LeetCode 1775: Equal Sum Arrays With Minimum Number of Operations Solution

Master LeetCode problem 1775 (Equal Sum Arrays With Minimum Number of Operations), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1775. Equal Sum Arrays With Minimum Number of Operations

Medium

Array Hash Table Greedy Counting

Problem Explanation

Explanation

To solve this problem, we need to find the minimum number of operations required to make the sum of values in nums1 equal to the sum of values in nums2. We can achieve this by calculating the sum of both arrays and then comparing them. If the sums are equal, then no operations are needed. If the sums are not equal, we need to find the absolute difference between them and then calculate the number of operations needed to make the sums equal.

Calculate the sum of nums1 and nums2.
If the sums are already equal, return 0 as no operations are required.
If the sums are not equal, calculate the absolute difference.
Count the occurrences of each value in both arrays.
Iterate over the possible values from 1 to 6 and calculate the operations needed to adjust the sums.
Return the minimum number of operations if it is possible to make the sums equal, otherwise return -1.

Time Complexity: O(n), where n is the size of the input arrays. Space Complexity: O(1) since we are using a constant amount of extra space.

Solution Code

class Solution {
    public int minOperations(int[] nums1, int[] nums2) {
        int sum1 = 0, sum2 = 0;
        int[] freq1 = new int[7];
        int[] freq2 = new int[7];
        
        for (int num : nums1) {
            sum1 += num;
            freq1[num]++;
        }
        
        for (int num : nums2) {
            sum2 += num;
            freq2[num]++;
        }
        
        if (sum1 == sum2) return 0;
        
        int diff = Math.abs(sum1 - sum2);
        int operations = 0;
        
        for (int i = 1; i <= 6; i++) {
            int need = Math.abs(sum1 - sum2) / (6 - i) + (Math.abs(sum1 - sum2) % (6 - i) == 0 ? 0 : 1);
            if (sum1 > sum2) {
                if (i < 6) operations += Math.min(need, freq1[i]);
            } else {
                if (i > 1) operations += Math.min(need, freq2[i]);
            }
        }
        
        return operations;
    }
}

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