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

class Solution:
    def chalkReplacer(self, chalk: List[int], k: int) -> int:
        s = sum(chalk)  # Total chalk consumption per cycle
        k %= s           # Remaining chalk after full cycles
 
        for i, x in enumerate(chalk): # Iterate through students
            if k < x:  # Check if current student needs to replace
                return i
            k -= x     # Update remaining chalk if sufficient

class Solution {
    public int chalkReplacer(int[] chalk, int k) {
        long s = 0; // Use long to handle potential overflow
        for (int x : chalk) {
            s += x;
        }
        k %= s; 
 
        for (int i = 0; ; ++i) {  // Iterate through students
            if (k < chalk[i]) {
                return i;
            }
            k -= chalk[i];
        }
    }
}

class Solution {
public:
    int chalkReplacer(vector<int>& chalk, int k) {
        long long s = accumulate(chalk.begin(), chalk.end(), 0LL); // Use long long to avoid overflow
        k %= s;
 
        for (int i = 0; ; ++i) {
            if (k < chalk[i]) {
                return i;
            }
            k -= chalk[i];
        }
    }
};