Intuitive approach. Easy
Intuition\n Describe your first thoughts on how to solve this problem. \nSum of the element of each array and add the number of zeroes. This is your minimum sum
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nSum of the element of each array and add the number of zeroes. This is your minimum sum\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nIn code\n\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nO(n)\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\nO(1)\n\n# Code\n```\nclass Solution:\n    def minSum(self, nums1: List[int], nums2: List[int]) -> int:\n        \n\n        #get the sum for the each array and their potential minimum. This can be found by sum(arr) + sum(number of zeroes)\n        #each zero can be represented by 1 to get the minimum answer\n        #return -1 if the array with a lower minEqual has no zeroes to fill up \n        \n\n        def countZeroes(arr):\n            return len([x for x in arr if x == 0])\n\n        zeroes1 = countZeroes(nums1)\n        zeroes2 = countZeroes(nums2)\n        minEqualSum1 = sum(nums1) + zeroes1\n        minEqualSum2 = sum(nums2) + zeroes2\n\n        if minEqualSum1 > minEqualSum2:\n            if zeroes2 ==0: return -1\n\n            else: return minEqualSum1\n\n        elif minEqualSum2 > minEqualSum1:\n            if zeroes1 ==0: return -1\n\n            else: return minEqualSum2\n\n        else:\n            return minEqualSum2\n            \n\n```
['Array', 'Python3']