How do you solve Minimum Cost to Equalize Arrays Using Swaps on LeetCode?

Minimum Cost to Equalize Arrays Using Swaps (LeetCode #3868) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Frequencies matter more than positions.

What is the time complexity of Minimum Cost to Equalize Arrays Using Swaps?

The optimal solution for Minimum Cost to Equalize Arrays Using Swaps runs in O(n) time and uses O(n) space. The complexity is linear due to counting frequencies and checking conditions.

What is the brute force solution for Minimum Cost to Equalize Arrays Using Swaps?

The brute force approach for Minimum Cost to Equalize Arrays Using Swaps is Brute Force, with O(n²) time complexity and O(1) space complexity. We can try all possible swaps between the two arrays to see if they can be made identical. This approach is inefficient as it checks every combination. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Cost to Equalize Arrays Using Swaps?

Minimum Cost to Equalize Arrays Using Swaps uses the following concepts: Array, Hash Table, Greedy, Counting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Minimum Cost to Equalize Arrays Using Swaps?

Explain your thought process clearly. Consider edge cases early. Practice similar problems for familiarity.

What are common mistakes when solving Minimum Cost to Equalize Arrays Using Swaps?

Forgetting to check odd total frequencies. Not using a hash map for counting.

#3868

Minimum Cost to Equalize Arrays Using Swaps

Medium

ArrayHash TableGreedyCountingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

Count the frequency of each number in both arrays. If any number has an odd total frequency, return -1. Calculate the minimum swaps needed based on the frequency differences.

⚙️

Algorithm

3 steps

1Step 1: Count the frequency of each number in both arrays using a hash map.
2Step 2: Check if the total frequency of any number is odd; if so, return -1.
3Step 3: Calculate the total number of swaps needed based on the frequency differences.

solution.py9 lines

1from collections import Counter
2
3def minCost(nums1, nums2):
4    count1, count2 = Counter(nums1), Counter(nums2)
5    for num in set(count1) | set(count2):
6        if (count1[num] + count2[num]) % 2 != 0:
7            return -1
8    swaps = sum(abs(count1[num] - count2[num]) // 2 for num in count1)
9    return swaps // 2

ℹ

Complexity note: The complexity is linear due to counting frequencies and checking conditions.

1Frequencies matter more than positions.
2Odd total frequencies indicate impossibility.

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