
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 innums2
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:
- Maximizing the sum of selected elements from
nums1
. - Maximizing the minimum value from the corresponding indices in
nums2
.
Here's a strategy we can consider based on example observations:
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 innums1
is also high.
- 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
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 firstk
values of indices sorted this way to maximize the minimum innums2
. - Similarly, from the selected indices, we sum the
k
elements innums1
to maximize the product.
- One promising approach could be to sort the indices based on values in
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.
- Maintain a priority queue to help manage the largest sums dynamically as we consider each subset of indices based on their
Iterative Combination Generation:
- Iteratively generate combinations of indices, calculate their scores, and track the maximum.
Optimization Thoughts:
- Since directly generating all combinations for large
n
can be computationally intensive (since it involves combinations ofn
selectk
), focus on heuristic or greedy approaches that make logical sense based on the constraints of maximizing the core components in the score formula.
- Since directly generating all combinations for large
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
- Java
- Python
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;
}
}
The provided Java solution aims to identify the maximum score of a subsequence within two integer arrays, constrained by a specified limit. The problem involves pairing the elements from the two arrays, sorting the pairs based on the second array's values in descending order, and then calculating scores to identify the maximal result under given constraints using efficient data structures and algorithms.
Here’s an explanation of how the solution works:
- Initialize a 2D array
sortedPairs
to store elements fromarray1
andarray2
as pairs. - Sort
sortedPairs
by the second element of each pair in descending order. This prioritization ensures that the most significant multipliers fromarray2
are considered first for higher score potential. - Utilize a
PriorityQueue
(minHeap
) to maintain the smallest elements fromarray1
that form part of the maximum score calculation. This heap helps efficiently find and remove the smallest element when adding a new one fromsortedPairs
. - Calculate the initial potential max score using the top elements up to the specified limit from the sorted pairs.
- Iterate through the rest of the pairs in
sortedPairs
. For each element:- Update the score by removing the smallest element in the heap and adding the current element.
- Calculate the potential result with the updated score and the current multiplier from
array2
. - Update the result to store the maximum score found so far.
- Finally, the maximum score found, as per given constraints, is returned.
This method leverages sorting and a min-heap to efficiently manage and update the elements contributing to the score, ensuring optimal time complexity and performance for large inputs.
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
The solution provided aims to calculate the maximum subsequence score by considering two integer lists, list1
and list2
, along with a given integer count
. The method maxScoreCalculation
achieves this through a series of strategic operations:
- Pairs elements from
list1
andlist2
into tuples, sorts these tuples based on the elements fromlist2
in descending order. - Utilizes a min-heap data structure to maintain the
count
largest elements derived fromlist1
values, creating a sum of these maximal elements termedheap_sum
. - Computes an initial potential maximum score using the smallest value among the top
count
elements oflist2
. - Iterates through elements of
list2
, starting from thecount
index to the end. In each iteration:- Adjusts the heap by replacing the smallest element with the current element from
list1
, ensuring it continuously reflects thecount
largest values. - Recalculates the maximum score using the updated
heap_sum
and the current element fromlist2
.
- Adjusts the heap by replacing the smallest element with the current element from
The process concludes by returning the max_result
, which holds the highest score computed across all iterations. This approach efficiently harnesses sorting and heap operations to ascertain the optimal scoring sequence based on the criteria set by list1
, list2
, and count
.
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