How do you solve Minimum Total Cost to Make Arrays Unequal on LeetCode?

Minimum Total Cost to Make Arrays Unequal (LeetCode #2499) 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: Greedily swap elements where nums1[i] == nums2[i] to minimize costs.

What is the brute force solution for Minimum Total Cost to Make Arrays Unequal?

The brute force approach for Minimum Total Cost to Make Arrays Unequal is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach tries all possible swaps in the array to ensure that nums1[i] is not equal to nums2[i]. This is straightforward but inefficient, as it checks every combination of swaps. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Total Cost to Make Arrays Unequal?

Minimum Total Cost to Make Arrays Unequal 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 Total Cost to Make Arrays Unequal?

Clarify the problem requirements and constraints before diving into coding. Think about edge cases, such as arrays with all equal elements. Discuss your thought process and approach with the interviewer.

What are common mistakes when solving Minimum Total Cost to Make Arrays Unequal?

Not checking if nums1[i] == nums2[i] before attempting a swap. Failing to account for cases where no swaps are possible.

#2499

Minimum Total Cost to Make Arrays Unequal

Hard

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)

The optimal solution uses a greedy approach to minimize the total cost by focusing on the indices where nums1 matches nums2. It ensures that we only swap when necessary and selects the least costly swaps.

⚙️

Algorithm

4 steps

1Step 1: Create a list of indices where nums1[i] == nums2[i].
2Step 2: For each index in this list, find the minimum cost swap with an index that has a different value in nums1.
3Step 3: If no valid swaps can be made and there are still indices where nums1[i] == nums2[i], return -1.
4Step 4: Sum the costs of all swaps made.

solution.py17 lines

1def minCost(nums1, nums2):
2    n = len(nums1)
3    indices = [i for i in range(n) if nums1[i] == nums2[i]]
4    total_cost = 0
5    if len(indices) == 0:
6        return 0
7    for i in indices:
8        swap_found = False
9        for j in range(n):
10            if nums1[j] != nums2[i]:
11                nums1[i], nums1[j] = nums1[j], nums1[i]
12                total_cost += i + j
13                swap_found = True
14                break
15        if not swap_found:
16            return -1
17    return total_cost

ℹ

Complexity note: The time complexity is O(n) because we only traverse the arrays a limited number of times, and the space complexity is O(n) due to storing indices of swaps.

1Greedily swap elements where nums1[i] == nums2[i] to minimize costs.
2If no valid swaps can be made, return -1.

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