How do you solve Find the Student that Will Replace the Chalk on LeetCode?

Find the Student that Will Replace the Chalk (LeetCode #1894) 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: Understanding the total chalk usage helps in reducing unnecessary iterations.

What is the time complexity of Find the Student that Will Replace the Chalk?

The optimal solution for Find the Student that Will Replace the Chalk runs in O(n) time and uses O(1) space. The time complexity is O(n) because we only need to sum the chalk array once and then iterate through it again to find the student, making it efficient.

What is the brute force solution for Find the Student that Will Replace the Chalk?

The brute force approach for Find the Student that Will Replace the Chalk is Brute Force, with O(n²) time complexity and O(1) space complexity. We can simulate the process of students using chalk until we run out. This approach is straightforward but inefficient, as it may require multiple passes through the list of students. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Find the Student that Will Replace the Chalk?

Find the Student that Will Replace the Chalk uses the following concepts: Array, Binary Search, Simulation, Prefix Sum. The recommended patterns to study are: Prefix Sum, Simulation.

What are the interview tips for Find the Student that Will Replace the Chalk?

Always consider edge cases, such as when k is much larger than the total chalk. Explain your thought process clearly, especially when transitioning from brute force to optimal solutions. Practice simulating the problem with small inputs to build intuition before coding.

What are common mistakes when solving Find the Student that Will Replace the Chalk?

Not considering the case where k is larger than the total chalk, leading to incorrect results. Failing to reset the index properly when looping through the students.

#1894

Find the Student that Will Replace the Chalk

Medium

ArrayBinary SearchSimulationPrefix SumPrefix SumSimulation

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 simulating each student's turn, we can calculate the total chalk needed for one full round of students and then determine how many complete rounds can be done with the available chalk. This reduces unnecessary iterations.

⚙️

Algorithm

3 steps

1Step 1: Calculate the total chalk needed for one complete round by summing all elements in the chalk array.
2Step 2: If `k` is greater than or equal to the total chalk, reduce `k` by the total chalk until `k` is less than the total chalk.
3Step 3: Iterate through the chalk array to find the first student where `k` is less than chalk[i] and return that index.

solution.py9 lines

1chalk = [5, 1, 5]
2k = 22
3total_chalk = sum(chalk)
4if k >= total_chalk:
5    k %= total_chalk
6for i in range(len(chalk)):
7    if k < chalk[i]:
8        print(i)
9        break

ℹ

Complexity note: The time complexity is O(n) because we only need to sum the chalk array once and then iterate through it again to find the student, making it efficient.

1Understanding the total chalk usage helps in reducing unnecessary iterations.
2Using modulo operation allows us to efficiently determine remaining chalk after complete rounds.

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