2321. Maximum Score Of Spliced Array


Problem Description

In this problem, we are given two integer arrays, nums1 and nums2, both with the same length n. We have the option to perform a single operation: choose two indices left and right such that 0 <= left <= right < n and swap the subarray nums1[left...right] with the subarray nums2[left...right]. This operation can be done only once or we may choose not to perform any swap at all.

The goal is to maximize the score of the arrays after the swap operation. The score is defined as the maximum sum of either nums1 or nums2. The objective is to determine the maximum possible score. A subarray is a contiguous sequence of elements within an array, and arr[left...right] represents the subarray that contains all elements in nums from index left to index right, inclusive.

Intuition

To arrive at the solution, we need to determine if performing the swap operation would be beneficial and if so, which indices to choose for the left and right boundaries of the subarrays to swap. Since we are trying to maximize the sum of the higher-scoring array, we should look for the subarray within nums1 and nums2 that, when swapped, will lead to the greatest increase in sum for the resulting array.

The intuition here is to calculate the difference between corresponding elements of nums1 and nums2. By finding the contiguous subsequence with the maximum sum (max subarray sum) which can be positive or negative, we can identify the impact of swapping this particular subarray on the overall score.

We implement a helper function f(nums1, nums2) that calculates this maximum sum for the differences between nums1 and nums2. This is similar to the classic maximum subarray problem, which can be solved using Kadane's algorithm. For the chosen subarray, we add this difference to the sum of nums2, and similarly, for the opposite operation, we add the difference to the sum of nums1 when considering the swap from nums2 to nums1.

In the provided solution, the maximumSplicedArray function performs this comparison and returns the higher value achieved by either adding the maximum contiguous sum difference of nums1 to the sum of nums2, or vice versa. The maximum of these two gives us the maximum possible score after performing the optimal swap operation, and that's what we return as the solution.

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Solution Approach

The implementation of the solution involves a function f(nums1, nums2) that employs the idea of Kadane's algorithm to find the maximum subarray sum difference between the two input arrays nums1 and nums2. Kadane's algorithm is a classic dynamic programming approach to solve the maximum subarray problem.

Here's a step-by-step breakdown of the algorithm and patterns used in the implementation:

  1. Difference Calculation: We start by calculating the difference array d where each element d[i] is the result of nums1[i] - nums2[i]. This array is critical because it represents the change in the sum when we swap corresponding elements from nums1 and nums2.

  2. Kadane's Algorithm: The helper function f(nums1, nums2) runs a modified version of Kadane's algorithm on the difference array d. The algorithm iterates through the array and looks for the subarray with the maximum sum, which indicates the best subarray to swap (if beneficial):

    • Initialize two variables t and mx with the value of the first element in the array d.
    • Iterate through the difference array starting from the second element, updating t with the sum of t and the current element if t is positive; otherwise, reset t to the current element's value.
    • At each step, update mx to be the maximum of mx and t.
    • After iterating through the entire array, mx holds the maximum subarray sum difference.
  3. Sum Calculation and Comparison: We calculate s1 and s2 as the sums of nums1 and nums2 respectively. The final score is computed by adding the maximum subarray sum difference to the sum of the opposite array.

    • To consider the swap from nums1 to nums2, compute s2 + f(nums1, nums2) which represents the score if we swap the most beneficial subarray from nums1 to nums2.
    • To consider the swap from nums2 to nums1, compute s1 + f(nums2, nums1) which represents the score if we swap the most beneficial subarray from nums2 to nums1.
  4. Maximization: Since we want the maximum score, we take the maximum value from the two calculated scores mentioned above. This gives us the maximum possible score after considering whether a swap would be advantageous or not.

The final call max(s2 + f(nums1, nums2), s1 + f(nums2, nums1)) evaluates both scenarios and returns the higher score as the solution to the problem.

This implementation is efficient because Kadane's algorithm runs in linear time O(n), and the rest of the operations are simple sums and comparisons, also done in linear time. Thus, the overall time complexity of the solution is O(n).

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Example Walkthrough

Let's illustrate the solution approach with a small example:

Suppose nums1 = [1, 3, 5, 7, 9] and nums2 = [2, 4, 6, 8, 10] with n = 5.

Step 1: Difference Calculation

First, we compute the difference array d by subtracting each element of nums1 from nums2:

d = [2 - 1, 4 - 3, 6 - 5, 8 - 7, 10 - 9] d = [1, 1, 1, 1, 1]

Step 2: Kadane's Algorithm

Now, let's apply Kadane's algorithm to find the maximum subarray sum in d.

  • Initialize t = d[0] = 1 and mx = d[0] = 1.
  • Move through the difference array starting from index 1:
    • At each element d[i], if t > 0, then update t = t + d[i].
    • Otherwise, set t = d[i].
    • Update mx to the larger of mx or t at each step.

After iterating through the array d, the maximum subarray sum mx turns out to be 5.

Step 3: Sum Calculation and Comparison

Calculate the sum of the original arrays:

  • s1 = sum(nums1) = 1 + 3 + 5 + 7 + 9 = 25
  • s2 = sum(nums2) = 2 + 4 + 6 + 8 + 10 = 30

We consider the scores after the swap operation:

  • Swapping a beneficial subarray from nums1 to nums2 gives us s2 + mx = 30 + 5 = 35.
  • Swapping a beneficial subarray from nums2 to nums1 gives us s1 + mx = 25 + 5 = 30.

Step 4: Maximization

Finally, we choose the higher of the two calculated scores:

  • max(35, 30) = 35

Hence, the operation that gives us the maximum score is swapping the beneficial subarray from nums1 to nums2, and the maximum score achievable is 35. This is our final solution.

Solution Implementation

1from typing import List
2
3class Solution:
4    def maximumSplicedArray(self, nums1: List[int], nums2: List[int]) -> int:
5        # Helper function to calculate the maximum difference between subarrays
6        def max_subarray_difference(nums1, nums2):
7            # Calculate the differences between elements of the two arrays
8            differences = [a - b for a, b in zip(nums1, nums2)]
9          
10            # Initialize the current and maximum subarray sum with the first difference element
11            current_sum = max_sum = differences[0]
12          
13            # Loop through the rest of the differences
14            for value in differences[1:]:
15                # If current_sum is positive, add the value to it; else, start a new subarray sum with the value
16                if current_sum > 0:
17                    current_sum += value
18                else:
19                    current_sum = value
20                # Update max_sum with the maximum sum found so far
21                max_sum = max(max_sum, current_sum)
22          
23            # Return the maximum subarray sum found
24            return max_sum
25
26        # Calculate the total sums of both input arrays
27        total_sum1, total_sum2 = sum(nums1), sum(nums2)
28      
29        # Calculate the maximum possible sum by splicing subarrays from one array to another
30        # Two cases: splicing from nums1 to nums2, and from nums2 to nums1
31        # We add the subarray_sum to the total sum of the opposite array
32        return max(
33            total_sum2 + max_subarray_difference(nums1, nums2),
34            total_sum1 + max_subarray_difference(nums2, nums1)
35        )
36
37# The Solution class can now be used to find the maximum spliced array sum
38
1class Solution {
2    public int maximumsSplicedArray(int[] nums1, int[] nums2) {
3        int sum1 = 0, sum2 = 0; // Initialize sums of both arrays to 0
4        int length = nums1.length; // Length of the arrays
5        // Calculate the sum of each array
6        for (int i = 0; i < length; ++i) {
7            sum1 += nums1[i];
8            sum2 += nums2[i];
9        }
10        // Return the maximum sum possible by splicing. Calculate twice, swapping the arrays, to ensure all possibilities are examined.
11        return Math.max(sum2 + delta(nums1, nums2), sum1 + delta(nums2, nums1));
12    }
13
14    // Helper function to calculate the maximum difference subarray when splicing nums2 into nums1, represented as 'delta'.
15    private int delta(int[] nums1, int[] nums2) {
16        int temporarySum = nums1[0] - nums2[0];
17        int maxDiff = temporarySum; // Maximum difference found so far
18
19        // Iterate over the arrays starting from the second element
20        for (int i = 1; i < nums1.length; ++i) {
21            int valueDifference = nums1[i] - nums2[i];
22
23            // If the current temporary sum is positive, continue the subarray
24            if (temporarySum > 0) {
25                temporarySum += valueDifference;
26            } else {
27                // Else start a new subarray
28                temporarySum = valueDifference;
29            }
30            // Update the maximum difference if the new temporary sum is greater
31            maxDiff = Math.max(maxDiff, temporarySum);
32        }
33
34        return maxDiff; // Return the maximum difference found
35    }
36}
37
1class Solution {
2public:
3    // Function to find the maximum sum we can obtain by splicing two arrays
4    int maximumsSplicedArray(vector<int>& nums1, vector<int>& nums2) {
5        // Initialize the sum for each array
6        int sumNums1 = 0, sumNums2 = 0;
7        int n = nums1.size(); // Get the size of the arrays
8
9        // Calculate the sum of elements for each array
10        for (int i = 0; i < n; ++i) {
11            sumNums1 += nums1[i];
12            sumNums2 += nums2[i];
13        }
14
15        // The maximum sum is the maximum of splicing in both directions
16        return max(sumNums2 + getMaxDiff(nums1, nums2), sumNums1 + getMaxDiff(nums2, nums1));
17    }
18
19    // Helper function to calculate the maximum difference splicing sequence
20    int getMaxDiff(vector<int>& nums1, vector<int>& nums2) {
21        // Starting difference between the first elements of both arrays
22        int currentDiff = nums1[0] - nums2[0];
23        int maxDiff = currentDiff; // Initialize maxDiff with the first difference
24
25        // Iterate through the arrays starting from the second element
26        for (int i = 1; i < nums1.size(); ++i) {
27            int diff = nums1[i] - nums2[i]; // Calculate the difference for the current index
28
29            // If the accumulated difference is positive, continue the sequence
30            // Otherwise, start a new sequence with the current difference
31            currentDiff = (currentDiff > 0) ? currentDiff + diff : diff;
32
33            // Update maxDiff if the current accumulated difference is larger
34            maxDiff = max(maxDiff, currentDiff);
35        }
36
37        // Return the maximum difference found
38        return maxDiff;
39    }
40};
41
1// Function to calculate the sum of the elements of an array
2function sumArray(array: number[]): number {
3  return array.reduce((acc, value) => acc + value, 0);
4}
5
6// Function to find the maximum sum we can obtain by splicing two arrays
7function maximumSplicedArray(nums1: number[], nums2: number[]): number {
8  // Calculate the sum of elements for each array using the sumArray function
9  const sumNums1: number = sumArray(nums1);
10  const sumNums2: number = sumArray(nums2);
11
12  // The maximum sum is the maximum of splicing in both directions
13  return Math.max(
14    sumNums1 + getMaxDiff(nums2, nums1),
15    sumNums2 + getMaxDiff(nums1, nums2)
16  );
17}
18
19// Helper function to calculate the maximum difference splicing sequence
20function getMaxDiff(nums1: number[], nums2: number[]): number {
21  // Starting difference between the first elements of both arrays
22  let currentDiff: number = nums1[0] - nums2[0];
23  let maxDiff: number = currentDiff; // Initialize maxDiff with the first difference
24
25  // Iterate through the arrays starting from the second element
26  for (let i: number = 1; i < nums1.length; i++) {
27    const diff: number = nums1[i] - nums2[i]; // Calculate the difference for the current index
28
29    // If the accumulated difference is positive, continue the sequence
30    // Otherwise, start a new sequence with the current difference
31    currentDiff = currentDiff > 0 ? currentDiff + diff : diff;
32
33    // Update maxDiff if the current accumulated difference is larger
34    maxDiff = Math.max(maxDiff, currentDiff);
35  }
36
37  // Return the maximum difference found
38  return maxDiff;
39}
40
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Time and Space Complexity

The given Python code defines a method maximumsSplicedArray which compares the sums of two arrays after possibly splicing in continuous subarray sections from one to the other to maximize the sum. The code makes use of a helper function f which computes the maximum subarray sum difference.

Time Complexity

The time complexity of the function maximumsSplicedArray is determined by the helper function f, which is called twice, and the sum computation for each array.

Here is the breakdown of the time complexity:

  • Computing the sum for nums1 and nums2 each takes O(n) time where n is the length of each array.
  • The function f is called twice, each call involves:
    • Creating a difference array d which takes O(n) time.
    • Iterating over the difference array excluding the first element to find the maximum subarray sum difference, which also takes O(n) time due to the use of the Kadane's algorithm approach.

So the overall time complexity of the function is O(n) for the sum computations plus 2 * O(n) for the two calls to function f, resulting in a total time complexity of O(n).

Space Complexity

As for space complexity:

  • The difference array d requires O(n) space.
  • The variables t, mx, and the sums s1 and s2 use constant space O(1).

Hence, the space complexity of the function maximumsSplicedArray is O(n).

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