2499. Minimum Total Cost to Make Arrays Unequal LeetCode Solution

Last updated: February 27, 2025

In this guide, you will get 2499. Minimum Total Cost to Make Arrays Unequal LeetCode Solution with the best time and space complexity. The solution to Minimum Total Cost to Make Arrays Unequal problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Table of Contents

  1. Problem Statement
  2. Complexity Analysis
  3. Minimum Total Cost to Make Arrays Unequal solution in C++
  4. Minimum Total Cost to Make Arrays Unequal solution in Java
  5. Minimum Total Cost to Make Arrays Unequal solution in Python
  6. Additional Resources
2499. Minimum Total Cost to Make Arrays Unequal LeetCode Solution image

Problem Statement of Minimum Total Cost to Make Arrays Unequal

You are given two 0-indexed integer arrays nums1 and nums2, of equal length n.
In one operation, you can swap the values of any two indices of nums1. The cost of this operation is the sum of the indices.
Find the minimum total cost of performing the given operation any number of times such that nums1[i] != nums2[i] for all 0 <= i <= n – 1 after performing all the operations.
Return the minimum total cost such that nums1 and nums2 satisfy the above condition. In case it is not possible, return -1.

Example 1:

Input: nums1 = [1,2,3,4,5], nums2 = [1,2,3,4,5]
Output: 10
Explanation:
One of the ways we can perform the operations is:
– Swap values at indices 0 and 3, incurring cost = 0 + 3 = 3. Now, nums1 = [4,2,3,1,5]
– Swap values at indices 1 and 2, incurring cost = 1 + 2 = 3. Now, nums1 = [4,3,2,1,5].
– Swap values at indices 0 and 4, incurring cost = 0 + 4 = 4. Now, nums1 =[5,3,2,1,4].
We can see that for each index i, nums1[i] != nums2[i]. The cost required here is 10.
Note that there are other ways to swap values, but it can be proven that it is not possible to obtain a cost less than 10.

Example 2:

Input: nums1 = [2,2,2,1,3], nums2 = [1,2,2,3,3]
Output: 10
Explanation:
One of the ways we can perform the operations is:
– Swap values at indices 2 and 3, incurring cost = 2 + 3 = 5. Now, nums1 = [2,2,1,2,3].
– Swap values at indices 1 and 4, incurring cost = 1 + 4 = 5. Now, nums1 = [2,3,1,2,2].
The total cost needed here is 10, which is the minimum possible.

Example 3:

Input: nums1 = [1,2,2], nums2 = [1,2,2]
Output: -1
Explanation:
It can be shown that it is not possible to satisfy the given conditions irrespective of the number of operations we perform.
Hence, we return -1.

Constraints:

n == nums1.length == nums2.length
1 <= n <= 105
1 <= nums1[i], nums2[i] <= n

Complexity Analysis

  • Time Complexity: O(n)
  • Space Complexity: O(n)

2499. Minimum Total Cost to Make Arrays Unequal LeetCode Solution in C++

class Solution {
 public:
  long long minimumTotalCost(vector<int>& nums1, vector<int>& nums2) {
    const int n = nums1.size();
    long ans = 0;
    int maxFreq = 0;
    int maxFreqNum = 0;
    int shouldBeSwapped = 0;
    vector<int> conflictedNumCount(n + 1);

    // Collect the indices i s.t. nums1[i] == nums2[i] and record their
    // `maxFreq` and `maxFreqNum`.
    for (int i = 0; i < n; ++i)
      if (nums1[i] == nums2[i]) {
        const int conflictedNum = nums1[i];
        if (++conflictedNumCount[conflictedNum] > maxFreq) {
          maxFreq = conflictedNumCount[conflictedNum];
          maxFreqNum = conflictedNum;
        }
        ++shouldBeSwapped;
        ans += i;
      }

    // Collect the indices with nums1[i] != nums2[i] that contribute less cost.
    // This can be greedily achieved by iterating from 0 to n - 1.
    for (int i = 0; i < n; ++i) {
      // Since we have over `maxFreq * 2` spaces, `maxFreqNum` can be
      // successfully distributed, so no need to collectextra spaces.
      if (maxFreq * 2 <= shouldBeSwapped)
        break;
      if (nums1[i] == nums2[i])
        continue;
      // The numbers == `maxFreqNum` worsen the result since they increase the
      // `maxFreq`.
      if (nums1[i] == maxFreqNum || nums2[i] == maxFreqNum)
        continue;
      ++shouldBeSwapped;
      ans += i;
    }

    return maxFreq * 2 > shouldBeSwapped ? -1 : ans;
  }
};
/* code provided by PROGIEZ */

2499. Minimum Total Cost to Make Arrays Unequal LeetCode Solution in Java

class Solution {
  public long minimumTotalCost(int[] nums1, int[] nums2) {
    final int n = nums1.length;
    long ans = 0;
    int maxFreq = 0;
    int maxFreqNum = 0;
    int shouldBeSwapped = 0;
    int[] conflictedNumCount = new int[n + 1];

    // Collect the indices i s.t. nums1[i] == nums2[i] and record their `maxFreq`
    // and `maxFreqNum`.
    for (int i = 0; i < n; ++i)
      if (nums1[i] == nums2[i]) {
        final int conflictedNum = nums1[i];
        if (++conflictedNumCount[conflictedNum] > maxFreq) {
          maxFreq = conflictedNumCount[conflictedNum];
          maxFreqNum = conflictedNum;
        }
        ++shouldBeSwapped;
        ans += i;
      }

    // Collect the indices with nums1[i] != nums2[i] that contribute less cost.
    // This can be greedily achieved by iterating from 0 to n - 1.
    for (int i = 0; i < n; ++i) {
      // Since we have over `maxFreq * 2` spaces, `maxFreqNum` can be
      // successfully distributed, so no need to collectextra spaces.
      if (maxFreq * 2 <= shouldBeSwapped)
        break;
      if (nums1[i] == nums2[i])
        continue;
      // The numbers == `maxFreqNum` worsen the result since they increase the
      // `maxFreq`.
      if (nums1[i] == maxFreqNum || nums2[i] == maxFreqNum)
        continue;
      ++shouldBeSwapped;
      ans += i;
    }

    return maxFreq * 2 > shouldBeSwapped ? -1 : ans;
  }
}
// code provided by PROGIEZ

2499. Minimum Total Cost to Make Arrays Unequal LeetCode Solution in Python

class Solution:
  def minimumTotalCost(self, nums1: list[int], nums2: list[int]) -> int:
    n = len(nums1)
    ans = 0
    maxFreq = 0
    maxFreqNum = 0
    shouldBeSwapped = 0
    conflictedNumCount = [0] * (n + 1)

    # Collect the indices i s.t. num1 == num2 and record their `maxFreq`
    # and `maxFreqNum`.
    for i, (num1, num2) in enumerate(zip(nums1, nums2)):
      if num1 == num2:
        conflictedNum = num1
        conflictedNumCount[conflictedNum] += 1
        if conflictedNumCount[conflictedNum] > maxFreq:
          maxFreq = conflictedNumCount[conflictedNum]
          maxFreqNum = conflictedNum
        shouldBeSwapped += 1
        ans += i

    # Collect the indices with num1 != num2 that contribute less cost.
    # This can be greedily achieved by iterating from 0 to n - 1.
    for i, (num1, num2) in enumerate(zip(nums1, nums2)):
      # Since we have over `maxFreq * 2` spaces, `maxFreqNum` can be
      # successfully distributed, so no need to collectextra spaces.
      if maxFreq * 2 <= shouldBeSwapped:
        break
      if num1 == num2:
        continue
      # The numbers == `maxFreqNum` worsen the result since they increase the
      # `maxFreq`.
      if num1 == maxFreqNum or num2 == maxFreqNum:
        continue
      shouldBeSwapped += 1
      ans += i

    return -1 if maxFreq * 2 > shouldBeSwapped else ans
 # code by PROGIEZ

Additional Resources

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