1742. Maximum Number of Balls in a Box-盒子中小球的最大数量

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 in python:

class Solution:
    def countBalls(self, lowLimit: int, highLimit: int) -> int:
        # 9+9+9+9+9 = 45
        arr = [0 for i in range(46)]
        def numSum(n):
            total = 0
            while n > 0:
                n, r = divmod(n, 10)
                total += r 
            return total
        for i in range(lowLimit, highLimit+1):
            index = numSum(i)
            arr[index] += 1
        return max(arr)