2918.minimum-equal-sum-of-two-arrays-after-replacing-zeros.cpp

//  Tag: Array, Greedy
//  Time: O(N)
//  Space: O(1)
//  Ref: -
//  Note: -
//  Video: https://youtu.be/o5nHLgCQXG8

//  You are given two arrays nums1 and nums2 consisting of positive integers.
//  You have to replace all the 0's in both arrays with strictly positive integers such that the sum of elements of both arrays becomes equal.
//  Return the minimum equal sum you can obtain, or -1 if it is impossible.
//   
//  Example 1:
//  
//  Input: nums1 = [3,2,0,1,0], nums2 = [6,5,0]
//  Output: 12
//  Explanation: We can replace 0's in the following way:
//  - Replace the two 0's in nums1 with the values 2 and 4. The resulting array is nums1 = [3,2,2,1,4].
//  - Replace the 0 in nums2 with the value 1. The resulting array is nums2 = [6,5,1].
//  Both arrays have an equal sum of 12. It can be shown that it is the minimum sum we can obtain.
//  
//  Example 2:
//  
//  Input: nums1 = [2,0,2,0], nums2 = [1,4]
//  Output: -1
//  Explanation: It is impossible to make the sum of both arrays equal.
//  
//   
//  Constraints:
//  
//  1 <= nums1.length, nums2.length <= 105
//  0 <= nums1[i], nums2[i] <= 106
//  
//  

class Solution {
public:
    long long minSum(vector<int>& nums1, vector<int>& nums2) {
        long long zero1 = count(nums1.begin(), nums1.end(), 0);
        long long zero2 = count(nums2.begin(), nums2.end(), 0);
        long long s1 = accumulate(nums1.begin(), nums1.end(), zero1);
        long long s2 = accumulate(nums2.begin(), nums2.end(), zero2);
        if ((s1 < s2 && zero1 == 0) || (s2 < s1 && zero2 == 0)) {
            return -1;
        }

        return max(s1, s2);
    }
};