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2499. Minimum Total Cost to Make Arrays Unequal

Description

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

Solutions

Solution 1

Thinking

Positions where the two arrays already match must be swapped away; their indices are paid first. If one value occupies more than half of those positions, they cannot pair among themselves and extra swaps with other indices are required.

Count equal positions. If the leading frequency \(v\) has \(2v>same\), \(2v-same\) extras must go outside. Then take \(a\ne b\) positions that avoid that leading value. Leftover extras mean impossible.

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class Solution:
    def minimumTotalCost(self, nums1: List[int], nums2: List[int]) -> int:
        ans = same = 0
        cnt = Counter()
        for i, (a, b) in enumerate(zip(nums1, nums2)):
            if a == b:
                same += 1
                ans += i
                cnt[a] += 1

        m = lead = 0
        for k, v in cnt.items():
            if v * 2 > same:
                m = v * 2 - same
                lead = k
                break
        for i, (a, b) in enumerate(zip(nums1, nums2)):
            if m and a != b and a != lead and b != lead:
                ans += i
                m -= 1
        return -1 if m else ans

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class Solution {
    public long minimumTotalCost(int[] nums1, int[] nums2) {
        long ans = 0;
        int same = 0;
        int n = nums1.length;
        int[] cnt = new int[n + 1];
        for (int i = 0; i < n; ++i) {
            if (nums1[i] == nums2[i]) {
                ans += i;
                ++same;
                ++cnt[nums1[i]];
            }
        }
        int m = 0, lead = 0;
        for (int i = 0; i < cnt.length; ++i) {
            int t = cnt[i] * 2 - same;
            if (t > 0) {
                m = t;
                lead = i;
                break;
            }
        }
        for (int i = 0; i < n; ++i) {
            if (m > 0 && nums1[i] != nums2[i] && nums1[i] != lead && nums2[i] != lead) {
                ans += i;
                --m;
            }
        }
        return m > 0 ? -1 : ans;
    }
}

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class Solution {
public:
    long long minimumTotalCost(vector<int>& nums1, vector<int>& nums2) {
        long long ans = 0;
        int same = 0;
        int n = nums1.size();
        int cnt[n + 1];
        memset(cnt, 0, sizeof cnt);
        for (int i = 0; i < n; ++i) {
            if (nums1[i] == nums2[i]) {
                ans += i;
                ++same;
                ++cnt[nums1[i]];
            }
        }
        int m = 0, lead = 0;
        for (int i = 0; i < n + 1; ++i) {
            int t = cnt[i] * 2 - same;
            if (t > 0) {
                m = t;
                lead = i;
                break;
            }
        }
        for (int i = 0; i < n; ++i) {
            if (m > 0 && nums1[i] != nums2[i] && nums1[i] != lead && nums2[i] != lead) {
                ans += i;
                --m;
            }
        }
        return m > 0 ? -1 : ans;
    }
};

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func minimumTotalCost(nums1 []int, nums2 []int) (ans int64) {
    same, n := 0, len(nums1)
    cnt := make([]int, n+1)
    for i, a := range nums1 {
        b := nums2[i]
        if a == b {
            same++
            ans += int64(i)
            cnt[a]++
        }
    }
    var m, lead int
    for i, v := range cnt {
        if t := v*2 - same; t > 0 {
            m = t
            lead = i
            break
        }
    }
    for i, a := range nums1 {
        b := nums2[i]
        if m > 0 && a != b && a != lead && b != lead {
            ans += int64(i)
            m--
        }
    }
    if m > 0 {
        return -1
    }
    return ans
}

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