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

Fatboy Slim
2 min readFeb 2, 2021

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:

  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

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