2014. Longest Subsequence Repeated k Times

Description

You are given a string s of length n, and an integer k. You are tasked to find the longest subsequence repeated k times in string s.

A subsequence is a string that can be derived from another string by deleting some or no characters without changing the order of the remaining characters.

A subsequence seq is repeated k times in the string s if seq * k is a subsequence of s, where seq * k represents a string constructed by concatenating seq k times.

• For example, "bba" is repeated 2 times in the string "bababcba", because the string "bbabba", constructed by concatenating "bba" 2 times, is a subsequence of the string "bababcba".

Return the longest subsequence repeated k times in string s. If multiple such subsequences are found, return the lexicographically largest one. If there is no such subsequence, return an empty string.

 

Example 1:

Input: s = "letsleetcode", k = 2
Output: "let"
Explanation: There are two longest subsequences repeated 2 times: "let" and "ete".
"let" is the lexicographically largest one.

Example 2:

Input: s = "bb", k = 2
Output: "b"
Explanation: The longest subsequence repeated 2 times is "b".

Example 3:

Input: s = "ab", k = 2
Output: ""
Explanation: There is no subsequence repeated 2 times. Empty string is returned.

 

Constraints:

• n == s.length
• 2 <= n, k <= 2000
• 2 <= n < k * 8
• s consists of lowercase English letters.

Solution

longest-subsequence-repeated-k-times.py

class Solution:
    def longestSubsequenceRepeatedK(self, s: str, k: int) -> str:

        def isSubsequence(s, t):
            t = iter(t)

            return all(c in t for c in s)

        hot = "".join(key * (value // k) for key, value in Counter(s).items())
        combs = set()

        for i in range(len(hot) + 1):
            for comb in combinations(hot, i):
                for perm in permutations(comb):
                    combs.add("".join(perm))

        combs = sorted(combs, key = lambda x: (len(x), x), reverse = 1)

        for comb in combs:
            if isSubsequence(comb * k, s):
                return comb