1775. Equal Sum Arrays With Minimum Number of Operations

Problem Description

You are given two arrays of integers, nums1 and nums2, which may not be of equal length. Each element in both arrays will be an integer between 1 and 6, inclusive. Your task is to equalize the sum of both arrays using the minimum number of operations. An operation consists of changing any element's value in either array to any integer between 1 and 6 (also inclusive). You need to determine the least number of such operations needed to make the sums of both arrays equal, or return -1 if equalizing the sums is impossible.

Intuition

To solve this problem intuitively, one should start by comparing the sums of both arrays. If they are already equal, no operations are needed, so the answer is 0. If the sums are not equal, operations will entail either increasing some numbers in the smaller-summed array or decreasing some in the larger-summed array.

There are limit cases where it’s impossible to make the sums equal. One occurs when the smaller array is so small and full of sixes, and the larger array is so large and full of ones that even maxing out the values in the smaller one won't help. Another occurs when the larger array is full of sixes, and the smaller is full of ones, and decreasing values in the larger array is equally futile.

Approaching the array with the smaller sum, we look at how much we can increase each of its elements (up to a maximum of 6). For the larger sum array, we calculate how much we can decrease each element (down to a minimum of 1). The resulting potential changes are combined into a single list, sorted by the magnitude of potential change in descending order.

We then iterate through each potential change from largest to smallest. With every iteration, we subtract the value of the potential change from the sum difference between the arrays. If at any point the sum difference drops to 0 or below, we know that we've made enough changes to equalize the arrays, and we return the number of operations taken thus far. If we exhaust the list of potential changes and the sum difference remains positive, we conclude that equalizing the sums is impossible and return -1.

Learn more about Greedy patterns.

Solution Approach

The implementation groups a clear algorithmic approach with efficient use of data structures, mainly lists and the sum function. Here's the breakdown:

1. First, calculate the sums of both input arrays, nums1 and nums2, using the sum function. We need these sums to decide our next step and to calculate the difference between the two.

2. If the sums of both arrays are already equal (s1 == s2), return 0 as no operations are required.

3. Verify if the sum of nums1 is greater than that of nums2. If this is true, call the function recursively with the arrays flipped because it's more practical to begin with the array that currently has the smaller sum.

4. To generate the list of potential changes, subtract each value in the smaller-summed array (nums1) from 6 to see the maximum increase possible for each element. For the larger-summed array (nums2), subtract 1 from each value to see the maximum decrease possible. Combine these two lists.

5. Sort this combined list in reverse order (from the largest to the smallest potential change) because you want the most significant changes to be considered first.

6. Initialize a variable d with the value of the difference between s2 and s1 (the sum of nums2 and nums1).

7. Iterate over the sorted list of potential changes. During each iteration:

   • Subtract the current potential change from d.
   • Increment a counter to keep track of how many changes we've used.
   • If d is less or equal to 0 after the subtraction, return the number of operations (the current counter value) as you've managed to equalize the sums.

8. If you complete the iteration without d dropping to or below 0, return -1 as it confirms that equalizing the sums is not possible with the given constraints.

This algorithm effectively utilizes greedy strategy by sorting and consuming potential changes from greatest to least effect. It assures the minimum number of operations as each step makes the largest possible contribution to equalizing the array sums.

Example Walkthrough

Let's consider the following example to illustrate the solution approach:

• nums1 = [1, 2, 5] and nums2 = [3, 1, 6, 1]

Now, let's walk through the solution steps:

1. Calculate the sums of nums1 and nums2:

   • sum(nums1) = 1+2+5 = 8
   • sum(nums2) = 3+1+6+1 = 11

2. The sums are not equal (8 != 11), so we proceed with the next steps.

3. Since the sum of nums2 is greater than that of nums1, we recursively call the function with nums2 and nums1 swapped. Now our goal is to increase the sum of nums1 or decrease the sum of nums2.

4. Calculate the potential changes:

   • For nums1, we can increase 1 to 6 (gaining 5), 2 to 6 (gaining 4), and we cannot increase 5 any further.
   • For nums2, we can decrease 3 to 1 (gaining 2), 6 to 1 (gaining 5), and we do not gain from 1s.
   • The potential changes list is [5, 4, 2, 5].

5. Sort the potential changes list into [5, 5, 4, 2], favoring the biggest changing numbers.

6. Initialize d with the difference between sums: d = 11 - 8 = 3.

7. Iterate over the sorted list of potential changes:

   • Using 5 from the first array brings d to -2 (3 - 5 = -2).
   • Since d is now less than 0, we know that we have equalized the arrays. It took us 1 operation to achieve this.

Therefore, in this example, the smallest number of operations needed to equalize the sums of both arrays is 1.

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the code can be broken down into the following parts:

• Summation of both arrays nums1 and nums2: This has a time complexity of O(N + M), where N is the length of nums1, and M is the length of nums2, since each array is iterated over once.

• The recursive call self.minOperations(nums2, nums1) doesn't change the time complexity since it happens at most once and just swaps the role of nums1 and nums2.

• Creation of the combined array arr: This involves iterating over both nums1 and nums2 and thus has a complexity of O(N + M).

• Sorting the arr: The sorting operation has a complexity of O((N + M) * log(N + M)), because it sorts a list that has a length equal to the sum of lengths of nums1 and nums2.

• Iterating over the sorted array to find the minimum number of operations: In the worst case, this is O(N + M) when we iterate over the entire combined list.

Therefore, the dominant factor for time complexity here is the sorting operation. The overall time complexity of the function is O((N + M) * log(N + M)).

Space Complexity

The space complexity of the code can be analyzed as follows:

• Extra space for the sum operations is O(1) since we just accumulate values.

• Extra space for the combined array arr: This takes up O(N + M) space.

• Space required for the sorting of arr: Sorting typically requires O(N + M) space.

Given that the combined array arr and its sorting are the steps that require the most space, the overall space complexity is O(N + M).

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