EASY, In-Depth, O(N)/O(1) Explanation || Python
Intuition\nThe problem requires us to equalize the sum of two arrays after replacing all zeros with strictly positive integers. Since the task is to achieve the
# Intuition\nThe problem requires us to equalize the sum of two arrays after replacing all zeros with strictly positive integers. Since the task is to achieve the minimum possible equal sum, the best approach is to replace all zeros with 1 (the smallest possible positive integer) and then compare the sums. If either array has no zeros left to adjust the sum of the other array, it becomes impossible to equalize the sums, and we should return -1.\n\n# Approach\n1. **Initialize Counters**: First, count the number of zeros in both arrays (zero_count1 and zero_count2).\n2. **Calculate the Sums**: Calculate the sum of both arrays, including zeros initially.\n3. **Replace Zeros with 1 for New Sum Calculations**: Recalculate the sums, this time increasing each of them by 1 for each zero in each array. 1 is the smallest possible integer, and we must replace ***all*** zeros in the array according to the problem description\n4. **Compare Sums**: If the sum of one array is larger than the other, check if the smaller array contains zeros. If it does, return the larger sum as the minimum equal sum. If there are no zeros to adjust the sum, return -1 because it is impossible to equalize the sums.\n5. **Return the Equal Sum**: If both sums are already equal after replacing zeros, return that sum as the result.\n# Complexity\n- **Time complexity**: The time complexity is $$O(n)$$, where $$n$$ is the total number of elements in nums1 and nums2. We loop through both arrays once to calculate the sums and count zeros, which takes linear time.\n\n- **Space complexity**: The space complexity is $$O(1)$$ since we only use a few variables to store sums and zero counts, and no additional data structures are used.\n\n# Code\n```python3 []\nclass Solution:\n    def minSum(self, nums1: List[int], nums2: List[int]) -> int:\n        # Initialize zero counters for nums1 and nums2\n        # These will keep track of how many 0s need to be replaced in each array\n        zero_count1, zero_count2 = 0, 0 \n        \n        # Initialize sums for both arrays\n        sum1, sum2 = 0, 0\n\n        # Iterate through the first array nums1\n        for num in nums1:\n            if num == 0:  # If a zero is found, increment zero_count1\n                zero_count1 += 1\n            # Add current element to sum1, zeroes included for now\n            sum1 += num\n\n        # Iterate through the second array nums2\n        for num in nums2:\n            if num == 0:  # If a zero is found, increment zero_count2\n                zero_count2 += 1\n            # Add current element to sum2, zeroes included for now\n            sum2 += num\n        \n        # Replace all zeroes in nums1 with 1 (the smallest positive integer)\n        # This is the minimum sum that sum1 can be\n        sum1 += 1 * zero_count1\n\n        # Replace all zeroes in nums2 with 1 (the smallest positive integer)\n        # This is the minimum sum that sum2 can be\n        sum2 += 1 * zero_count2\n\n        # Now, compare the two sums and determine the result\n        if sum1 > sum2:\n            # If sum1 is greater, check if it\'s possible to make sum2 equal by replacing zeros in nums2\n            # If nums2 has zeroes to replace, return sum1 as the minimum equal sum, else return -1\n            return sum1 if zero_count2 else -1\n        elif sum2 > sum1:\n            # If sum2 is greater, check if it\'s possible to make sum1 equal by replacing zeros in nums1\n            # If nums1 has zeroes to replace, return sum2 as the minimum equal sum, else return -1\n            return sum2 if zero_count1 else -1\n        \n        # If both sums are already equal, return the sum (this is the minimum equal sum)\n        return sum1\n```
['Python3']