What data structures are used to solve Maximum Subsequence Score?

This problem uses Array, Greedy, Sorting, Heap (Priority Queue). Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Maximum Subsequence Score?

Insight 1: Sorting the pairs (nums1[i], nums2[i]) based on nums2[i] in descending order allows us to iterate through potential minimum values for nums2 and greedily select the largest nums1 values. Insight 2: Using a min-heap of size k to track the k largest nums1 values encountered so far ensures that at any point, we have the k largest nums1 elements for a given minimum value (nums2[i]).

Which companies ask LeetCode 2542. Maximum Subsequence Score in interviews?

This problem is frequently asked at DE Shaw.

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Maximum Subsequence Score - LeetCode 2542 Solution

Q: What is the time complexity of Maximum Subsequence Score?

The optimal solution has a time complexity of O(n log n), where n is the length of the arrays. Sorting the pairs takes O(n log n) time. The iteration through the pairs takes O(n) time, and each heap operation takes O(log k) time. However, since we are performing heap operation at most n times, the time complexity contributed by the heap operations is O(n log k). Since k <= n, O(n log k) <= O(n log n). Therefore, the dominant factor is the sorting step making the total time complexity O(n log n)..

Q: What is the space complexity of Maximum Subsequence Score?

The space complexity is O(n), where n is the length of the arrays. The pairs array takes O(n) space. The min-heap takes O(k) space, and since k <= n, the heap space is bounded by O(n)..

Maximum Subsequence Score - Complete Solution Guide

Maximum Subsequence Score is LeetCode problem 2542, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

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 i 0 , i 1 , ..., 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[i 0 ] + nums1[i 1 ] +...+ nums1[i k - 1 ]) * min(nums2[i 0 ] , nums2[i 1 ], ... ,nums2[i k - 1 ]) . Return the maxi

Detailed Explanation

The problem asks us to find the maximum possible score achievable by selecting a subsequence of length `k` from `nums1`, where the score is defined as the sum of the selected elements from `nums1` multiplied by the minimum of the selected elements from `nums2`. We are given two arrays, `nums1` and `nums2`, of equal length `n`, and an integer `k`. We need to select `k` indices and calculate the score for each possible subsequence and return the maximum score among them.

Solution Approach

The solution uses a greedy approach combined with sorting and a min-heap. First, we create pairs of (nums1[i], nums2[i]) and sort them in descending order based on the values in `nums2`. This allows us to consider potential minimums of `nums2` in decreasing order. Then, we iterate through the sorted pairs, maintaining a min-heap of size `k` to store the `nums1` values encountered so far. The `current_sum` variable stores the sum of elements in the min-heap. At each step, the algorithm updates the `max_score` by considering the current element's `nums2` value as the minimum and `current_sum` as the sum of corresponding `nums1` values of the k largest `nums1` values so far.

Step-by-Step Algorithm

Step 1: Create an array of pairs (nums1[i], nums2[i]).
Step 2: Sort the pairs in descending order based on the nums2 values.
Step 3: Initialize a min-heap, current_sum, and max_score to 0.
Step 4: Iterate through the sorted pairs:
Step 5: Add the current nums1 value to the min-heap and update the current_sum.
Step 6: If the min-heap size exceeds k, remove the smallest element from the heap and subtract it from current_sum.
Step 7: If the min-heap size is equal to k, calculate the current score (current_sum * current nums2 value) and update max_score if the current score is greater.
Step 8: Return the max_score.

Key Insights

Insight 1: Sorting the pairs (nums1[i], nums2[i]) based on nums2[i] in descending order allows us to iterate through potential minimum values for nums2 and greedily select the largest nums1 values.
Insight 2: Using a min-heap of size k to track the k largest nums1 values encountered so far ensures that at any point, we have the k largest nums1 elements for a given minimum value (nums2[i]).
Insight 3: The core idea is that we iterate through the `nums2` values from largest to smallest. Each value we encounter is a potential minimum for our subsequence. We then want to maximize the sum of `nums1` values included in the subsequence, given that `nums2[i]` is the minimum. Maintaining a min-heap of the largest `k` `nums1` values allows us to efficiently update the sum.

Complexity Analysis

Time Complexity: O(n log n), where n is the length of the arrays. Sorting the pairs takes O(n log n) time. The iteration through the pairs takes O(n) time, and each heap operation takes O(log k) time. However, since we are performing heap operation at most n times, the time complexity contributed by the heap operations is O(n log k). Since k <= n, O(n log k) <= O(n log n). Therefore, the dominant factor is the sorting step making the total time complexity O(n log n).

Space Complexity: O(n), where n is the length of the arrays. The pairs array takes O(n) space. The min-heap takes O(k) space, and since k <= n, the heap space is bounded by O(n).

Topics

This problem involves: Array, Greedy, Sorting, Heap (Priority Queue).

Companies

Asked at: DE Shaw.

Maximum Subsequence Score | LeetCode 2542

Medium

Array Greedy Sorting Heap (Priority Queue)

Time: O(n log n), where n is the length of the arrays. Sorting the pairs takes O(n log n) time. The iteration through the pairs takes O(n) time, and each heap operation takes O(log k) time. However, since we are performing heap operation at most n times, the time complexity contributed by the heap operations is O(n log k). Since k <= n, O(n log k) <= O(n log n). Therefore, the dominant factor is the sorting step making the total time complexity O(n log n).Space: O(n), where n is the length of the arrays. The pairs array takes O(n) space. The min-heap takes O(k) space, and since k <= n, the heap space is bounded by O(n).

Solution Code

Available in:

import java.util.Arrays;
import java.util.PriorityQueue;

class Solution {
    public long maxScore(int[] nums1, int[] nums2, int k) {
        int n = nums1.length;
        int[][] pairs = new int[n][2];
        for (int i = 0; i < n; i++) {
            pairs[i][0] = nums1[i];
            pairs[i][1] = nums2[i];
        }

        Arrays.sort(pairs, (a, b) -> b[1] - a[1]);

        PriorityQueue<Integer> minHeap = new PriorityQueue<>();
        long currentSum = 0;
        long maxScore = 0;

        for (int[] pair : pairs) {
            int n1 = pair[0];
            int n2 = pair[1];

            minHeap.offer(n1);
            currentSum += n1;

            if (minHeap.size() > k) {
                currentSum -= minHeap.poll();
            }

            if (minHeap.size() == k) {
                maxScore = Math.max(maxScore, currentSum * n2);
            }
        }

        return maxScore;
    }
}

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Problem Info

Description

Problem 2542: Maximum Subsequence Score - Detailed explanation and solution approaches.

Topics

Array Greedy Sorting Heap (Priority Queue)

Companies

DE Shaw

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2542

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

8 steps

1
Step 1: Create an array of pairs (nums1[i], nums2[i]).
2
Step 2: Sort the pairs in descending order based on the nums2 values.
3
Step 3: Initialize a min-heap, current_sum, and max_score to 0.
4
Step 4: Iterate through the sorted pairs:
5
Step 5: Add the current nums1 value to the min-heap and update the current_sum.
6
Step 6: If the min-heap size exceeds k, remove the smallest element from the heap and subtract it from current_sum.
7
Step 7: If the min-heap size is equal to k, calculate the current score (current_sum * current nums2 value) and update max_score if the current score is greater.
8
Step 8: Return the max_score.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n log n), where n is the length of the arrays. Sorting the pairs takes O(n log n) time. The iteration through the pairs takes O(n) time, and each heap operation takes O(log k) time. However, since we are performing heap operation at most n times, the time complexity contributed by the heap operations is O(n log k). Since k <= n, O(n log k) <= O(n log n). Therefore, the dominant factor is the sorting step making the total time complexity O(n log n).

🚀 Logarithmic - Very efficient! Halves the problem each step.✨ Log-linear - Great for sorting algorithms!

💾

Space Complexity

O(n), where n is the length of the arrays. The pairs array takes O(n) space. The min-heap takes O(k) space, and since k <= n, the heap space is bounded by O(n).

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is dominated by sorting the pairs which takes O(n log n) time. The space complexity is determined by the storage of pairs and the min-heap, both of which are bounded by O(n).

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Brute Force

Generate all possible subsequences of size k, compute the score for each, and return the maximum score. This is highly inefficient.

Time: O(C(n, k) * k), where C(n, k) is the binomial coefficient (n choose k). This simplifies to O(n!/(k!(n-k)!) * k).Space: O(k)

Trade-offs: The brute force approach is simple to understand but computationally expensive and impractical for larger input sizes. The provided solution using sorting and a heap is significantly more efficient.

✨

Key Insights

1Insight 1: Sorting the pairs (nums1[i], nums2[i]) based on nums2[i] in descending order allows us to iterate through potential minimum values for nums2 and greedily select the largest nums1 values.
2Insight 2: Using a min-heap of size k to track the k largest nums1 values encountered so far ensures that at any point, we have the k largest nums1 elements for a given minimum value (nums2[i]).
3Insight 3: The core idea is that we iterate through the `nums2` values from largest to smallest. Each value we encounter is a potential minimum for our subsequence. We then want to maximize the sum of `nums1` values included in the subsequence, given that `nums2[i]` is the minimum. Maintaining a min-heap of the largest `k` `nums1` values allows us to efficiently update the sum.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Greedy Algorithm

Making the best choice at each step without looking ahead.

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: When k = 1, the algorithm will select the pair with the highest product of nums1[i] * nums2[i]. The solution handles this correctly because it iterates through all pairs, maintaining the maximum score.
Edge case 2: When all values in `nums1` or `nums2` are zero, the algorithm will still correctly return 0 because the multiplication will always result in 0.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not sorting the pairs based on nums2 leads to an incorrect greedy selection because the algorithm doesn't prioritize potential minimum values.

✗Common mistake 2: Not using a min-heap (or similar data structure) to track the k largest elements efficiently makes it difficult to maintain the k largest nums1 values.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Resource allocation: Suppose `nums1` represents the value of different projects, and `nums2` represents the risk associated with each project. The problem can be seen as selecting `k` projects to maximize the total value, considering that we want to minimize the risk (represented by the minimum risk among chosen projects).

🏦Real-world scenario 2: Portfolio optimization: Imagine `nums1` as potential return on investment for some asset, and `nums2` as associated volatility. We would want to maximize the sum of returns while being aware of the lowest volatility amongst our selected assets. The goal here is to select a set of `k` assets to maximize gains while minimizing risk exposure.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: June 23, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 13

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I verify if a greedy approach is correct?

Prove the greedy choice property: the locally optimal choice leads to globally optimal solution. Use exchange argument: show that swapping any element with the greedy choice doesn't improve the solution. If you can't prove it, test extensively or consider DP.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Maximum Subsequence Score"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether greedy techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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