LeetCode-Longest Substring with At Least K Repeating Characters

Find the length of the longest substring T of a given string (consists of lowercase letters only) such that every character in T appears no less than k times.

Example 1:

Input: s = "aaabb", k = 3 Output: 3 The longest substring is "aaa", as 'a' is repeated 3 times.

Example 2:

Input: s = "ababbc", k = 2 Output: 5 The longest substring is "ababb", as 'a' is repeated 2 times and 'b' is repeated 3 times.

Analysis:

Given a string s, find out all chars that are invalid (i.e., count < k). The longest substring must reside in one of the substrings divided by those invalid chars. We find out all those possible substrings and recursively address each of them.

NOTE: repeatedly using s.charAt() is actually very slow. If we use charAt() in the following code, the runtime becomes 6ms, doubled!

Solution:

public class Solution {
public int longestSubstring(String s, int k) {
return longestSubstringRecur(s,k);
}

public int longestSubstringRecur(String s, int k){
if (s.length()<k) return 0;

int[] charCounts = new int[26];
for (int i=0;i<s.length();i++){
char c = s.charAt(i);
charCounts[c-'a']++;
}

// Early termination:
//  1. invalid: no char in s appears >= k times.
//  Or 2. Good enough: every char appears >= k times.
boolean valid = false, goodEnough = true;
for (int count : charCounts){
if (count>=k){
valid = true;
}
if (count>0 && count <k){
goodEnough = false;
}
if (valid && !goodEnough) break;
}
if (!valid) return 0;
if (goodEnough) return s.length();

// Address every valid substring, i.e., substring between two invalid chars.
int p1=0, p2=-1, maxLen=0;
while (p1<s.length()){
p2++;
while (p2<s.length() && charCounts[s.charAt(p2)-'a']>=k) p2++;
int curMaxLen = longestSubstringRecur(s.substring(p1,p2),k);
maxLen = Math.max(maxLen,curMaxLen);
p1 = p2+1;
}
return maxLen;

}
}

bstring T of a given string (consists of lowercase letters only) such that every character in T appears