159. Longest Substring with At M

Given a string, find the length of the longest substring T that contains at most 2 distinct characters.
For example, Given s = “eceba”, T is "ece" which its length is 3.

Task is that we should find the longest substring which only has two characters but get the two appear as many times as possible.

First Thought

The most important idea in solving this kind of questions is "how to update the "start" pointer".

int lengthOfLongestSubstringTwoDistinct(string s) {
  if(s.length() <= 2) return s.length();
  vector<int> dict(256, -1);
  int count = 0, maxLength = 0, start = 0;
  char prev;
  for(int i = 0; i < s.length(); i++){
    if(prev != s[i]){ //chars not the same as the current one
      if(count < 2){ // found new chars in substr
        count++;
        dict[s[i]] = i;
        maxLength = max(maxLength, i-start+1);
      }else if(count == 2 && dict[s[i]] >= 0){ // char already exist in substr
        dict[s[i]] = i;
        maxLength = max(maxLength, i-start+1);
      }else if(dict[s[i]] < 0){ // third char
        maxLength = max(maxLength, i-start);
        start = dict[prev];
        dict[s[dict[prev]-1]] = -1;
        dict[s[i]] = i;
      }
      prev = s[i];
    }else{
      maxLength = max(maxLength, i-start+1);
    }
  }
  return maxLength;
}

Second Solution

Two Pointers. Still O(N), but hard to extend to k distinct characters substring.

int lengthOfLongestSubstringTwoDistinct(string s){
  int first = 0, second = -1;
  int maxLength = 0;
  for(int i = 1; i < s.length(); i++){
    if(s[i-1] == s[i]) continue;
    if(second > -1 && s[i] != s[second]){
      maxLength = max(maxLength, i-first);
      first = second+1;
    }
    second = i - 1;
  }
  return maxLength > (s.size() - first) ? maxLength : s.size() - first;
}