README_EN.md

1894. Find the Student that Will Replace the Chalk

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Description

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.

 

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

Solutions

PreSum and Binary search.

Python3

class Solution:
    def chalkReplacer(self, chalk: List[int], k: int) -> int:
        pre_sum = list(itertools.accumulate(chalk))
        k %= pre_sum[-1]
        left, right = 0, len(chalk) - 1
        while left < right:
            mid = (left + right) >> 1
            if pre_sum[mid] > k:
                right = mid
            else:
                left = mid + 1
        return left

Java

class Solution {
    public int chalkReplacer(int[] chalk, int k) {
        int n = chalk.length;
        long[] preSum = new long[n + 1];
        for (int i = 0; i < n; ++i) {
            preSum[i + 1] = preSum[i] + chalk[i];
        }
        k %= preSum[n];
        int left = 0, right = n - 1;
        while (left < right) {
            int mid = (left + right) >> 1;
            if (preSum[mid + 1] > k) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return left;
    }
}

C++

class Solution {
public:
    int chalkReplacer(vector<int>& chalk, int k) {
        int n = chalk.size();
        vector<long long> preSum(n + 1);
        for (int i = 0; i < n; ++i) {
            preSum[i + 1] = preSum[i] + chalk[i];
        }
        k %= preSum[n];
        int left = 0, right = n - 1;
        while (left < right) {
            int mid = left + (right - left >> 1);
            if (preSum[mid + 1] > k) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return left;
    }
};

Go

func chalkReplacer(chalk []int, k int) int {
	n := len(chalk)
	preSum := make([]int, n+1)
	for i := 0; i < n; i++ {
		preSum[i+1] = preSum[i] + chalk[i]
	}
	k %= preSum[n]
	left, right := 0, n-1
	for left < right {
		mid := left + ((right - left) >> 1)
		if preSum[mid+1] > k {
			right = mid
		} else {
			left = mid + 1
		}
	}
	return left
}

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