[Leetcode] 1742. 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.

 

Constraints:

• 1 <= lowLimit <= highLimit <= 105

 

### Solution 

class Solution:
    def countBalls(self, lowLimit: int, highLimit: int) -> int:
        
        # make dictionary 
        box_num = [i for i in range(1,highLimit+1)]
        length = [0 for _ in range(len(box_num))]

        dic = { key : value for key,value in zip(box_num,length) }

        for n in range(lowLimit, highLimit+1):
            list_num = list(map(int, str(n)))

            sum_list_num = sum(list_num)
            print(sum_list_num)

            dic[sum_list_num] += 1

        # result 
        return max(dic.values())

 

# tip 

 

- input 으로 들어온 int 값들을 각각의 자리별로 분해하기. 

- e.g. 321 => 3,2,1 

n = input 

map(int, str(n))

 

- 두 개의 list 로 dict 만들기 

dic = { key : value for key,value in zip( list_A, list_B ) }