Maximum Subsequence Score

Problem Statement

In this problem, you are provided with two integer arrays, nums1 and nums2, both having the same length n. Additionally, you are given a positive integer k. Your task is to select a subsequence of indices of length k from nums1. The term "subsequence" here means a subset of the original index set {0, 1, ..., n-1} which can contain zero or more deletions but no reordering.

For any chosen subsequence of indices i0, i1, ..., ik - 1, you need to calculate the score defined as:

• The sum of the elements in nums1 at these indices multiplied by the minimum of the corresponding elements in nums2 at these indices.
• Mathematically: (nums1[i0] + nums1[i1] + ... + nums1[ik - 1]) * min(nums2[i0], nums2[i1], ..., nums2[ik - 1]).

The goal is to find and return the maximum possible score by selecting an optimal subsequence of indices.

Examples

Example 1

Input:

nums1 = [1,3,3,2], nums2 = [2,1,3,4], k = 3

Output:

12

Explanation:

The four possible subsequence scores are:
- We choose the indices 0, 1, and 2 with score = (1+3+3) * min(2,1,3) = 7.
- We choose the indices 0, 1, and 3 with score = (1+3+2) * min(2,1,4) = 6.
- We choose the indices 0, 2, and 3 with score = (1+3+2) * min(2,3,4) = 12.
- We choose the indices 1, 2, and 3 with score = (3+3+2) * min(1,3,4) = 8.
Therefore, we return the max score, which is 12.

Example 2

Input:

nums1 = [4,2,3,1,1], nums2 = [7,5,10,9,6], k = 1

Output:

30

Explanation:

Choosing index 2 is optimal: nums1[2] * nums2[2] = 3 * 10 = 30 is the maximum possible score.

Constraints

• n == nums1.length == nums2.length
• 1 <= n <= 105
• 0 <= nums1[i], nums2[j] <= 105
• 1 <= k <= n

Approach and Intuition

To understand the optimal approach for solving this problem, consider the fact that maximizing the score depends on two major components:

1. Maximizing the sum of selected elements from nums1.
2. Maximizing the minimum value from the corresponding indices in nums2.

Here's a strategy we can consider based on example observations:

1. Basic Strategy and Insight from Examples:

   • From the examples, it can be observed that to maximize the score, it is often beneficial to select a combination of indices where the minimum value in nums2 (among the selected indices) is relatively high, as well as ensuring the sum of corresponding indices in nums1 is also high.

2. Sorting for Potential Optimal Solutions:

   • One promising approach could be to sort the indices based on values in nums2 in ascending order and then pick the first k values of indices sorted this way to maximize the minimum in nums2.
   • Similarly, from the selected indices, we sum the k elements in nums1 to maximize the product.

3. Using a Priority Queue:

   • Maintain a priority queue to help manage the largest sums dynamically as we consider each subset of indices based on their nums2 values.
   • By pushing and popping elements in and out of the queue based on nums1 values, we can manage the largest sum efficiently.

4. Iterative Combination Generation:

   • Iteratively generate combinations of indices, calculate their scores, and track the maximum.

5. Optimization Thoughts:

   • Since directly generating all combinations for large n can be computationally intensive (since it involves combinations of n select k), focus on heuristic or greedy approaches that make logical sense based on the constraints of maximizing the core components in the score formula.

By considering these approaches, especially focusing on the interplay between maximizing the sum from nums1 and maximizing the corresponding minimum from nums2 based on sorting or the use of a priority queue, you can develop a strategy that efficiently finds an optimal or near-optimal solution.

Solutions

class Solution {
    public long maximumScore(int[] array1, int[] array2, int limit) {
        int length = array1.length;
        int[][] sortedPairs = new int[length][2];
        for (int i = 0; i < length; ++i) {
            sortedPairs[i] = new int[]{array1[i], array2[i]};
        }
        Arrays.sort(sortedPairs, (x, y) -> y[1] - x[1]);
            
        PriorityQueue<Integer> minHeap = new PriorityQueue<>(limit, (a, b) -> a - b);
        long maxSum = 0;
        for (int i = 0; i < limit; ++i) {
            maxSum += sortedPairs[i][0];
            minHeap.add(sortedPairs[i][0]);
        }
            
        long result = maxSum * sortedPairs[limit - 1][1];
            
        for (int i = limit; i < length; ++i) {
            maxSum += sortedPairs[i][0] - minHeap.poll();
            minHeap.add(sortedPairs[i][0]);
            result = Math.max(result, maxSum * sortedPairs[i][1]);
        }
            
        return result;
    }
}

class Solution:
    def maxScoreCalculation(self, list1: List[int], list2: List[int], count: int) -> int:
        # Organize pairs by second element of each pair in descending order.
        combined = [(elem1, elem2) for elem1, elem2 in zip(list1, list2)]
        combined.sort(key=lambda element: -element[1])
            
        # Utilize a min heap for the largest 'count' elements based on list1's values.
        heap = [v[0] for v in combined[:count]]
        heap_sum = sum(heap)
        heapq.heapify(heap)
            
        # Initial calculation for maximum score using the smallest of kth elements from list2.
        max_result = heap_sum * combined[count - 1][1]
            
        # Review each value from list2 considering it as the minimum in the product score.
        for index in range(count, len(list1)):
            # Update the heap by replacing the smallest element and adding the current element from list1.
            heap_sum -= heapq.heappop(heap)
            heap_sum += combined[index][0]
            heapq.heappush(heap, combined[index][0])
                
            # Recalculate the potential max result.
            max_result = max(max_result, heap_sum * combined[index][1])
            
        return max_result