Simple Greedy Approach (Sum + Zeros)....
Approach
Calculate the sum and count the zeros

Iterate over nums1 and compute s1, the sum of non-zero elements, while counting the number of zeroes z1.
Do the
<!-- # Intuition -->
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# Approach
<!-- Describe your approach to solving the problem. -->
* Calculate the sum and count the zeros
  * Iterate over nums1 and compute s1, the sum of non-zero elements, while counting the number of zeroes z1.
  * Do the same for nums2, computing s2 and z2.
* Handle Edge Cases
  * If nums1 has no zero (z1 == 0), but s1 is strictly less than s2 + z2, it means we cannot make the sum of nums1 at least as large as nums2 even after replacing zeroes, so return -1.
  * Similarly, if nums2 has no zero (z2 == 0), but s2 is strictly less than s1 + z1, return -1.
* Calculate the Result
  * If s1 + z1 (maximum possible sum of nums1 after replacement) is smaller than s2 + z2, return s2 + z2.
  * Otherwise, return s1 + z1.

# Complexity
- Time complexity: O(N + M)
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- Space complexity: O(1)
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# Code
```cpp []
class Solution {
public:
    long long minSum(vector<int>& nums1, vector<int>& nums2) {
        int n = nums1.size(), m = nums2.size();

        long long s1 = 0, s2 = 0;
        int z1 = 0, z2 = 0;
        for(int i = 0; i < n; i++){
            if(nums1[i] != 0) s1 += nums1[i];
            else z1++;
        }
        for(int i = 0; i < m; i++){
            if(nums2[i] != 0) s2 += nums2[i];
            else z2++;
        }

        if((z1 == 0 && s1 < (s2 + z2)) || (z2 == 0 && s2 < (s1 + z1))) return -1;
        else{
            if((s1 + z1) < (s2 + z2)) return s2 + z2;
            else if((s1 + z1) >= (s2 + z2)) return s1 + z1;
        }
        return -1;
    }
};
```
['Array', 'Greedy', 'C++']