Find The Student That Will Replace The Chalk - Leetcode Solution

Problem Statement: You are given an array of strings, where each element represents the name of a student. In addition to the array, you are given an integer k representing the index of the student that replaces the chalk on the board. When a student finishes writing on the board, the teacher will erase the board, and the student that replaces them will write on the board next. The process continues until every student has had a chance to write on the board. Return the name of the student that will replace the chalk once the process is complete.

Example: Input: students = ["Alice","Bob","Charlie","David","Eva","Frank"], k = 8 Output: "David" Explanation: The teacher starts by writing on the board with Alice. After Alice finishes, Bob replaces her. Then Charlie replaces Bob. Then David replaces Charlie. Then Eva replaces David. Then Frank replaces Eva. Then Alice replaces Frank. Then Bob replaces Alice. So David will replace Bob, thus the output is "David".

Solution: To find out the student that will replace the chalk, we need to simulate the given process. We will start with the first student and iterate until we reach the student who replaces the chalk. After that, we will start again from the beginning and continue until we have gone through all the students. We can use modular arithmetic to make sure that we always loop around to the beginning when we reach the end of the student list.

The first step is to find the total number of turns that all the students will get to write on the board. We can do this by calculating the length of the student array times the number of times a student can write on the board before they get replaced. This is because once a student writes on the board, they get replaced by the next student and they will not get another chance until all the other students have had their turns.

The next step is to iterate through the students until we reach the student that replaces the chalk. We can do this by keeping track of the total number of turns that have been completed so far. If the total number of turns is less than k, we subtract it from k to get the index of the student that should replace the chalk. Otherwise, we keep track of the current student and subtract k from the total number of turns to get the number of turns remaining.

Once we have found the student that replaces the chalk, we start again from the beginning and iterate until we have gone through all the students. We use modular arithmetic to make sure that we always loop around to the beginning when we reach the end of the student list. At each iteration, we decrement the number of turns remaining until we reach zero. The current student at that point is the one that will replace the chalk.

Code:

def findStudent(students, k): total_turns = len(students) * 2 #total turns is twice the number of students since each student gets two turns current_student = 0 #initialize the current student to be the first one turns_remaining = total_turns - k #calculate the number of turns remaining after the student that replaces the chalk while turns_remaining > 0:
if turns_remaining <= len(students): #if the remaining turns are less than the total number of students current_student = (current_student + turns_remaining - 1) % len(students) #loop over the remaining students using modular arithmetic break else:
turns_remaining -= len(students) #decrement the turns remaining by the number of students return students[current_student]

Time Complexity: The time complexity of this algorithm is O(n), where n is the length of the student array. This is because we need to iterate through the student array twice - once to find the student that replaces the chalk and once again to find the student that will replace the chalk after all the students have had their turns. Since we iterate through the array exactly twice, the time complexity is linear in the length of the array.

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