Find shortest substring containing pattern string distinct characters - mock interview - August 23, 2020, 2:00 PM - as an interviewer. The code was reviewed, and I removed line from 51 to 53

FindShortestSubStringContainingPatternDistinctChars.cpp

/* 
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"

string minWindow(string s, string t) {        
    }             
 */

//
// test your own code
// 
string minWindow(string s, string t) {
  
  int lenT = t.length();
  int lenS = s.length();
  
  if(lenT == 0 || lenS == 0) return "";
  
  if(lenS < lenT) return "";
  
  string res;
  unordered_map<char,int> tMap;   // assuming that all are distinct
  for(auto ch:t){
    
    tMap[ch] = 0;  // 1 ++; tMap[ch]++; 
  }  // keep count - sliding window - 
  
  
  int cnt =0;
  int j = 0;  // left variable 
  int unq_cnt =0;
  
  // pretend to be an itnerviewee from Facebook, google
  // 
  for(int i=0; i < lenS; ++i){    
    
    if(tMap.find(s[i]) == tMap.end()) continue;  // skip 
    
    if(tMap[s[i]] == 0) unq_cnt++; // add count 
    
    tMap[s[i]]++; // how 
    
    while(unq_cnt == lenT){ // find all chars - not enough
      int tempLen = i-j+1;
      
      /*
      if(tempLen == lenT){        
        return s.substr(j, tempLen);  // return?  -   ????? break for loop on line 101 - will not visit all chars in original string 
      }*/
      
      if(res == "" || tempLen < res.size()){ // res - final minimum one        
        res = s.substr(j, tempLen);               
      }
      
      if(tMap.find(s[j]) != tMap.end()){ // left char - in pattern string         
        if(tMap[s[j]]-1 == 0){ // only one - decrement           
          unq_cnt-=1;          
        }
        
        tMap[s[j]]--;  // decrement one 
      }
      
      j = j+1;  // left pointer - slide window left pointer - continiuoly
    }    
  }
  
  return res;
}