Short and Simple code with intuitive approach! (❁´◡`❁)
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n1. calculate current su
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. calculate current sum and number of 0s in both the arrays\n2. find the minimum sum value that can be produced from both the arrays \n\n3. now check if the mini sum value of 2nd array can be made equal to mini sum value of first array ... if yessss , return minisum value of first array , else -1\n\n4. now check if the minimum sum value of 1st array can be made equal to minimum sum value of the second array... if yessss , return minimum sum value of second array , else -1\n\n5. now we are left with the condition.. both arrays have equal minisum so return either of the two minisum! \uD83D\uDE38\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ --> O(n+m)\nwhere n & m are the size of arrays\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --> O(1)\n\n# Code\n```\nclass Solution:\n    def minSum(self, nums1: List[int], nums2: List[int]) -> int:\n        cs1 , c1 = sum(nums1) , nums1.count(0)\n        cs2 , c2 = sum(nums2) , nums2.count(0)\n        print(cs1 ,c1 ,cs2 , c2)\n        ms1 = cs1 + c1      \n        ms2 = cs2 + c2 \n        if ms1 > ms2:\n            if c2 > 0:\n                return ms1\n            else:\n                return -1\n        elif ms2 > ms1:\n            if c1 > 0:\n                return ms2\n            else:\n                return -1\n        return ms2\n```
['Greedy', 'Python3']