README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Easy	https://github.com/doocs/leetcode/edit/main/solution/1700-1799/1742.Maximum%20Number%20of%20Balls%20in%20a%20Box/README_EN.md	1277	Weekly Contest 226 Q1	Hash Table Math Counting

1742. Maximum Number of Balls in a Box

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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 data range of the problem, the maximum number of the ball does not exceed $10^5$, so the maximum value of the sum of each digit of the number is less than $50$. Therefore, we can directly create an array $cnt$ with a length of $50$ to count the number of each digit sum of each number.

The answer is the maximum value in the array $cnt$.

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

Python3

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)

Java

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();
    }
}

C++

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;
    }
};

Go

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
}

TypeScript

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);
}