Solution Explanation

This problem asks to find the minimum number of operations needed to make the sum of two arrays equal. An operation involves changing any integer in either array to any value between 1 and 6.

The solution employs a greedy approach. The core idea is to prioritize larger potential changes. We can efficiently achieve this by sorting the potential changes.

Algorithm:

1. Calculate sums: Calculate the sums (s1, s2) of nums1 and nums2.

2. Handle trivial cases: If the sums are already equal, return 0. If s1 > s2, recursively call the function with nums2, nums1 to ensure s2 >= s1.

3. Calculate potential changes: Create an array (arr) containing potential changes: 6 - nums1[i] (increase nums1[i] to 6), and nums2[i] - 1 (decrease nums2[i] to 1).

4. Sort: Sort arr in descending order. This ensures we address the largest potential changes first.

5. Greedy iteration: Iterate through arr in descending order. For each potential change, subtract it from the difference d = s2 - s1. If d becomes less than or equal to 0, the number of iterations performed is the minimum number of operations required.

6. Handle impossible cases: If the loop completes without d becoming less than or equal to 0, it's impossible to make the sums equal. Return -1.

Time Complexity Analysis:

• Summing the arrays: O(m + n), where m and n are the lengths of nums1 and nums2, respectively.
• Creating the arr array: O(m + n)
• Sorting arr: O((m+n) log(m+n))
• Iterating and subtracting: O(m + n) in the worst case (when all changes are needed).

Therefore, the overall time complexity is dominated by sorting, resulting in O((m+n) log(m+n)).

Space Complexity Analysis:

• The space used is primarily for the arr array, which has a size of m + n.
• In the recursive approach (Solution 1), the recursive call stack can have a maximum depth of min(m,n) in the worst case.

Thus, the space complexity is O(m + n). (The space complexity of solution 2 is O(1) since it uses a constant size counter array.)

Code Explanation (Python Solution 1):

class Solution:
    def minOperations(self, nums1: List[int], nums2: List[int]) -> int:
        s1, s2 = sum(nums1), sum(nums2)
        if s1 == s2:
            return 0
        if s1 > s2:
            return self.minOperations(nums2, nums1)
        arr = [6 - v for v in nums1] + [v - 1 for v in nums2]
        d = s2 - s1
        for i, v in enumerate(sorted(arr, reverse=True), 1):
            d -= v
            if d <= 0:
                return i
        return -1

The Python code directly implements the algorithm described above. The enumerate function is used to efficiently track the number of iterations. The sorted(arr, reverse=True) ensures elements are sorted in descending order for the greedy approach. The recursive call handles the case where s1 > s2.

The second solution uses a counting array to achieve the same result with a slightly improved space complexity of O(1). The time complexity remains the same.