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

The optimal solution for Check if Strings Can be Made Equal With Operations I runs in O(1) time and uses O(1) space. The complexity is O(1) because we are only performing a constant number of comparisons, regardless of string length.

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

The brute force approach for Check if Strings Can be Made Equal With Operations I is Brute Force, with O(n²) time complexity and O(1) space complexity. We can try all possible combinations of swaps to see if we can transform s1 into s2. Since the strings are small, this approach is feasible. The optimal Optimal Solution approach improves this to O(1).

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

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

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

Clarify the constraints of the problem before jumping into coding. Think about edge cases, such as identical strings or strings with all same characters. Practice explaining your thought process clearly while coding.

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

Not considering the specific indices that can be swapped. Assuming that any permutation of characters is valid.

#2839

Check if Strings Can be Made Equal With Operations I

Easy

StringHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(1)
Space	O(1)	O(1)

💡

Intuition

Time O(1)Space O(1)

Instead of generating permutations, we can analyze the positions of characters. The only characters that can swap are at indices 0 and 2, and 1 and 3. We can check if the characters at these positions can match.

⚙️

Algorithm

3 steps

1Step 1: Check if the characters at indices 0 and 2 of s1 match with those in s2.
2Step 2: Check if the characters at indices 1 and 3 of s1 match with those in s2.
3Step 3: If both conditions are satisfied, return true; otherwise, return false.

solution.py5 lines

1# Full working Python code
2
3def can_make_equal(s1, s2):
4    return (s1[0] == s2[0] and s1[2] == s2[2] and s1[1] == s2[1] and s1[3] == s2[3]) or 
5           (s1[0] == s2[2] and s1[2] == s2[0] and s1[1] == s2[3] and s1[3] == s2[1])

ℹ

Complexity note: The complexity is O(1) because we are only performing a constant number of comparisons, regardless of string length.

1The only possible swaps are between specific pairs of indices.
2Character frequency must match for the strings to be equal.

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