[LC] 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).

Example:

Input: S = "ADOBECODEBANC", T = "ABC"
Output: "BANC"

Note:

• If there is no such window in S that covers all characters in T, return the empty string "".
• If there is such window, you are guaranteed that there will always be only one unique minimum window in S.

Time: O(N)

class Solution:
    def minWindow(self, s: str, t: str) -> str:
        res = ''
        my_dict = {}
        for char in t:
            freq = my_dict.get(char, 0)
            my_dict[char] = freq + 1
        count, start, end, min_len = len(my_dict), 0, 0, sys.maxsize
        res_end, res_start = 0, 0
        while end != len(s):
            char = s[end]
            if char in my_dict:
                my_dict[char] -= 1
                if my_dict[char] == 0:
                    count -= 1
            end += 1

            while count == 0:
                start_char = s[start]
                if start_char in my_dict:
                    my_dict[start_char] += 1
                    if my_dict[start_char] > 0:
                        count += 1            

                if min_len > end - start:
                    min_len = end - start
                    res_end = end
                    res_start = start
                start += 1

        return s[res_start: res_end]
            

class Solution {
    public String minWindow(String s, String t) {
        Map<Character, Integer> map = new HashMap<>();
        int minLen = Integer.MAX_VALUE;
        char[] tCharArray = t.toCharArray();
        for (char c : tCharArray) {
            map.put(c, map.getOrDefault(c, 0) + 1);
        }

        int count = map.size(), start = 0, globStart = 0, i = 0;
        char[] sCharArray = s.toCharArray();
        while (i < sCharArray.length) {
            char cur = sCharArray[i];
            if (map.containsKey(cur)) {
                map.put(cur, map.get(cur) - 1);
                if (map.get(cur) == 0) {
                    count -= 1;
                }
            }
            i += 1;
            while (count == 0) {
                char sChar = sCharArray[start];
                if (map.containsKey(sChar)) {
                    map.put(sChar, map.get(sChar) + 1);
                    if (map.get(sChar) > 0) {
                        count += 1;
                    }
                }
                if (i - start < minLen) {
                    globStart = start;
                    minLen = i - start;
                }
                start += 1;
            }
        }
        return minLen == Integer.MAX_VALUE ? "" : s.substring(globStart, globStart + minLen);
    }
}