Find the Student that Will Replace the Chalk

1894. Find the Student that Will Replace the Chalk

Difficulty: Medium

Topics: Array, Binary Search, Simulation, Prefix Sum

There are n students in a class numbered from 0 to n - 1. The teacher will give each student a problem starting with the student number 0, then the student number 1, and so on until the teacher reaches the student number n - 1. After that, the teacher will restart the process, starting with the student number 0 again.

You are given a 0-indexed integer array chalk and an integer k. There are initially k pieces of chalk. When the student number i is given a problem to solve, they will use chalk[i] pieces of chalk to solve that problem. However, if the current number of chalk pieces is strictly less than chalk[i], then the student number i will be asked to replace the chalk.

Return the index of the student that will replace the chalk pieces.

Example 1:

• Input: chalk = [5,1,5], k = 22
• Output: 0
• Explanation: The students go in turns as follows:
  • Student number 0 uses 5 chalk, so k = 17.
  • Student number 1 uses 1 chalk, so k = 16.
  • Student number 2 uses 5 chalk, so k = 11.
  • Student number 0 uses 5 chalk, so k = 6.
  • Student number 1 uses 1 chalk, so k = 5.
  • Student number 2 uses 5 chalk, so k = 0.
  • Student number 0 does not have enough chalk, so they will have to replace it.

Example 2:

• Input: chalk = [3,4,1,2], k = 25
• Output: 1
• Explanation: The students go in turns as follows:
  • Student number 0 uses 3 chalk so k = 22.
  • Student number 1 uses 4 chalk so k = 18.
  • Student number 2 uses 1 chalk so k = 17.
  • Student number 3 uses 2 chalk so k = 15.
  • Student number 0 uses 3 chalk so k = 12.
  • Student number 1 uses 4 chalk so k = 8.
  • Student number 2 uses 1 chalk so k = 7.
  • Student number 3 uses 2 chalk so k = 5.
  • Student number 0 uses 3 chalk so k = 2.
  • Student number 1 does not have enough chalk, so they will have to replace it.

Constraints:

• chalk.length == n
• 1 <= n <= 10₅
• 1 <= chalk[i] <= 10₅
• 1 <= k <= 10₉

Hint:

1. Subtract the sum of chalk from k until k is less than the sum of chalk.
2. Now iterate over the array. If chalk[i] is less than k, this is the answer. Otherwise, subtract chalk[i] from k and continue.

Solution:

Let's break down the problem step by step:

Approach:

1. Total Chalk Consumption:
   First, calculate the total amount of chalk needed for one complete round (from student 0 to student n-1). This will help us reduce the value of k by taking into account how many complete rounds can be covered by k pieces of chalk.

2. Reduce k by Modulo:
   If k is larger than the total chalk required for one complete round, we can simplify the problem by taking k % total_chalk. This operation will give us the remaining chalk after as many full rounds as possible, leaving us with a smaller problem to solve.

3. Find the Student Who Runs Out of Chalk:
   Iterate through each student's chalk consumption, subtracting it from k until k becomes less than the current student's chalk requirement. The index of this student is our answer.

Example Walkthrough:

Let's take an example chalk = [3, 4, 1, 2] and k = 25:

1. Total Chalk Consumption:

   text{total_chalk} = 3 + 4 + 1 + 2 = 10

1. Reduce k:

   k % 10 = 25 % 10 = 5

Now we have k = 5 after subtracting as many full rounds as possible.

1. Find the Student:
   • Student 0 uses 3 chalk, so k = 5 - 3 = 2.
   • Student 1 requires 4 chalk, but k = 2, which is less than 4.
   • Therefore, student 1 will be the one who needs to replace the chalk.

Let's implement this solution in PHP: 1894. Find the Student that Will Replace the Chalk

  
  
  Explanation:




Total Chalk Sum: We sum up all the chalk requirements to get the total for one complete round.

Modulo Operation: Using modulo with k, we get the effective number of chalks to distribute after full rounds.

Find the Student: We then iterate through the students, checking if the remaining chalk is sufficient. The first time it's insufficient, that student's index is the answer.



  
  
  Complexity:




Time Complexity: O(n) — we sum the array and then iterate through it once.

Space Complexity: O(1) — only a few variables are used, independent of the input size.


This approach ensures that the problem is solved efficiently even for large inputs.

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