maximum_number_of_balls_in_a_box.py

# https://leetcode.com/problems/maximum-number-of-balls-in-a-box/description/
# 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

class Solution:
    def countBalls(self, lowLimit: int, highLimit: int) -> int:
        # 1 <= lowLimit <= highLimit <= 105
        cnt = [0] * (9+9+9+9+9)
        for x in range(lowLimit, highLimit + 1):
            y = 0
            while x:
                y += x % 10
                x //= 10
            cnt[y] += 1
        return max(cnt)

lowLimit = 1, highLimit = 10
# Output: 2
lowLimit = 5, highLimit = 15
# Output: 2
lowLimit = 19, highLimit = 28
# Output: 2
Solution().countBalls(lowLimit, highLimit)