A trick to solve most "substring" problem

I will first give the solution then show you the magic template.

The code of solving this problem is below. It might be the shortest among all solutions provided in Discuss.

string minWindow(string s, string t) {
        vector map(128,0);
        for(auto c: t) map[c]++;
        int counter=t.size(), begin=0, end=0, d=INT_MAX, head=0;
        while(end0) counter--; //in t
            while(counter==0){ //valid
                if(end-begin

Here comes the template.

For most substring problem, we are given a string and need to find a substring of it which satisfy some restrictions. A general way is to use a hashmap assisted with two pointers. The template is given below.

int findSubstring(string s){
        vector map(128,0);
        int counter; // check whether the substring is valid
        int begin=0, end=0; //two pointers, one point to tail and one  head
        int d; //the length of substring

        for() { /* initialize the hash map here */ }

        while(end

One thing needs to be mentioned is that when asked to find maximum substring, we should update maximum after the inner while loop to guarantee that the substring is valid. On the other hand, when asked to find minimum substring, we should update minimum inside the inner while loop.

The code of solving Longest Substring with At Most Two Distinct Characters is below:

int lengthOfLongestSubstringTwoDistinct(string s) {
        vector map(128, 0);
        int counter=0, begin=0, end=0, d=0; 
        while(end2) if(map[s[begin++]]--==1) counter--;
            d=max(d, end-begin);
        }
        return d;
    }

The code of solving Longest Substring Without Repeating Characters is below:

Update 01.04.2016, thanks @weiyi3 for advise.

int lengthOfLongestSubstring(string s) {
        vector map(128,0);
        int counter=0, begin=0, end=0, d=0; 
        while(end0) counter++; 
            while(counter>0) if(map[s[begin++]]-->1) counter--;
            d=max(d, end-begin); //while valid, update d
        }
        return d;
    }

I think this post deserves some upvotes! : )