# [LeetCode]#1742. Maximum Number of Balls in a Box

**Environment: Python 3.8**

**Key technique:collections**,** %**, **//**

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.

**Analysis:**

- Calculate dig (%10) and num (//10) util (//10) is zero.
- boxes is {box : final dig}
- Return max value in boxes.

**Solution:**

`class Solution:`

def countBalls(self, lowLimit: int, highLimit: int) -> int:

ans = 0

boxes = collections.defaultdict(int)

for num in range(lowLimit, highLimit+1):

box = 0

while num:

digit = num%10

num = num//10

box += digit

boxes[box] +=1

ans = max(ans, boxes[box])

return ans

**Submissions:**

**Reference:**