# 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 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 $m = highLimit$.

• 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;
}
};

• 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)


• 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);
}