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. 

 

---

Sliding window [left, right]

技巧：

用hash table来存target string里每个字符出现的次数。

用count来数sliding window里包含的满足要求的字符的个数：

1. 当检查right pointer的时候，如果hash table里该字符的hash value >0，说明，这个字符满足要求（该字符出现在t里面，并且未达到t里该字符出现的次数）， count 加一;如果该字符的hash value<=0, 说明 1）该字符没有出现在t里面，或者2）出现在t里面，可是sliding window里该字符出现的次数已经满足，现在的是多余的。然后更新hash table里该字符的hash value： 减一。

2. 当检查left pointer的时候，首先更新hash table里该字符的hash value：加一。如果该字符更新后的hash value>0，说明，该字符是符合条件的字符，当移动left point（将该字符从sliding window里去除），count减一。

 

class Solution {
public:
    string minWindow(string s, string t) {
        string result;
        if(s.size()<t.size()){
            return result;
        }
        unordered_map<char,int> freq;
        //hash table to count frequency for each letter
        for(char c: t){
            freq[c]++;
        }
        
        int left=0;
        int count=0;
        int length = INT_MAX;
        
        //sliding window [left, i]
        for(int i=0;i<s.size();i++){ //move right pointer
            if(freq[s[i]]-->0){ //s[i], current letter is in T
                count++;
            }
            while(count==t.size()){ //sliding window meet the requirement (contains all charaters from T)
                if(i-left+1<length){
                    length = i -left+1;
                    result=s.substr(left,length);
                }
                
                if(++freq[s[left]]>0){ //s[left], current letter is in T
                    count--;
                }
                
                left++; //move left pointer
            }
        }
        
        return result;
    }
};
//The question is how to easily check whether the slicing window contains all characters in T?