## Description

https://leetcode.com/problems/maximum-number-of-balls-in-a-box/

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 = 10Output:2Explanation: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 = 15Output:2Explanation: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 = 28Output:2Explanation: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.

## Explanation

Convert ball number into strings and summing up by digits. Counter the occurrences of digits sums.

## Python Solution

```
class Solution:
def countBalls(self, lowLimit: int, highLimit: int) -> int:
counter = {}
for ball in range(lowLimit, highLimit + 1):
ball_str = str(ball)
digits_sum = 0
for c in ball_str:
digits_sum += int(c)
counter[digits_sum] = counter.get(digits_sum, 0) + 1
return max(counter.values())
```

- Time Complexity: O(N).
- Space Complexity: O(N).