1742. Maximum Number of Balls in a Box

Description

You are working in a ball factory where you have n balls numbered from lowLimit up to highLimit inclusive (i.e., n == highLimit - lowLimit + 1), and an infinite number of boxes numbered from 1 to infinity.

Your job at this factory is to put each ball in the box with a number equal to the sum of digits of the ball's number. For example, the ball number 321 will be put in the box number 3 + 2 + 1 = 6 and the ball number 10 will be put in the box number 1 + 0 = 1.

Given two integers lowLimit and highLimit, return the number of balls in the box with the most balls.

Example 1:

Input: lowLimit = 1, highLimit = 10
Output: 2
Explanation:
Box Number:  1 2 3 4 5 6 7 8 9 10 11 ...
Ball Count:  2 1 1 1 1 1 1 1 1 0  0  ...
Box 1 has the most number of balls with 2 balls.

Example 2:

Input: lowLimit = 5, highLimit = 15
Output: 2
Explanation:
Box Number:  1 2 3 4 5 6 7 8 9 10 11 ...
Ball Count:  1 1 1 1 2 2 1 1 1 0  0  ...
Boxes 5 and 6 have the most number of balls with 2 balls in each.

Example 3:

Input: lowLimit = 19, highLimit = 28
Output: 2
Explanation:
Box Number:  1 2 3 4 5 6 7 8 9 10 11 12 ...
Ball Count:  0 1 1 1 1 1 1 1 1 2  0  0  ...
Box 10 has the most number of balls with 2 balls.

Constraints:

1 <= lowLimit <= highLimit <= 10⁵

Solutions

Solution 1: Array + Simulation

Observing the problem's data range, the maximum number of balls does not exceed \(10^5\), so the maximum sum of the digits of each number is less than \(50\). Therefore, we can directly create an array \(\textit{cnt}\) of length \(50\) to count the number of occurrences of each digit sum.

The answer is the maximum value in the array \(\textit{cnt}\).

The time complexity is \(O(n \times \log_{10}m)\). Here, \(n = \textit{highLimit} - \textit{lowLimit} + 1\), and \(m = \textit{highLimit}\).

Python3JavaC++GoTypeScriptRustJavaScriptC#

class Solution:
    def countBalls(self, lowLimit: int, highLimit: int) -> int:
        cnt = [0] * 50
        for x in range(lowLimit, highLimit + 1):
            y = 0
            while x:
                y += x % 10
                x //= 10
            cnt[y] += 1
        return max(cnt)

class Solution {
    public int countBalls(int lowLimit, int highLimit) {
        int[] cnt = new int[50];
        for (int i = lowLimit; i <= highLimit; ++i) {
            int y = 0;
            for (int x = i; x > 0; x /= 10) {
                y += x % 10;
            }
            ++cnt[y];
        }
        return Arrays.stream(cnt).max().getAsInt();
    }
}

class Solution {
public:
    int countBalls(int lowLimit, int highLimit) {
        int cnt[50] = {0};
        int ans = 0;
        for (int i = lowLimit; i <= highLimit; ++i) {
            int y = 0;
            for (int x = i; x; x /= 10) {
                y += x % 10;
            }
            ans = max(ans, ++cnt[y]);
        }
        return ans;
    }
};

func countBalls(lowLimit int, highLimit int) (ans int) {
    cnt := [50]int{}
    for i := lowLimit; i <= highLimit; i++ {
        y := 0
        for x := i; x > 0; x /= 10 {
            y += x % 10
        }
        cnt[y]++
        if ans < cnt[y] {
            ans = cnt[y]
        }
    }
    return
}

function countBalls(lowLimit: number, highLimit: number): number {
    const cnt: number[] = Array(50).fill(0);
    for (let i = lowLimit; i <= highLimit; ++i) {
        let y = 0;
        for (let x = i; x; x = Math.floor(x / 10)) {
            y += x % 10;
        }
        ++cnt[y];
    }
    return Math.max(...cnt);
}

impl Solution {
    pub fn count_balls(low_limit: i32, high_limit: i32) -> i32 {
        let mut cnt = vec![0; 50];
        for x in low_limit..=high_limit {
            let mut y = 0;
            let mut n = x;
            while n > 0 {
                y += n % 10;
                n /= 10;
            }
            cnt[y as usize] += 1;
        }
        *cnt.iter().max().unwrap()
    }
}

/**
 * @param {number} lowLimit
 * @param {number} highLimit
 * @return {number}
 */
var countBalls = function (lowLimit, highLimit) {
    const cnt = Array(50).fill(0);
    for (let i = lowLimit; i <= highLimit; ++i) {
        let y = 0;
        for (let x = i; x; x = Math.floor(x / 10)) {
            y += x % 10;
        }
        ++cnt[y];
    }
    return Math.max(...cnt);
};

public class Solution {
    public int CountBalls(int lowLimit, int highLimit) {
        int[] cnt = new int[50];
        for (int x = lowLimit; x <= highLimit; x++) {
            int y = 0;
            int n = x;
            while (n > 0) {
                y += n % 10;
                n /= 10;
            }
            cnt[y]++;
        }
        return cnt.Max();
    }
}

1742. Maximum Number of Balls in a Box

Description

Solutions

Solution 1: Array + Simulation

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