How do you solve Frequencies of Shortest Supersequences on LeetCode?

Frequencies of Shortest Supersequences (LeetCode #3435) 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 generates all possible combinations of characters from the input strings and checks if they can form a valid supersequence. This method is straightforward but inefficient due to the exponential growth of combinations.

What is the time complexity of Frequencies of Shortest Supersequences?

The optimal solution for Frequencies of Shortest Supersequences runs in undefined time and uses undefined space. undefined

What algorithm or data structure is used to solve Frequencies of Shortest Supersequences?

Frequencies of Shortest Supersequences uses the following concepts: Array, String, Bit Manipulation, Graph Theory, Topological Sort, Enumeration. The recommended patterns to study are: .

#3435

Frequencies of Shortest Supersequences

Hard

ArrayStringBit ManipulationGraph TheoryTopological SortEnumeration

LeetCode ↗

Approaches

💡

Intuition

Time Space

The brute-force approach generates all possible combinations of characters from the input strings and checks if they can form a valid supersequence. This method is straightforward but inefficient due to the exponential growth of combinations.

⚙️

Algorithm

3 steps

1Step 1: Generate all possible combinations of characters from the input strings, allowing each character to appear at most twice.
2Step 2: For each combination, check if it contains all input strings as subsequences.
3Step 3: If valid, calculate the frequency of each character and store it in the result.

solution.py13 lines

1from itertools import combinations
2
3def get_frequencies(words):
4    freq_set = set()
5    for chars in combinations(set(''.join(words)), 2):
6        for count in range(1, 3):
7            candidate = ''.join(c * count for c in chars)
8            if all(word in candidate for word in words):
9                freq = [0] * 26
10                for c in candidate:
11                    freq[ord(c) - ord('a')] += 1
12                freq_set.add(tuple(freq))
13    return [list(freq) for freq in freq_set]

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