[LeetCode] 76. Minimum Window Substring

Given a string S and a string T, find the minimum window in S which will contain all the characters in T in complexity O(n).
For example,
S = "ADOBECODEBANC"
T = "ABC"
Minimum window is "BANC".
Note:
If there is no such window in S that covers all characters in T, return the empty string "".
If there are multiple such windows, you are guaranteed that there will always be only one unique minimum window in S.

Thought process:
Hash map + two pointers. Use a hash map to keep track of the occurrence of T's letters in S. The hash map maps a character in T to the remaining number of it to be matched. The value can also be negative, which means that there is one extra such character in the current window.
  1. Iterate through T. Populate map<letter, count>. This map represents the remaining characters in T to be matched.
  2. Iterate through S. Initialize variable count = t.length, window = empty string. Pointer i is the left end of the string. Pointer j is the right end of the string.
    1. If s[j] is in T, decrement map[s[j]]. Decrement count if map[s[j]] > 0.
    2. If count == 0, update minimum window.
    3. Increment i until map[s[i]] == 0. When I see an s[i] in T, increment map[s[i]].

Solution:
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class Solution {
    public String minWindow(String s, String t) {
        int count = t.length();
        int i = 0;
        Map<Character, Integer> map = countCharacters(t);
        String window = "";
        
        for (int j = 0; j < s.length(); j++) {
            char c1 = s.charAt(j);
            if (map.containsKey(c1)) {
                if (map.get(c1) > 0) {
                    count--;
                }
                map.put(c1, map.get(c1) - 1);
            }
            
            if (count == 0) {
                char c2 = s.charAt(i);
                while (!map.containsKey(c2) || map.get(c2) < 0) {
                    if (map.containsKey(c2)) {
                        map.put(c2, map.get(c2) + 1);
                    }
                    i++;
                    c2 = s.charAt(i);
                }
                
                if (window.length() == 0 || j - i + 1 < window.length()) {
                    window = s.substring(i, j + 1);
                }
            }
        }
        return window;
    }
    
    private Map<Character, Integer> countCharacters(String s) {
        Map<Character, Integer> map = new HashMap<>();
        for (char c : s.toCharArray()) {
            int count = map.getOrDefault(c, 0);
            map.put(c, count + 1);
        }
        return map;
    }
}

Time complexity: O(n).
Location: Philadelphia, PA, USA

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