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

4. References

来源：力扣（LeetCode） 链接：https://leetcode-cn.com/problems/find-the-student-that-will-replace-the-chalk 著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

5. Solutions

前缀和+二分

class Solution {
    public int chalkReplacer(int[] chalk, int k) {
        int n = chalk.length;
        long[] arr = new long[n];
        arr[0] = chalk[0];
        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;
    }
}

取总和，取余，遍历

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;
    }
}