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.

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.

class Solution {
    public String minWindow(String s, String t) {
                HashMap<Character,Integer> map = new HashMap<>();
        for(int i = 0; i < t.length(); i++){
            map.put(t.charAt(i),map.getOrDefault(t.charAt(i),0)+1);
        }

        int i = 0;
        int j = 0;
        int count = map.size();
        int minimumI = 0;
        int minimumJ = 0;
        int minimum = Integer.MAX_VALUE;

        while(j < s.length()) {

//            Calculations for j
            if (map.containsKey(s.charAt(j))) {
                map.put(s.charAt(j), map.get(s.charAt(j)) - 1);
                if (map.get(s.charAt(j)) == 0) {
                    count--;
                }
            }

//            Check if count is greater than 0. Count will never be less than 0
            if (count > 0) {
                j++;
            }

//            Check if condition meets k
            else{
                while (count == 0) {
                    if (minimum > j - i + 1) {
                        minimum = j - i + 1;
                        minimumI = i;
// As endIndex of substring is exclusive and if we do +1 in return 
// So, if empty return is to be returned then too it will return 1st character of the string. 
// So if we find the answer then only we will mark it as j+1.
// If answer is not found by our calculations, then as minimumI and minimumJ are zero only, so it will return empty substring.

                        minimumJ = j + 1;
                    }
                    if(map.containsKey(s.charAt(i))){
                        map.put(s.charAt(i), map.get(s.charAt(i)) + 1);
                        if(map.get(s.charAt(i))== 1){
                          count++;
                        }
                    }
                    i++;
                }
                j++;
            }
        }
        return s.substring(minimumI,minimumJ);
    }
}