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1775. Equal Sum Arrays With Minimum Number of Operations

Description

You are given two arrays of integers nums1 and nums2, possibly of different lengths. The values in the arrays are between 1 and 6, inclusive.

In one operation, you can change any integer's value in any of the arrays to any value between 1 and 6, inclusive.

Return the minimum number of operations required to make the sum of values in nums1 equal to the sum of values in nums2. Return -1​​​​​ if it is not possible to make the sum of the two arrays equal.

 

Example 1:

Input: nums1 = [1,2,3,4,5,6], nums2 = [1,1,2,2,2,2]
Output: 3
Explanation: You can make the sums of nums1 and nums2 equal with 3 operations. All indices are 0-indexed.
- Change nums2[0] to 6. nums1 = [1,2,3,4,5,6], nums2 = [6,1,2,2,2,2].
- Change nums1[5] to 1. nums1 = [1,2,3,4,5,1], nums2 = [6,1,2,2,2,2].
- Change nums1[2] to 2. nums1 = [1,2,2,4,5,1], nums2 = [6,1,2,2,2,2].

Example 2:

Input: nums1 = [1,1,1,1,1,1,1], nums2 = [6]
Output: -1
Explanation: There is no way to decrease the sum of nums1 or to increase the sum of nums2 to make them equal.

Example 3:

Input: nums1 = [6,6], nums2 = [1]
Output: 3
Explanation: You can make the sums of nums1 and nums2 equal with 3 operations. All indices are 0-indexed. 
- Change nums1[0] to 2. nums1 = [2,6], nums2 = [1].
- Change nums1[1] to 2. nums1 = [2,2], nums2 = [1].
- Change nums2[0] to 4. nums1 = [2,2], nums2 = [4].

 

Constraints:

  • 1 <= nums1.length, nums2.length <= 105
  • 1 <= nums1[i], nums2[i] <= 6

Solutions

  • class Solution {
    public int minOperations(int[] nums1, int[] nums2) {
        int s1 = Arrays.stream(nums1).sum();
        int s2 = Arrays.stream(nums2).sum();
        if (s1 == s2) {
            return 0;
        }
        if (s1 > s2) {
            return minOperations(nums2, nums1);
        }
        int d = s2 - s1;
        int[] arr = new int[nums1.length + nums2.length];
        int k = 0;
        for (int v : nums1) {
            arr[k++] = 6 - v;
        }
        for (int v : nums2) {
            arr[k++] = v - 1;
        }
        Arrays.sort(arr);
        for (int i = 1, j = arr.length - 1; j >= 0; ++i, --j) {
            d -= arr[j];
            if (d <= 0) {
                return i;
            }
        }
        return -1;
    }
}

  • class Solution {
public:
    int minOperations(vector<int>& nums1, vector<int>& nums2) {
        int s1 = accumulate(nums1.begin(), nums1.end(), 0);
        int s2 = accumulate(nums2.begin(), nums2.end(), 0);
        if (s1 == s2) return 0;
        if (s1 > s2) return minOperations(nums2, nums1);
        int d = s2 - s1;
        int arr[nums1.size() + nums2.size()];
        int k = 0;
        for (int& v : nums1) arr[k++] = 6 - v;
        for (int& v : nums2) arr[k++] = v - 1;
        sort(arr, arr + k, greater<>());
        for (int i = 0; i < k; ++i) {
            d -= arr[i];
            if (d <= 0) return i + 1;
        }
        return -1;
    }
};

  • class Solution:
    def minOperations(self, nums1: List[int], nums2: List[int]) -> int:
        s1, s2 = sum(nums1), sum(nums2)
        if s1 == s2:
            return 0
        if s1 > s2:
            return self.minOperations(nums2, nums1)
        arr = [6 - v for v in nums1] + [v - 1 for v in nums2]
        d = s2 - s1
        for i, v in enumerate(sorted(arr, reverse=True), 1):
            d -= v
            if d <= 0:
                return i
        return -1


  • func minOperations(nums1 []int, nums2 []int) int {
	s1, s2 := sum(nums1), sum(nums2)
	if s1 == s2 {
		return 0
	}
	if s1 > s2 {
		return minOperations(nums2, nums1)
	}
	d := s2 - s1
	arr := []int{}
	for _, v := range nums1 {
		arr = append(arr, 6-v)
	}
	for _, v := range nums2 {
		arr = append(arr, v-1)
	}
	sort.Sort(sort.Reverse(sort.IntSlice(arr)))
	for i, v := range arr {
		d -= v
		if d <= 0 {
			return i + 1
		}
	}
	return -1
}

func sum(nums []int) (s int) {
	for _, v := range nums {
		s += v
	}
	return
}

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