Problem Description

You have two arrays nums1 and nums2 containing integers between 1 and 6 (inclusive). The arrays can have different lengths.

In each operation, you can pick any number from either array and change it to any value between 1 and 6.

Your goal is to make the sum of all numbers in nums1 equal to the sum of all numbers in nums2 using the minimum number of operations.

Return the minimum number of operations needed to achieve equal sums. If it's impossible to make the sums equal, return -1.

Example scenarios:

• If nums1 = [1,2,3,4,5,6] and nums2 = [1,1,2,2,2,2], you need to find how many numbers to change to make sum(nums1) = 21 equal to sum(nums2) = 10.
• To maximize the impact of each operation, you'd want to decrease large numbers in the array with the larger sum (change 6→1, 5→1, etc.) or increase small numbers in the array with the smaller sum (change 1→6, 2→6, etc.).

Key insight: Each operation can change the total sum difference by at most 5 (changing a 1 to 6 or a 6 to 1). The solution calculates the maximum possible change for each number in both arrays, sorts these potential changes in descending order, and greedily applies them until the sum difference is eliminated.

Intuition

When we need to make two sums equal, we're essentially trying to eliminate the difference between them. Let's say sum(nums1) < sum(nums2), so we have a gap of d = sum(nums2) - sum(nums1) to close.

To close this gap efficiently, we want each operation to have maximum impact. For the smaller sum array (nums1), we want to increase values as much as possible - changing a 1 to 6 gives us +5, changing a 2 to 6 gives us +4, and so on. For the larger sum array (nums2), we want to decrease values as much as possible - changing a 6 to 1 gives us -5 (which helps close the gap by 5), changing a 5 to 1 gives us -4, etc.

The key realization is that we can treat both types of changes uniformly:

• Increasing a value v in nums1 to 6 contributes 6 - v toward closing the gap
• Decreasing a value v in nums2 to 1 contributes v - 1 toward closing the gap

By collecting all these potential contributions into a single array and sorting them in descending order, we can greedily pick the most impactful changes first. We keep applying changes until the cumulative contribution meets or exceeds the gap d.

The greedy approach works because each change is independent - we're always better off making the change with the largest impact first. If the gap can be closed at all, this greedy strategy will find the minimum number of operations needed.

The impossibility check (-1 case) happens implicitly: if we've exhausted all possible changes and still haven't closed the gap, it means the gap is too large to bridge even with all values set to their extremes.

Learn more about Greedy patterns.

Solution Approach

The implementation follows the greedy strategy we identified:

1. Calculate the initial difference: First, compute s1 = sum(nums1) and s2 = sum(nums2). If they're already equal, return 0.

2. Normalize the problem: To simplify the logic, we ensure s1 < s2 by swapping the arrays if necessary. This way, we always work with the same scenario - trying to increase the smaller sum or decrease the larger sum.

3. Build the contribution array: Create an array arr that contains all possible contributions:

   • For each value v in nums1 (the smaller sum array), add 6 - v to represent how much we can increase the sum by changing v to 6
   • For each value v in nums2 (the larger sum array), add v - 1 to represent how much we can decrease the sum by changing v to 1

4. Greedy selection: Sort arr in descending order to prioritize the most impactful changes. The difference to close is d = s2 - s1.

5. Apply changes until gap is closed: Iterate through the sorted contributions, subtracting each from d. Keep a counter i for the number of operations. Once d <= 0, we've closed the gap and return i.

6. Handle impossible cases: If we exhaust all possible changes (finish the loop) without closing the gap, return -1.

The time complexity is O((m + n) log(m + n)) where m and n are the lengths of the arrays, dominated by the sorting step. The space complexity is O(m + n) for storing the contribution array.

The elegance of this solution lies in treating increases and decreases uniformly as "contributions" toward closing the gap, allowing us to use a simple greedy approach rather than complex decision-making about which array to modify.

Example Walkthrough

Let's walk through a small example to illustrate the solution approach.

Example: nums1 = [1, 6, 6] and nums2 = [1, 1, 1]

Step 1: Calculate initial sums

• s1 = 1 + 6 + 6 = 13
• s2 = 1 + 1 + 1 = 3
• Since s1 > s2, we swap the arrays to ensure the first array has the smaller sum
• After swapping: nums1 = [1, 1, 1] (sum = 3) and nums2 = [1, 6, 6] (sum = 13)
• Gap to close: d = 13 - 3 = 10

Step 2: Build the contribution array For each element, calculate how much it can contribute to closing the gap:

• From nums1 = [1, 1, 1] (we want to increase these):
  • 6 - 1 = 5 (changing 1→6 adds 5 to the sum)
  • 6 - 1 = 5 (changing 1→6 adds 5 to the sum)
  • 6 - 1 = 5 (changing 1→6 adds 5 to the sum)
• From nums2 = [1, 6, 6] (we want to decrease these):
  • 1 - 1 = 0 (changing 1→1 contributes nothing)
  • 6 - 1 = 5 (changing 6→1 subtracts 5 from the sum)
  • 6 - 1 = 5 (changing 6→1 subtracts 5 from the sum)
• Combined contributions: arr = [5, 5, 5, 0, 5, 5]

Step 3: Sort contributions in descending order

• Sorted arr = [5, 5, 5, 5, 5, 0]

Step 4: Greedily apply changes

• Start with gap d = 10
• Operation 1: Use contribution of 5, d = 10 - 5 = 5
• Operation 2: Use contribution of 5, d = 5 - 5 = 0
• Gap closed! We need 2 operations

Verification: We could either:

• Change two 1s in the original nums1 to 6s: [1, 6, 6] → [6, 6, 6], sum becomes 18
• Change both 6s in the original nums2 to 1s: [1, 1, 1] → [1, 1, 1], sum stays 3 (Wait, that doesn't work since we need sum = 8 for both)

Actually, let me recalculate - if we change one 1→6 in the smaller array and one 6→1 in the larger array:

• Original nums1 = [1, 1, 1] becomes [6, 1, 1], sum = 8
• Original nums2 = [1, 6, 6] becomes [1, 1, 6], sum = 8
• Both sums equal 8, achieved in 2 operations ✓

Solution Implementation

Time and Space Complexity

Time Complexity: O(n log n + m log m) where n is the length of nums1 and m is the length of nums2.

• Calculating the sums s1 and s2 takes O(n) + O(m) time.
• Creating the arr list with list comprehensions takes O(n) + O(m) time.
• Sorting the combined array arr of size n + m takes O((n + m) log(n + m)) time.
• The iteration through the sorted array takes at most O(n + m) time.
• The recursive call (if triggered) doesn't add to the complexity as it only swaps the parameters once and won't recurse again.
• Overall, the dominant operation is sorting, giving us O((n + m) log(n + m)) which simplifies to O(n log n + m log m) when n and m are of similar magnitude.

Space Complexity: O(n + m)

• The arr list stores n + m elements, requiring O(n + m) space.
• The sorting operation may require O(n + m) additional space depending on the implementation.
• The recursion depth is at most 1 (only one recursive call possible), adding O(1) to the call stack.
• Therefore, the total space complexity is O(n + m).

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Forgetting to Check if Equal Sums are Achievable

A critical pitfall is not verifying whether it's mathematically possible to make the sums equal before attempting the greedy approach. This can lead to incorrect results or unnecessary computation.

The Problem: Consider extreme cases like nums1 = [1] and nums2 = [6,6,6,6,6,6]. Here:

• sum(nums1) = 1 (can be at most 6 after changes)
• sum(nums2) = 36 (can be at least 6 after changes)

Even with all possible changes, the maximum achievable sum for nums1 is 6, while the minimum achievable sum for nums2 is 6. These ranges don't overlap sufficiently to reach equality.

The Solution: Add a feasibility check at the beginning:

class Solution:
    def minOperations(self, nums1: List[int], nums2: List[int]) -> int:
        # Check if it's possible to make sums equal
        len1, len2 = len(nums1), len(nums2)
      
        # Maximum possible sum for each array is length * 6
        # Minimum possible sum for each array is length * 1
        if 6 * len1 < len2 or 6 * len2 < len1:
            return -1
      
        # Rest of the original solution continues...
        sum1, sum2 = sum(nums1), sum(nums2)
        # ... (rest of the code)

2. Inefficient Array Swapping

The recursive call return self.minOperations(nums2, nums1) when sum1 > sum2 creates unnecessary overhead and potential stack usage.

The Solution: Instead of recursion, simply swap the references:

class Solution:
    def minOperations(self, nums1: List[int], nums2: List[int]) -> int:
        sum1, sum2 = sum(nums1), sum(nums2)
      
        if sum1 == sum2:
            return 0
      
        # Swap references instead of recursive call
        if sum1 > sum2:
            nums1, nums2 = nums2, nums1
            sum1, sum2 = sum2, sum1
      
        # Continue with the logic...

3. Not Handling Edge Cases with Empty or Single-Element Arrays

While the problem states arrays contain integers between 1 and 6, it's good practice to handle edge cases explicitly, especially when array lengths might be very different.

Enhanced Solution with All Fixes:

class Solution:
    def minOperations(self, nums1: List[int], nums2: List[int]) -> int:
        # Early feasibility check
        len1, len2 = len(nums1), len(nums2)
        if 6 * min(len1, len2) < max(len1, len2):
            return -1
      
        sum1, sum2 = sum(nums1), sum(nums2)
      
        if sum1 == sum2:
            return 0
      
        # Ensure sum1 < sum2 without recursion
        if sum1 > sum2:
            nums1, nums2 = nums2, nums1
            sum1, sum2 = sum2, sum1
      
        # Calculate possible changes
        max_changes = [6 - v for v in nums1] + [v - 1 for v in nums2]
      
        difference = sum2 - sum1
      
        # Greedily apply largest changes
        for operation_count, change in enumerate(sorted(max_changes, reverse=True), 1):
            difference -= change
            if difference <= 0:
                return operation_count
      
        return -1

This enhanced version avoids recursion overhead, checks feasibility early to avoid unnecessary computation, and handles the problem more efficiently.