direct answer to the question and relevant excerpts and sources from the documentation.

Use this mechanism when the answer is not explicitly present in the current page, you need clarification or additional context, or you want to retrieve related documentation sections.