- Difficulty: Medium
- Tags: LeetCode, Medium, Array, Binary Search, Prefix Sum, Simulation, leetcode-1894, Greedy, O(n), O(1)
Problem
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