76. Minimum Window Substring

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.

Solution

public class Solution {
    public String minWindow(String s, String t) {
       Map<Character, Integer> patternMap = new HashMap<>();
        for (Character c : t.toCharArray()) {
            patternMap.put(c, patternMap.containsKey(c) ? patternMap.get(c) + 1 : 1);
        }

        Map<Character, Integer> runnerMap = new HashMap<>();
        int lengthRunner = 0;
        int slow = 0;
        int result = Integer.MAX_VALUE;
        String finalResult = "";
        for (int i = 0; i < s.length(); i++) {
            Character currentChar = s.charAt(i);
            if (patternMap.containsKey(currentChar)) {
                runnerMap.put(currentChar, runnerMap.containsKey(currentChar) ? runnerMap.get(currentChar) + 1: 1);
                if (runnerMap.get(currentChar) <= patternMap.get(currentChar)) lengthRunner++;

                if (lengthRunner >= t.length()) {
                    char slowChar = s.charAt(slow);
                    while (!patternMap.containsKey(slowChar) || runnerMap.get(slowChar) > patternMap.get(slowChar)) {
                        if (runnerMap.containsKey(slowChar)) runnerMap.put(slowChar, runnerMap.get(slowChar) - 1);
                        slow++;
                                                slowChar = s.charAt(slow);

                    }

                    if (i - slow + 1 < result) {
                        finalResult = s.substring(slow, i + 1);
                        result = i - slow + 1;
                    }


                }
            }
        }

        return finalResult;
    }
}