Maximum Subsequence Score

Medium

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ArraysGreedy AlgorithmsDynamic ProgrammingStacks

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, ..., i_{k - 1}, your score is defined as:

The sum of the selected elements from nums1 multiplied with the minimum of the selected elements from nums2.
It can defined simply as: (nums1[i₀] + nums1[i₁] +...+ nums1[i_{k - 1}]) * min(nums2[i₀] , nums2[i₁], ... ,nums2[i_{k - 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.length
1 <= n <= 10⁵
0 <= nums1[i], nums2[j] <= 10⁵
1 <= k <= n

Maximum Subsequence Score

Medium

Asked by:

8 views

Topics:

ArraysGreedy AlgorithmsDynamic ProgrammingStacks

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, ..., i_{k - 1}, your score is defined as:

The sum of the selected elements from nums1 multiplied with the minimum of the selected elements from nums2.
It can defined simply as: (nums1[i₀] + nums1[i₁] +...+ nums1[i_{k - 1}]) * min(nums2[i₀] , nums2[i₁], ... ,nums2[i_{k - 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.length
1 <= n <= 10⁵
0 <= nums1[i], nums2[j] <= 10⁵
1 <= k <= n

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the size constraints for the input arrays `nums1` and `nums2`? What is the range of values within these arrays?
Can `nums1` and `nums2` contain negative numbers, zeros, or duplicate values?
If multiple subsequences achieve the maximum score, can I return any one of them, or is there a specific criteria for choosing among them?
Is `k` guaranteed to be a valid value, meaning `1 <= k <= min(len(nums1), len(nums2))`?
What should I return if `nums1` and `nums2` are empty, null, or if `k` is invalid?

Brute Force Solution

Approach

The brute force method for finding the maximum subsequence score is all about trying out every possible combination. We explore all possible selections from two groups of numbers and compute the score for each combination to find the highest one. It's an exhaustive approach where we leave no stone unturned.

Here's how the algorithm would work step-by-step:

Imagine we have two bags of numbers. We need to pick the same number of numbers from each bag and calculate a score based on our choices.
Start by picking one number from each bag. Calculate the score based on these two numbers.
Next, try picking two numbers from each bag. There are many different ways to choose two numbers from each bag, so we try every single one.
Calculate the score for each of these pairs of selections.
Continue this process, picking three numbers, then four, and so on, until we've tried picking as many numbers as there are in the smaller bag.
For each number of selections, make sure to try every single possible combination of numbers from both bags.
Each time you calculate a score, remember to keep track of the highest score you've seen so far.
Once you have considered all possible combinations and their scores, the highest score you've kept track of is the answer.

Code Implementation

def maximum_subsequence_score_brute_force(numbers1, numbers2, k):
    maximum_score_found = 0
    length_of_numbers1 = len(numbers1)
    length_of_numbers2 = len(numbers2)

    # Iterate through all possible subsequence lengths
    for subsequence_length in range(1, min(length_of_numbers1, length_of_numbers2) + 1):

        # Iterate through all combinations of indices for numbers1
        for first_combination_indices in combinations(range(length_of_numbers1), subsequence_length):

            # Iterate through all combinations of indices for numbers2
            for second_combination_indices in combinations(range(length_of_numbers2), subsequence_length):
                
                # Calculate the current score for this combination
                minimum_value = float('inf')
                sum_of_numbers = 0

                for index in range(subsequence_length):
                    sum_of_numbers += numbers1[first_combination_indices[index]]
                    minimum_value = min(minimum_value, numbers2[second_combination_indices[index]])

                current_score = sum_of_numbers * minimum_value

                # Update the maximum score if necessary
                maximum_score_found = max(maximum_score_found, current_score)

    return maximum_score_found

from itertools import combinations

def maximumSubsequenceScore(numbers1, numbers2, k):
    return maximum_subsequence_score_brute_force(numbers1, numbers2, k)

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Big(O) Analysis

Time Complexity

O(n!) – The brute force approach described iterates through all possible subsequences of size k, where k ranges from 1 to the length of the smaller input array. For each k, it explores all combinations of selecting k elements from nums1 and nums2. The number of such combinations grows factorially with n (the length of the input arrays), as we are essentially considering all possible subsets. Thus, the total number of operations is dominated by the generation and evaluation of these combinations, resulting in a time complexity of O(n!).

Space Complexity

O(1) – The brute force approach described exhaustively explores all possible combinations without using any significant auxiliary data structures. It keeps track of the maximum score found so far, which requires constant space. No temporary lists, hash maps, or recursion are involved in the described process. Therefore, the space complexity is O(1).

Optimal Solution

Approach

The goal is to pick numbers from two lists to maximize a score. We can find the best score by focusing on the smallest number in one of the lists and strategically choosing the corresponding largest numbers from the other list.

Here's how the algorithm would work step-by-step:

Pair up numbers from the two lists based on their position. Imagine connecting matching positions between the two lists.
Sort these pairs based on the numbers in the second list from largest to smallest. This means we'll consider the biggest multipliers first.
Start by considering the top `k` pairs (where `k` is a given number). Imagine a smaller group of connections, consisting of only the pairs where the multiplier from list two is in the top `k`.
Within this group of `k` pairs, find the `k` largest numbers from the *first* list. These will be the numbers from the first list we'll use to compute the score.
Calculate the score by multiplying the sum of these `k` largest numbers (from the first list) by the smallest number in the *second* list among the `k` pairs we initially considered.
Repeat this process, each time considering a different set of top `k` pairs (moving from largest to smallest in the second list). Track the highest score seen so far.
The highest score found across all these sets of top `k` pairs is the maximum possible subsequence score.

Code Implementation

import heapq

def maximum_subsequence_score(nums1, nums2, k):
    pairs = sorted(zip(nums1, nums2), key=lambda x: x[1], reverse=True)
    
    max_score = 0
    current_sum = 0
    min_heap = []

    for number1, number2 in pairs:
        heapq.heappush(min_heap, number1)
        current_sum += number1

        if len(min_heap) > k:
            # Maintain only the k largest elements from nums1.
            current_sum -= heapq.heappop(min_heap)

        if len(min_heap) == k:
            # Calculate score with current smallest number2.
            max_score = max(max_score, current_sum * number2)

    return max_score

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Big(O) Analysis

Time Complexity

O(n log n) – The dominant operation is sorting the pairs based on the values in nums2, which takes O(n log n) time, where n is the number of pairs (and the length of the input arrays). The subsequent loop iterates through the sorted pairs, and within the loop, a heap of size k is maintained to find the k largest elements from nums1. Heap operations take O(log k) which is less than O(log n) and the loop is O(n), so the overall contribution of the heap operations is O(n log k) which is <= O(n log n). Therefore, the total time complexity is dominated by the initial sorting step, resulting in O(n log n).

Space Complexity

O(N) – The algorithm pairs up numbers from two lists of size N, creating a list of N pairs. This list is then sorted, requiring O(N) space in many common sorting algorithms like merge sort or timsort. Additionally, a heap of size k (where k <= N) is implicitly maintained to find the k largest elements, contributing O(k) space. Since k is bounded by N, the overall auxiliary space complexity is dominated by the storage of the N pairs, which is O(N).

Edge Cases

Case	How to Handle
Empty nums1 or nums2 arrays	Return 0 immediately as there are no elements to form a subsequence.
k is greater than the length of nums1 or nums2	Return 0 because a subsequence of length k cannot be formed.
Arrays contain negative numbers	The algorithm should correctly handle negative numbers in both nums1 and nums2 since the score calculation involves multiplication which can be negative.
Arrays contain zero values	Zero values in nums2 will lead to a zero score regardless of nums1, and this will be automatically accounted for during max score calculation.
Large array sizes for nums1 and nums2 with small k	Using a min-heap (priority queue) to keep track of the k largest values from nums1 will help maintain O(n log k) complexity.
Integer overflow during score calculation (sum * min)	Use long data type for intermediate calculations of sum and score to prevent integer overflow.
All elements in nums2 are the same and small, and elements in nums1 are very large	The algorithm should still correctly identify the top k elements from nums1 regardless of the values in nums2, allowing it to calculate the maximum subsequence score.
k is 1	The algorithm should find the element in nums1 that corresponds to the largest element in nums2.