LeetCode 1742. Maximum Number of Balls in a Box

Description

https://leetcode.com/problems/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.

Explanation

Convert ball number into strings and summing up by digits. Counter the occurrences of digits sums.

Python Solution

class Solution:
    def countBalls(self, lowLimit: int, highLimit: int) -> int:
        counter = {}
        
        for ball in range(lowLimit, highLimit + 1):
            ball_str = str(ball)
            
            digits_sum = 0
            for c in ball_str:
                digits_sum += int(c)
                
            counter[digits_sum] = counter.get(digits_sum, 0) + 1
            
        
        return max(counter.values())
  • Time Complexity: O(N).
  • Space Complexity: O(N).

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