76. Minimum Window Substring My Submissions Question

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.

在实现中就是维护一个HashMap，一开始key包含字典中所有字符，value就是该字符的数量，然后遇到字典中字符时就将对应字符的数量减一。算法的时间复杂度是O(n),其中n是字符串的长度，因为每个字符再维护窗口的过程中不会被访问多于两次。空间复杂度则是O(字典的大小)，也就是代码中T的长度。

import java.util.HashMap;

/**
 * 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".
 * Tags: Hash Table Two Pointers String
 * Similar Problems:  (H) Substring with Concatenation of All Words (M) Minimum Size Subarray Sum (H) Sliding Window Maximum
 * 
 * Analysis:
 * 1.Store the char in string t as a key and times appearing as value into a hashmap;
 * 2.If char at s is contained in hashmap, value(times) in map - 1 and count ++;
 * 3.When count = t.length(), minLenght and minStart records the min number
 * 4.Slide window and let the left point to the first char in s ,which is in map  
 *        s                        t        
 * A D O B E C O D E B A N C      A B C
 * A D O B E C 
 * B E C O D E B A
 * C O D E B A
 * B A N C*/

public class MinimumWindowSubstring {
    public static String minWindow(String s, String t) {
        if(s.length() == 0 || t.length() ==0 || s.length() < t.length()) {
            return "";
        }
        HashMap<Character, Integer> map = new HashMap<Character, Integer>();
        for(char c : t.toCharArray()) {
            if(map.containsKey(c)) {
                map.put(c, map.get(c) + 1);
            }
            else {
                map.put(c, 1);
            }
        }
        int left = 0;
        int count = 0;
        int minStart = 0;
        int minLength = s.length() + 1;
        for(int right = 0; right < s.length(); right++) {
            if(map.containsKey(s.charAt(right))) {
                map.put(s.charAt(right), map.get(s.charAt(right)) - 1);
                if(map.get(s.charAt(right)) >= 0) { 
                    count++;
                }
                while(count == t.length()) {
                    if((right - left + 1) < minLength) {
                        minStart = left;
                        minLength = right - left + 1;
                    }
                    if(map.containsKey(s.charAt(left))) {
                        map.put(s.charAt(left), map.get(s.charAt(left)) + 1 );
                        if(map.get(s.charAt(left)) > 0) {
                            count--;
                        }
                    }
                    left ++;
                }
            }
        }
        if(minLength == s.length()+1) {
            return "";
        }
        return s.substring(minStart, minStart + minLength);
    }

    public static void main(String[] args) {
        System.out.print(minWindow("ADOBECODEBANC", "ABC"));
    }
}