Minimum substring window

1. /********************************************** 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. ***********************************/ class Solution { public: string minWindow(string s, string t) { int n = s.size(); int m = t.size(); //two pointers, and two hashtable unordered_map<char,int> mytable; unordered_map<char,int> currtable; int count = 0; string result = ""; int minLen = INT_MAX; for(int i = 0;i mytable[t[i]]++; int left = 0; int right = 0; for(int right = 0;right

{ char c = s[right]; //move the right pointer if(mytable.find(c) != mytable.end())//found { if(currtable[c]// make all number from t being counted count++; currtable[c]++; } if(count == m) //found { char b = s[left]; // pay more attention to the following logic, how to move left pointer. while (mytable.find(b)==mytable.end() || currtable[b]> mytable[b]) { //found the character currtable[b]--; //go to next character left++; b = s[left]; } if((right-left+1) { minLen = right-left+1; result = s.substr(left,right-left+1); } } } return result; } };