Description

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.

Solution

Divide and conquer, time O(n ^ 2), space O(n)

乍一看这道题完全没思路，只能想到O(n ^ 3)的暴力解法。用Two-pointer呢，也不知道该怎么移动。后来发现，用普通的分治思想就可以解决。

首先遍历arr，找到第一个illegal char的位置 x，然后将arr在x处分成两部分，然后递归解决即可。也算是DFS解法了。

For Example: bbcddefegaghfh and 2, so we shall dfs on "bb", "ddefeg", "ghfh", since a , c only appears1 for once.

class Solution {
    public int longestSubstring(String s, int k) {
        if (s == null || k < 1) {
            return 0;
        }
        
        return dfs(s.toCharArray(), 0, s.length(), k);
    }
    
    public int dfs(char[] arr, int start, int end, int k) {
        if (end - start < k) {
            return 0;
        }
        
        int[] count = new int[26];
        for (int i = start; i < end; ++i) {
            ++count[arr[i] - 'a'];
        }
        
        for (int i = 0; i < count.length; ++i) {
            if (count[i] == 0 || count[i] >= k) {
                continue;
            }
            // find the first illegal char between [start, end)
            for (int j = start; j < end; ++j) {
                if (arr[j] - 'a' != i) {
                    continue;
                }
                
                // split arr by the first illegal char
                int left = dfs(arr, start, j, k);
                int right = dfs(arr, j + 1, end, k);
                return Math.max(left, right);
            }
        }
        
        return end - start;
    }
}