How do you solve Minimum Swaps to Make Strings Equal on LeetCode?

Minimum Swaps to Make Strings Equal (LeetCode #1247) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Both strings must have the same number of 'x's and 'y’s for it to be possible to make them equal.

What is the time complexity of Minimum Swaps to Make Strings Equal?

The optimal solution for Minimum Swaps to Make Strings Equal runs in O(n) time and uses O(1) space. This complexity is linear because we only traverse each string once to count the characters.

What is the brute force solution for Minimum Swaps to Make Strings Equal?

The brute force approach for Minimum Swaps to Make Strings Equal is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves checking all possible pairs of swaps between the two strings to see if they can be made equal. This method is straightforward but inefficient, as it may require checking many combinations. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Swaps to Make Strings Equal?

Minimum Swaps to Make Strings Equal uses the following concepts: Math, String, Greedy. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Minimum Swaps to Make Strings Equal?

Always validate the input conditions before diving into the logic. Think about edge cases, such as strings that are already equal or impossible to equalize. Explain your thought process clearly while coding, as it helps the interviewer understand your approach.

What are common mistakes when solving Minimum Swaps to Make Strings Equal?

Failing to check if the total counts of 'x' and 'y' are even before proceeding. Not considering that swaps can only occur between different strings.

#1247

Minimum Swaps to Make Strings Equal

Medium

MathStringGreedyHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(1)

The optimal approach focuses on counting the mismatched characters and calculating the minimum swaps needed based on the counts of 'x' and 'y'. This is more efficient because it avoids unnecessary checks and directly computes the result.

⚙️

Algorithm

3 steps

1Step 1: Count the number of 'x's and 'y’s in both strings.
2Step 2: Check if the total counts of 'x' and 'y' are even; if not, return -1 (impossible to balance).
3Step 3: Calculate the number of swaps needed using the formula: swaps = (count_x_in_s1 - count_x_in_s2) / 2.

solution.py12 lines

1# Full working Python code
2
3def min_swaps(s1, s2):
4    count_x1 = s1.count('x')
5    count_y1 = s1.count('y')
6    count_x2 = s2.count('x')
7    count_y2 = s2.count('y')
8    if (count_x1 + count_x2) % 2 != 0:
9        return -1
10    swaps = abs(count_x1 - count_x2) // 2
11    return swaps
12

ℹ

Complexity note: This complexity is linear because we only traverse each string once to count the characters.

1Both strings must have the same number of 'x's and 'y’s for it to be possible to make them equal.
2The number of swaps needed can be derived from the difference in counts of 'x's and 'y’s.

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