You are given two 0-indexed integer arrays nums1 and nums2 of equal length n and a positive integer k. You must choose a subsequence of indices from nums1 of length k.
For chosen indices i0, i1, ..., ik - 1, your score is defined as:
nums1 multiplied with the minimum of the selected elements from nums2.(nums1[i0] + nums1[i1] +...+ nums1[ik - 1]) * min(nums2[i0] , nums2[i1], ... ,nums2[ik - 1]).Return the maximum possible score.
A subsequence of indices of an array is a set that can be derived from the set {0, 1, ..., n-1} by deleting some or no elements.
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.length1 <= n <= 1050 <= nums1[i], nums2[j] <= 1051 <= k <= nIn this approach, the idea is to pair up elements of nums1 and nums2 into tuples. Then, sort these pairs by nums2 in descending order. This way, we process elements with higher values in nums2 first. Using a priority queue (min-heap), we can efficiently keep track of the k largest nums1 values, which we use to calculate the score. The maximum score is computed by iterating through potential minimums, maintaining the current sum of the largest k values seen so far.
The solution sorts nums2 together with nums1. It processes from the largest nums2 value, maintaining a sum and a priority queue to track the largest k values of nums1. The score is calculated as the product of this sum and the current nums2 value in the loop, updating the maximum score as needed.
C++
Time Complexity: O(n log n) due to sorting and heap operations.
Space Complexity: O(k) for the min-heap storing k elements.
This approach emphasizes efficient subsequence selection by partitioning the array. Divide the array in such a way that potential results can be directly computed without redundant calculations. Use recursion to evaluate possible subsequences efficiently.
The Java implementation uses List to manage pairs and utilizes the built-in sorting method for ordering. It implements similar logic using priority queues to track k largest elements for scoring.
JavaScript
Time Complexity: O(n log n) primary by sorting and heap operations.
Space Complexity: O(k) for the priority queue.
| Approach | Complexity |
|---|---|
| Sorting with Priority Queue | Time Complexity: O(n log n) due to sorting and heap operations. |
| Divide and Conquer (Optimal Subsequence Selection) | Time Complexity: O(n log n) primary by sorting and heap operations. |
Maximum Subsequence Score - Leetcode 2542 - Python • NeetCodeIO • 27,505 views views
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