Leetcode: Minimum Window Substring (hash table)(two pointers)

PROBLEM:
https://leetcode.com/problems/minimum-window-substring/



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The general idea is to use map and two pointers. Use two pointer to form a sliding window while the value between is processed through a map, or hash table.

1. we build a map, tMap, for t;
2. using two pointers: left and right to form a windon:
3.  when moving right we do
       tMap[s[r]]--; 
    for each step, until all value in tMap is <=0;

4.  then moving left until any value in tMap reaches 1, which means the current minimal window is found.

5. repeat until the end.


  ////////////////////////////////////////////////////////////////////
// codes

class Solution {

public:

       string minWindow(string s, string t) {

              unordered_map<char, int> tMap;

              int ss = s.size(), st = t.size();

              //check input

              if (ss<st)return "";

              for (int i = 0; i<st; i++){

                     tMap[t[i]]++;

              }

              //search

              int l = 0, r = 0, lflag = 0, rflag = 1, count = 0, sm = tMap.size();

              int a[3] = { ss + 1, 0, ss + 1 };

              while (r<ss + 1 && count <= st){

                     //move right

                     if (rflag == 1){

                           tMap[s[r]]--;

                          

                           if (tMap[s[r]] == 0)count++;//counted when an item is removed

                           if (count == sm){//check whether all items in t have been found

                                  lflag = 1;

                                  rflag = 0;

                           }

            r++;

                     }

                     //move left

                     if (lflag == 1){

                           tMap[s[l]]++;

                     //when any value reaches 1, that means current minimal window is reached

                           if (tMap[s[l]] == 1){

                                  if (r - 1 - l + 1<a[2]){

                                         a[0] = l; a[1] = r - 1; a[2] = r - 1 - l + 1;

                                  }

                                  lflag = 0;

                                  rflag = 1;

                                  count--;

                           }

                           l++;

                     }

                     //when to jump out

                     if (r == ss && count<sm)break;

              }

              if (a[0] == ss + 1)return "";

              else{

                     string res(s, a[0], a[2]);

                     return res;

              }

       }

};