# [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 = 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.`

Analysis:

1. Calculate dig (%10) and num (//10) util (//10) is zero.
2. boxes is {box : final dig}
3. 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:

https://leetcode.com/problems/maximum-number-of-balls-in-a-box/discuss/1043081/Python-Dictionary-or-easy-to-understand

Interesting in any computer science.