How do you solve Longest Subsequence Repeated k Times on LeetCode?

Longest Subsequence Repeated k Times (LeetCode #2014) can be solved using Brute Force. The optimal approach is Brute Force with undefined time complexity and undefined space complexity. Key insight: The brute-force approach involves generating all possible subsequences of the string and checking each one to see if it can be repeated k times. This is simple but inefficient due to the exponential number of subsequences.

What is the time complexity of Longest Subsequence Repeated k Times?

The optimal solution for Longest Subsequence Repeated k Times runs in undefined time and uses undefined space. undefined

What algorithm or data structure is used to solve Longest Subsequence Repeated k Times?

Longest Subsequence Repeated k Times uses the following concepts: Hash Table, Two Pointers, String, Backtracking, Counting, Enumeration. The recommended patterns to study are: .

#2014

Longest Subsequence Repeated k Times

Hard

Hash TableTwo PointersStringBacktrackingCountingEnumeration

LeetCode ↗

Approaches

💡

Intuition

Time Space

The brute-force approach involves generating all possible subsequences of the string and checking each one to see if it can be repeated k times. This is simple but inefficient due to the exponential number of subsequences.

⚙️

Algorithm

3 steps

1Step 1: Generate all possible subsequences of the string s.
2Step 2: For each subsequence, check if it can be repeated k times in s.
3Step 3: Keep track of the longest valid subsequence found, and if there are ties, choose the lexicographically largest one.

solution.py25 lines

1from itertools import combinations
2
3def longest_repeated_subsequence(s, k):
4    n = len(s)
5    longest = ""
6    for length in range(1, n // k + 1):
7        for subseq in combinations(s, length):
8            candidate = ''.join(subseq)
9            if can_form_k_times(candidate, s, k):
10                if (len(candidate) > len(longest)) or (len(candidate) == len(longest) and candidate > longest):
11                    longest = candidate
12    return longest
13
14def can_form_k_times(seq, s, k):
15    count = 0
16    j = 0
17    for char in seq:
18        for i in range(j, len(s)):
19            if s[i] == char:
20                count += 1
21                j = i + 1
22                break
23        if count == k * len(seq):
24            return True
25    return False

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