Minimum Equal Sum of Two Arrays After Replacing Zeros - Greedy [JS]

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Description 

Solution: Greedy Logic

Count the sum and number of zeros in nums1 and nums2.
Get the base sum for both arrays, which is turning all the zeros into 1's, since that is the minimum.
From there, we can calculate the difference between the two base sums and take the maximum sum.
The only case where this is impossible, is if one side is less than the other, but there is no zero present that we can turn into a larger number.

n = length of nums1, m = length of nums2
Time Complexity: O(n + m)
Space Complexity: O(1)

var minSum = function(nums1, nums2) {
  let n = nums1.length, m = nums2.length;
  let sum1 = 0, zeros1 = 0;
  for (let i = 0; i < n; i++) {
    sum1 += nums1[i];
    zeros1 += nums1[i] === 0 ? 1 : 0;
  }
  let sum2 = 0, zeros2 = 0;
  for (let i = 0; i < m; i++) {
    sum2 += nums2[i];
    zeros2 += nums2[i] === 0 ? 1 : 0;
  }
  
  let baseSum1 = sum1 + zeros1, baseSum2 = sum2 + zeros2;
  if (baseSum1 > baseSum2 && zeros2 === 0) return -1;
  if (baseSum2 > baseSum1 && zeros1 === 0) return -1;
  return Math.max(baseSum1, baseSum2);
};
Location: Auckland, New Zealand

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