# 1894. Find the Student that Will Replace the Chalk

## 1. Question

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.

## 2. Examples

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.


## 3. Constraints

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

## 5. Solutions

class Solution {
public int chalkReplacer(int[] chalk, int k) {
int n = chalk.length;
long[] arr = new long[n];
arr = chalk;
for(int i = 1; i < n; i++) {
arr[i] = arr[i - 1] + chalk[i];
}
long total = arr[n - 1];
k %= total;
return binarySearch(arr, k);
}

private int binarySearch(long[] arr, int k) {
int left = 0;
int right = arr.length - 1;
while(left < right) {
int mid = (left + right) / 2;
if(arr[mid] < k) {
left = mid + 1;
} else if (arr[mid] > k) {
right = mid;
} else {
return mid + 1;
}
}
return left;
}
}


class Solution {
public int chalkReplacer(int[] chalk, int k) {
long sum = 0L;
for (int num : chalk) {
sum += num;
}
k = (int) (k % sum);
for (int i = 0; i < chalk.length; i++) {
if ((k = k - chalk[i]) < 0) {
return i;
}
}
return -1;
}
}