76.minimum-window-substring.java

/*
 * @lc app=leetcode id=76 lang=java
 *
 * [76] Minimum Window Substring
 *
 * https://leetcode.com/problems/minimum-window-substring/description/
 *
 * algorithms
 * Hard (34.75%)
 * Likes:    5026
 * Dislikes: 342
 * Total Accepted:    428.9K
 * Total Submissions: 1.2M
 * Testcase Example:  '"ADOBECODEBANC"\n"ABC"'
 *
 * 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.
 * 
 * 
 */

// @lc code=start
class Solution {
    public String minWindow(String s, String t) {
        if(t.length()> s.length()) return "";
        Map<Character, Integer> map = new HashMap<>();
        for(char c : t.toCharArray()){
            map.put(c, map.getOrDefault(c,0) + 1);
        }
        int counter = map.size();
        
        int begin = 0, end = 0;
        int head = 0;
        int len = Integer.MAX_VALUE;
        
        while(end < s.length()){
            char c = s.charAt(end);
            if( map.containsKey(c) ){
                map.put(c, map.get(c)-1);
                if(map.get(c) == 0) counter--;
            }
            end++;
            
            while(counter == 0){
                char tempc = s.charAt(begin);
                if(map.containsKey(tempc)){
                    map.put(tempc, map.get(tempc) + 1);
                    if(map.get(tempc) > 0){
                        counter++;
                    }
                }
                if(end-begin < len){
                    len = end - begin;
                    head = begin;
                }
                begin++;
            }
            
        }
        if(len == Integer.MAX_VALUE) return "";
        return s.substring(head, head+len);
    
    }
}
// @lc code=end