Equal Sum Arrays With Minimum Number of Operations LeetCode solution in Python

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class Solution:
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    def minOperations(self, nums1: List[int], nums2: List[int]) -> int:
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        # step1. determine the larger array and the smaller array, and get the difference on sum
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        sum1 = sum(nums1)
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        sum2 = sum(nums2)
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        if sum1 == sum2:
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            return 0
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        elif sum1 > sum2:
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            larger_sum_nums = nums1
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            smaller_sum_nums = nums2
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        else:
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            larger_sum_nums = nums2
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            smaller_sum_nums = nums1
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        sum_diff = abs(sum1 - sum2)
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        # step2. calculate the max "gain" at each position (how much difference we can reduce if operating on that position)
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        gains_in_larger_array = [num - 1 for num in larger_sum_nums]
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        gains_in_smaller_array = [6 - num for num in smaller_sum_nums]
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        # step3. sort the "gain" and check the least number of steps to reduce the difference to 0
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        gains = gains_in_larger_array + gains_in_smaller_array
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        gains.sort(reverse=True)
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        count = 0
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        target_diff = sum_diff
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        for i in range(len(gains)):
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            target_diff -= gains[i]
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            count += 1
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            if target_diff <= 0:
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                return count
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        # return -1 if the difference still cannot be reduced to 0 even after operating on all positions
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        return -1