How do you solve Maximum Subsequence Score on LeetCode?

Maximum Subsequence Score (LeetCode #2542) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Choosing the right indices is crucial for maximizing the score.

What is the time complexity of Maximum Subsequence Score?

The optimal solution for Maximum Subsequence Score runs in O(n log n) time and uses O(n) space. The time complexity is O(n log n) due to the sorting step, while the space complexity is O(n) for storing the paired elements and the min-heap.

What is the brute force solution for Maximum Subsequence Score?

The brute force approach for Maximum Subsequence Score is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible combinations of indices of length k from nums1, calculating the score for each combination, and returning the maximum score found. This method ensures we explore all possibilities, but it's inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Maximum Subsequence Score?

Maximum Subsequence Score uses the following concepts: Array, Greedy, Sorting, Heap (Priority Queue). The recommended patterns to study are: Greedy, Sorting, Heap (Priority Queue).

What are the interview tips for Maximum Subsequence Score?

Explain your thought process clearly when discussing your approach. Consider edge cases, such as when k equals 1 or n. Practice writing clean and efficient code, as interviewers often look for both correctness and efficiency.

What are common mistakes when solving Maximum Subsequence Score?

Not considering the impact of the minimum value on the score. Overlooking the need to manage the size of the selected elements.

#2542

Maximum Subsequence Score

Medium

ArrayGreedySortingHeap (Priority Queue)GreedySortingHeap (Priority Queue)

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(n)

💡

Intuition

Time O(n log n)Space O(n)

The optimal solution leverages sorting and a min-heap to efficiently find the maximum score. By sorting based on nums2, we can ensure that we always consider the largest possible minimum value while maintaining the sum of the largest k elements from nums1.

⚙️

Algorithm

4 steps

1Step 1: Pair elements of nums1 and nums2 and sort them based on nums2 in descending order.
2Step 2: Use a min-heap to keep track of the largest k elements from nums1 as we iterate through the sorted pairs.
3Step 3: Calculate the score using the sum of elements in the heap multiplied by the current nums2 value (which is the minimum for the current k elements).
4Step 4: Update the maximum score encountered during the iteration.

solution.py18 lines

1# Full working Python code
2import heapq
3
4def maxSubsequenceScore(nums1, nums2, k):
5    paired = sorted(zip(nums1, nums2), key=lambda x: -x[1])
6    max_score = 0
7    sum_nums1 = 0
8    min_heap = []
9
10    for num1, num2 in paired:
11        heapq.heappush(min_heap, num1)
12        sum_nums1 += num1
13        if len(min_heap) > k:
14            sum_nums1 -= heapq.heappop(min_heap)
15        if len(min_heap) == k:
16            max_score = max(max_score, sum_nums1 * num2)
17
18    return max_score

ℹ

Complexity note: The time complexity is O(n log n) due to the sorting step, while the space complexity is O(n) for storing the paired elements and the min-heap.

1Choosing the right indices is crucial for maximizing the score.
2Sorting based on the second array allows us to efficiently manage the minimum value.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.