LeetCode. 1775 Equal Sum Arrays With Minimum Number of Operations

Greedy

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 <= 10^5

• 1 <= nums1[i], nums2[i] <= 6

Solution:

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class Solution {
    
    int process(int* table1, int* table2, int diff) {
        int ans = 0;
        for (int k = 1; k <= 5 && diff > 0; k++) {
            int contrib = 6 - k;
            
            while (table2[k] > 0 && diff > 0) {
                diff -= contrib;
                ans++;
                table2[k]--;
            }
            while (table1[7 - k] > 0 && diff > 0) {
                diff -= contrib;
                ans++;
                table1[7 - k]--;
            }
        }
        
        return ans;
    }
    
public:
    int minOperations(vector<int>& nums1, vector<int>& nums2) {
        if (6 * nums1.size() < nums2.size()) return -1;
        if (6 * nums2.size() < nums1.size()) return -1;
        int sum1 = 0;
        int table1[7] = {0};
        int sum2 = 0;
        int table2[7] = {0};
        
        for (int num : nums1) {
            sum1 += num;
            table1[num]++;
        }
        for (int num : nums2) {
            sum2 += num;
            table2[num]++;
        }
        
        if (sum1 == sum2) return 0;
        else if (sum1 > sum2) return process(table1, table2, sum1 - sum2);
        else return process(table2, table1, sum2 - sum1);
    }
};