# LeetCode 1742. Maximum Number of Balls in a Box

## 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 = 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.```

## 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).