LeetCode 2014: Longest Subsequence Repeated k Times Solution

Master LeetCode problem 2014 (Longest Subsequence Repeated k Times), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2014. Longest Subsequence Repeated k Times

Hard

String Backtracking Greedy Counting Enumeration

Problem Explanation

Explanation:

To solve this problem, we can iterate through all possible subsequences of the given string s and check if each subsequence can be repeated k times in the original string. We can achieve this by using a recursive function to generate all subsequences and a helper function to check if a subsequence can be repeated k times.

Here are the detailed steps:

Create a recursive function to generate all subsequences of the given string s.
For each subsequence generated, check if it can be repeated k times in the original string using a helper function.
Keep track of the longest subsequence that can be repeated k times and return the lexicographically largest one.

Time Complexity:

The time complexity of this approach is O(2^n * n) where n is the length of the input string s.

Space Complexity:

The space complexity of this approach is O(n) for the recursive stack.

Solution Code

class Solution {
    String longestSubsequenceRepeatedK(String s, int k) {
        String res = "";
        generateSubsequences(s, k, "", 0, res);
        return res;
    }
    
    void generateSubsequences(String s, int k, String current, int index, String res) {
        if (current.length() > res.length() && isRepeatedKTimes(s, current, k)) {
            res = current;
        }
        for (int i = index; i < s.length(); i++) {
            generateSubsequences(s, k, current + s.charAt(i), i + 1, res);
        }
    }
    
    boolean isRepeatedKTimes(String s, String subseq, int k) {
        StringBuilder sb = new StringBuilder();
        for (int i = 0; i < k; i++) {
            sb.append(subseq);
        }
        int sIndex = 0;
        int subIndex = 0;
        while (sIndex < s.length() && subIndex < sb.length()) {
            if (s.charAt(sIndex) == sb.charAt(subIndex)) {
                subIndex++;
            }
            sIndex++;
        }
        return subIndex == sb.length();
    }
}

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