README_EN.md

1894. Find the Student that Will Replace the Chalk

中文文档

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:
        s = list(accumulate(chalk))
        k %= s[-1]
        return bisect_right(s, k)

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> s(n, chalk[0]);
        for (int i = 1; i < n; ++i) s[i] = s[i - 1] + chalk[i];
        k %= s[n - 1];
        return upper_bound(s.begin(), s.end(), k) - s.begin();
    }
};

Go

func chalkReplacer(chalk []int, k int) int {
	n := len(chalk)
	s := make([]int, n+1)
	for i := 0; i < n; i++ {
		s[i+1] = s[i] + chalk[i]
	}
	k %= s[n]
	return sort.Search(n, func(i int) bool { return s[i+1] > k })
}

Rust

impl Solution {
    pub fn chalk_replacer(chalk: Vec<i32>, k: i32) -> i32 {
        let pre_sum: Vec<i64> = chalk
            .into_iter()
            .map(|x| x as i64)
            .scan(0, |state, x| {
                *state += x;
                Some(*state)
            })
            .collect();

        pre_sum
            .binary_search(&(k as i64 % pre_sum.last().unwrap()))
            .map_or_else(|e| e, |v| v + 1) as i32
    }
}

...