What is the time complexity of Maximum Points You Can Obtain from Cards?

The optimal solution has a time complexity of O(k).

What data structures are used to solve Maximum Points You Can Obtain from Cards?

This problem uses Array, Sliding Window, Prefix Sum. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Maximum Points You Can Obtain from Cards?

Insight 1: Instead of trying all combinations of taking cards from the beginning and end, recognize that you're essentially choosing a window of `n - k` cards *not* to take from the middle of the array. Insight 2: The problem can be solved using a sliding window technique. Calculate the sum of `k` cards from the right, then slide the window by adding one card from the left and removing one from the right, updating the maximum score in each step.

Which companies ask LeetCode 1423. Maximum Points You Can Obtain from Cards in interviews?

This problem is frequently asked at DE Shaw, Flipkart.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

Maximum Points You Can Obtain from Cards - LeetCode 1423 Solution

Maximum Points You Can Obtain from Cards - Complete Solution Guide

Maximum Points You Can Obtain from Cards is LeetCode problem 1423, 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

There are several cards arranged in a row , and each card has an associated number of points. The points are given in the integer array cardPoints . In one step, you can take one card from the beginning or from the end of the row. You have to take exactly k cards. Your score is the sum of the points of the cards you have taken. Given the integer array cardPoints and the integer k , return the maximum score you can obtain. Example 1: Input: cardPoints = [1,2,3,4,5,6,1], k = 3 Output: 12 Explanati

Detailed Explanation

The problem presents an array `cardPoints` representing points associated with cards arranged in a row. You need to take exactly `k` cards, either from the beginning or the end of the row. The goal is to maximize the total score, which is the sum of the points of the cards taken. The input includes the `cardPoints` array and the integer `k`, and the output is the maximum achievable score.

Solution Approach

The solution uses a sliding window approach to efficiently find the maximum score. It initializes the `current_score` with the sum of the last `k` cards in the array, representing taking all `k` cards from the right. Then, it iterates `k` times, simulating moving one card from the right to the left. In each iteration, it adds the value of the next card from the left and subtracts the value of the card being removed from the right, updating the `current_score`. The `max_score` is updated in each iteration with the maximum `current_score` encountered so far. This avoids recalculating the sum from scratch for each possible combination and maintains a running sum.

Step-by-Step Algorithm

Step 1: Calculate the initial score by summing the last `k` elements of the `cardPoints` array. This represents taking all `k` cards from the right side.
Step 2: Initialize `max_score` to the `current_score` calculated in Step 1.
Step 3: Iterate `k` times, where `i` ranges from 0 to `k-1`.
Step 4: In each iteration, update `current_score` by adding the value of the `i`-th card (from the left) and subtracting the value of the card that is being removed from the right side (at index `n - k + i`).
Step 5: Update `max_score` to be the maximum of `max_score` and `current_score`.
Step 6: After the loop completes, return the `max_score`.

Key Insights

Insight 1: Instead of trying all combinations of taking cards from the beginning and end, recognize that you're essentially choosing a window of `n - k` cards *not* to take from the middle of the array.
Insight 2: The problem can be solved using a sliding window technique. Calculate the sum of `k` cards from the right, then slide the window by adding one card from the left and removing one from the right, updating the maximum score in each step.
Insight 3: The initial sum can be conveniently calculated by accumulating the values of the last k elements, rather than recalculating the total sum for each possible k-element combination.

Complexity Analysis

Time Complexity: O(k)

Space Complexity: O(1)

Topics

This problem involves: Array, Sliding Window, Prefix Sum.

Companies

Asked at: DE Shaw, Flipkart.

Maximum Points You Can Obtain from Cards | LeetCode 1423

Medium

Array Sliding Window Prefix Sum

Time: O(k)Space: O(1)

Solution Code

Available in:

class Solution {
    public int maxScore(int[] cardPoints, int k) {
        int n = cardPoints.length;
        
        // We start by considering the score from taking k cards from the right.
        // This is one of the k+1 possibilities.
        int current_score = 0;
        for (int i = n - k; i < n; i++) {
            current_score += cardPoints[i];
        }
        int max_score = current_score;
        
        // Now, we iterate k times. In each iteration, we simulate taking one more card
        // from the left and one less card from the right.
        for (int i = 0; i < k; i++) {
            // The new score is the previous score plus the new card from the left,
            // minus the card that is no longer part of our selection from the right.
            // The card being added is cardPoints[i].
            // The card being removed is cardPoints[n - k + i].
            current_score += cardPoints[i] - cardPoints[n - k + i];
            max_score = Math.max(max_score, current_score);
            
        }
        
        return max_score;
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 1423: Maximum Points You Can Obtain from Cards - Detailed explanation and solution approaches.

Topics

Array Sliding Window Prefix Sum

Companies

DE ShawFlipkart

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:1423

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

6 steps

1
Step 1: Calculate the initial score by summing the last `k` elements of the `cardPoints` array. This represents taking all `k` cards from the right side.
2
Step 2: Initialize `max_score` to the `current_score` calculated in Step 1.
3
Step 3: Iterate `k` times, where `i` ranges from 0 to `k-1`.
4
Step 4: In each iteration, update `current_score` by adding the value of the `i`-th card (from the left) and subtracting the value of the card that is being removed from the right side (at index `n - k + i`).
5
Step 5: Update `max_score` to be the maximum of `max_score` and `current_score`.
6
Step 6: After the loop completes, 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(k)

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The algorithm iterates `k` times, where `k` is the number of cards to take. Inside the loop, the operations are constant time additions and subtractions. Hence, the time complexity is O(k). The algorithm uses a few integer variables to store the current score and maximum score. Therefore, the space complexity is O(1), constant space.

📖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

Prefix Sum

Calculate the prefix sum of the `cardPoints` array. Then, iterate through all possible combinations of taking `i` cards from the left and `k-i` cards from the right, using the prefix sum to calculate the score for each combination. The maximum score can be found by considering all i values from 0 to k.

Time: O(k)Space: O(n)

Trade-offs: The prefix sum approach requires extra space O(n) to store the prefix sum array, while the sliding window approach uses constant O(1) space. However, both has O(k) time complexity. Sliding window performs better on space complexity.

✨

Key Insights

1Insight 1: Instead of trying all combinations of taking cards from the beginning and end, recognize that you're essentially choosing a window of `n - k` cards *not* to take from the middle of the array.
2Insight 2: The problem can be solved using a sliding window technique. Calculate the sum of `k` cards from the right, then slide the window by adding one card from the left and removing one from the right, updating the maximum score in each step.
3Insight 3: The initial sum can be conveniently calculated by accumulating the values of the last k elements, rather than recalculating the total sum for each possible k-element combination.

📚

Key Terms Explained

For Students

Array

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

Sliding Window

A moving "window" over data to process subsets efficiently.

Edge Cases

Edge case 1: `k` is equal to the length of `cardPoints`. In this case, you must take all cards, so the solution correctly calculates the sum of all elements.
Edge case 2: `cardPoints` has only one element. The loop will not execute and the intialized current_score will be returned as max_score which is correct

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Trying to explore all combinations of cards taken from both ends recursively. This leads to exponential time complexity and TLE (Time Limit Exceeded).

✗Common mistake 2: Incorrectly calculating the indices for adding cards from the left and removing cards from the right, leading to incorrect sum calculations.

✅

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: Optimizing resource allocation. Imagine you have a series of tasks with associated values, and you need to select a fixed number of tasks from either the beginning or end of the sequence to maximize the total value. This problem can be modeled as selecting the best cards in this LeetCode problem.

🏦Real-world scenario 2: Stock trading. You want to maximize profit by buying or selling a limited number of stocks either at the start or the end of a period based on their initial and final values. You need to select which stocks to buy or sell in the given time frame to maximize overall gain

💡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: November 11, 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

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

@CodeNinjaJul 13

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 6

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

What data structures complement sliding window?

Hash maps track character frequencies within the window. Deques enable O(1) access to both window boundaries. Heaps help find min/max within the window. Choose based on what property you need to maintain as the window slides.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Maximum Points You Can Obtain from Cards"?

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 sliding window techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?