What is the time complexity of Check if Strings Can be Made Equal With Operations II?

The optimal solution for Check if Strings Can be Made Equal With Operations II runs in O(n) time and uses O(n) space. The time complexity is O(n) because we traverse the strings once to count characters. The space complexity is O(n) due to the storage of frequency counts.

What is the brute force solution for Check if Strings Can be Made Equal With Operations II?

The brute force approach for Check if Strings Can be Made Equal With Operations II is Brute Force, with O(n!) time complexity and O(n) space complexity. The brute force approach would involve generating all possible permutations of one string and checking if any of them match the other string. This is straightforward but inefficient due to the factorial growth of permutations. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Check if Strings Can be Made Equal With Operations II?

Check if Strings Can be Made Equal With Operations II uses the following concepts: Hash Table, String, Sorting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Check if Strings Can be Made Equal With Operations II?

Always clarify the constraints and operations allowed before jumping into coding. Think about the properties of the problem (like parity in this case) to simplify your approach. Practice explaining your thought process as you code, as this helps in interviews.

What are common mistakes when solving Check if Strings Can be Made Equal With Operations II?

Overlooking the parity condition when swapping characters. Not checking the lengths of the strings before proceeding with character comparisons.

#2840

Check if Strings Can be Made Equal With Operations II

Medium

Hash TableStringSortingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n!)	O(n)
Space	O(n)	O(n)

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Intuition

Time O(n)Space O(n)

Instead of generating permutations, we can leverage the fact that we can only swap characters at even indices with each other and odd indices with each other. Therefore, we need to check if characters at even indices in s1 can match those in s2, and the same for odd indices.

⚙️

Algorithm

4 steps

1Step 1: Create two frequency counters for characters at even indices and odd indices for both strings.
2Step 2: Compare the frequency counters for even indices of s1 and s2.
3Step 3: Compare the frequency counters for odd indices of s1 and s2.
4Step 4: If both frequency counters match, return true; otherwise, return false.

solution.py11 lines

1# Full working Python code
2from collections import Counter
3
4def canMakeEqual(s1, s2):
5    if len(s1) != len(s2):
6        return False
7    even_count_s1 = Counter(s1[i] for i in range(0, len(s1), 2))
8    odd_count_s1 = Counter(s1[i] for i in range(1, len(s1), 2))
9    even_count_s2 = Counter(s2[i] for i in range(0, len(s2), 2))
10    odd_count_s2 = Counter(s2[i] for i in range(1, len(s2), 2))
11    return even_count_s1 == even_count_s2 and odd_count_s1 == odd_count_s2

ℹ

Complexity note: The time complexity is O(n) because we traverse the strings once to count characters. The space complexity is O(n) due to the storage of frequency counts.

1Characters can only be swapped between even indices and between odd indices.
2To make the strings equal, the frequency of characters at even and odd indices must match.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.