Find the Student that Will Replace the Chalk | 2 Ways | Leetcode 1894 | codestorywithMIK
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This is the 105th Video of our Playlist "Array 1D/2D : Popular Interview Problems" by codestorywithMIK
In this video we will try to solve a simple Simulation Array Problem : Find the Student that Will Replace the Chalk | 2 Ways | Leetcode 1894 | codestorywithMIK
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Problem Name : Find the Student that Will Replace the Chalk | 2 Ways | Leetcode 1894 | codestorywithMIK
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Summary :
Approach 1: Simulation (TLE)
Description: This approach simulates the chalk usage by iterating through the students in a loop and subtracting their chalk requirements from k until k becomes less than a student's chalk requirement. The simulation continues until the condition is met.
Time Complexity: O(k/sum * n) — where sum is the total sum of chalk requirements in one full iteration and n is the number of students. This can result in a Time Limit Exceeded (TLE) error for large inputs, as k might be very large.
Space Complexity: O(1) — The approach only uses a constant amount of extra space.
Approach 2: Remainder Optimization
Description: Instead of simulating the entire process, this approach calculates the total chalk required for one complete round of all students (totalChalkSum). It then computes the remainder of k when divided by totalChalkSum (remainChalk). This allows us to reduce the problem to a smaller equivalent by only considering the remaining chalk. Finally, it finds the first student whose chalk requirement exceeds remainChalk.
Time Complexity: O(n) — The total chalk sum is computed in one pass, and finding the student who runs out of chalk takes another pass.
Space Complexity: O(1) — This approach also only uses a constant amount of extra space.
✨ Timelines✨
00:00 - Introduction
0:17 - Problem Explanation
2:52 - Simulation Approach - TLE
6:52 - Improved Approach - Accepted
13:20 - Coding it up
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