LeetCode-Longest Substring with At Least K Repeating Characters

Posted 2020-08-08 LiBlog

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