How do you solve Design Exam Scores Tracker on LeetCode?

Design Exam Scores Tracker (LeetCode #3709) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using prefix sums allows for efficient range queries.

What is the time complexity of Design Exam Scores Tracker?

The optimal solution for Design Exam Scores Tracker runs in O(n) time and uses O(n) space. The prefix sum allows us to compute the total in constant time after linear time setup.

What is the brute force solution for Design Exam Scores Tracker?

The brute force approach for Design Exam Scores Tracker is Brute Force, with O(n²) time complexity and O(1) space complexity. This approach involves iterating through all recorded scores to sum those within the specified time range. It's straightforward but inefficient for large datasets. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Design Exam Scores Tracker?

Design Exam Scores Tracker uses the following concepts: Array, Binary Search, Design, Prefix Sum. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Design Exam Scores Tracker?

Explain your thought process clearly. Discuss edge cases, like no records. Practice coding without an IDE to simulate interview conditions.

What are common mistakes when solving Design Exam Scores Tracker?

Forgetting to update the prefix sum correctly. Misusing binary search boundaries.

#3709

Design Exam Scores Tracker

Medium

ArrayBinary SearchDesignPrefix SumHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

By maintaining a prefix sum array, we can quickly calculate the total score for any time range using binary search to find the relevant indices.

⚙️

Algorithm

3 steps

1Step 1: Store times and scores in separate lists.
2Step 2: Maintain a prefix sum array where each index holds the cumulative score up to that index.
3Step 3: Use binary search to find the indices corresponding to startTime and endTime, then return the difference of prefix sums.

solution.py15 lines

1class ExamTracker:
2    def __init__(self):
3        self.times = []
4        self.scores = []
5        self.prefix = []
6
7    def record(self, time, score):
8        self.times.append(time)
9        self.scores.append(score)
10        self.prefix.append((self.prefix[-1] if self.prefix else 0) + score)
11
12    def totalScore(self, startTime, endTime):
13        l = bisect.bisect_left(self.times, startTime)
14        r = bisect.bisect_right(self.times, endTime) - 1
15        return self.prefix[r] - (self.prefix[l-1] if l > 0 else 0)

ℹ

Complexity note: The prefix sum allows us to compute the total in constant time after linear time setup.

1Using prefix sums allows for efficient range queries.
2Binary search optimizes the search for time indices.

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