Formatted question description: https://leetcode.ca/all/1742.html

# 1742. Maximum Number of Balls in a Box

## Level

Easy

## 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 <= 10^5`

## Solution

Since `highLimit <= 10^5`

, the sum of digits of each number is at most 46. Create an array `sums`

where `sums[i]`

is the number of balls that have sum of digits equal to `i`

. Loop over all numbers from `lowLimit`

to `highLimit`

. For each number, calculate the sum of digits and update the array. Finally, loop over the array and obtain the maximum count.

```
class Solution {
public int countBalls(int lowLimit, int highLimit) {
int[] sums = new int[50];
for (int i = lowLimit; i <= highLimit; i++) {
int sum = getSumOfDigits(i);
sums[sum]++;
}
int maxCount = 0;
for (int i = 0; i < 50; i++)
maxCount = Math.max(maxCount, sums[i]);
return maxCount;
}
public int getSumOfDigits(int num) {
int sum = 0;
while (num > 0) {
sum += num % 10;
num /= 10;
}
return sum;
}
}
```