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1775. Equal Sum Arrays With Minimum Number of Operations

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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

Python3

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)
        freq = [0] * 6
        for x in nums1:
            freq[6 - x] += 1
        for x in nums2:
            freq[x - 1] += 1
        diff = s2 - s1
        ans, i = 0, 5
        while i > 0 and diff > 0:
            while freq[i] and diff > 0:
                diff -= i
                freq[i] -= 1
                ans += 1
            i -= 1
        return -1 if diff > 0 else ans

Java

class Solution {
    public int minOperations(int[] nums1, int[] nums2) {
        int s1 = sum(nums1);
        int s2 = sum(nums2);
        if (s1 == s2) {
            return 0;
        }
        if (s1 > s2) {
            return minOperations(nums2, nums1);
        }
        int[] freq = new int[6];
        for (int x : nums1) {
            ++freq[6 - x];
        }
        for (int x : nums2) {
            ++freq[x - 1];
        }
        int diff = s2 - s1;
        int ans = 0;
        for (int i = 5; i > 0 && diff > 0; --i) {
            while (freq[i] > 0 && diff > 0) {
                diff -= i;
                --freq[i];
                ++ans;
            }
        }
        return diff > 0 ? - 1 : ans;
    }

    private int sum(int[] nums) {
        int s = 0;
        for (int x : nums) {
            s += x;
        }
        return s;
    }
}

C++

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);
        vector<int> freq(6);
        for (int x : nums1) ++freq[6 - x];
        for (int x : nums2) ++freq[x - 1];
        int diff = s2 - s1;
        int ans = 0;
        for (int i = 5; i > 0 && diff > 0; --i)
        {
            while (freq[i] && diff > 0)
            {
                diff -= i;
                --freq[i];
                ++ans;
            }
        }
        return diff > 0 ? -1 : ans;
    }
};

Go

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)
	}
	freq := make([]int, 6)
	for _, x := range nums1 {
		freq[6-x]++
	}
	for _, x := range nums2 {
		freq[x-1]++
	}
	diff := s2 - s1
	ans := 0
	for i := 5; i > 0 && diff > 0; i-- {
		for freq[i] > 0 && diff > 0 {
			diff -= i
			freq[i]--
			ans++
		}
	}
	if diff > 0 {
		return -1
	}
	return ans
}

func sum(nums []int) int {
	s := 0
	for _, x := range nums {
		s += x
	}
	return s
}

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