How do you solve Maximum Points You Can Obtain from Cards on LeetCode?

Maximum Points You Can Obtain from Cards (LeetCode #1423) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: The problem can be transformed into finding the minimum sum of a sub-array that we don't take.

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

The optimal solution for Maximum Points You Can Obtain from Cards runs in O(n) time and uses O(1) space. The time complexity is O(n) because we traverse the array a constant number of times, making it much more efficient than the brute-force approach.

What is the brute force solution for Maximum Points You Can Obtain from Cards?

The brute force approach for Maximum Points You Can Obtain from Cards is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute-force approach, we explore all possible combinations of taking k cards from either end of the array. This means we will check every possible selection of k cards, which can be quite slow. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Maximum Points You Can Obtain from Cards?

Maximum Points You Can Obtain from Cards uses the following concepts: Array, Sliding Window, Prefix Sum. The recommended patterns to study are: Sliding Window, Prefix Sum.

What are the interview tips for Maximum Points You Can Obtain from Cards?

Clearly explain your thought process and why you choose a particular approach. Consider edge cases and test your solution against them. Practice explaining your code as you write it, as this is often part of the interview.

What are common mistakes when solving Maximum Points You Can Obtain from Cards?

Not considering edge cases where k equals the length of the array. Failing to optimize the approach and sticking to brute-force due to lack of understanding.

#1423

Maximum Points You Can Obtain from Cards

Medium

ArraySliding WindowPrefix SumSliding WindowPrefix Sum

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(1)

Instead of checking all combinations, we can use the concept of removing a sub-array from the total points. By calculating the total points and subtracting the minimum points of the sub-array we don't take, we can efficiently find the maximum score.

⚙️

Algorithm

4 steps

1Step 1: Calculate the total points from all cards.
2Step 2: Identify the size of the sub-array to remove, which is n - k.
3Step 3: Use a sliding window to find the minimum sum of this sub-array.
4Step 4: Subtract this minimum sum from the total points to get the maximum score.

solution.py11 lines

1# Full working Python code
2
3def maxScore(cardPoints, k):
4    n = len(cardPoints)
5    total_pts = sum(cardPoints)
6    min_subarray_sum = sum(cardPoints[:n - k])
7    current_sum = min_subarray_sum
8    for i in range(n - k, n):
9        current_sum += cardPoints[i] - cardPoints[i - (n - k)]
10        min_subarray_sum = min(min_subarray_sum, current_sum)
11    return total_pts - min_subarray_sum

ℹ

Complexity note: The time complexity is O(n) because we traverse the array a constant number of times, making it much more efficient than the brute-force approach.

1The problem can be transformed into finding the minimum sum of a sub-array that we don't take.
2Using a sliding window approach allows us to efficiently calculate the minimum sum.

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