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  1. Maximum Number of Balls in a Box (LeetCode):

You are working in a ball factory where you have n balls numbered from lowLimit 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 whose number is equal to the sum of the digits of the ball’s number. For example, the ball number 321 will belong to the box number 3 + 2 + 1 = 6.

Given two integers lowLimit and highLimit, return the number of balls in the box with the most balls.

Clarifying Questions

  1. Is there any upper limit to the values of lowLimit and highLimit, or can they be any integer within Python’s integer range?
    • The values will be within a reasonable constraint based on standard problem constraints (0 ≤ lowLimit, highLimit ≤ 10^5).
  2. Do the boxes have any specific constraints or limitations?
    • No, the boxes are numbered as needed based on the sum of the digits, which means the box numbers will typically be much smaller than the ball numbers.

Strategy

To solve this problem, we can follow these steps:

  1. Digit Sum Calculation: Write a helper function to calculate the sum of the digits of a number.
  2. Counting Balls in Boxes: Iterate over each ball from lowLimit to highLimit, compute the box number using the digit sum, and count how many balls go into each box.
  3. Finding the Maximum: Determine the maximum number of balls in any box after counting.

Code

def countBalls(lowLimit: int, highLimit: int) -> int:
    from collections import defaultdict

    def digit_sum(num: int) -> int:
        return sum(int(digit) for digit in str(num))

    box_count = defaultdict(int)

    for ball in range(lowLimit, highLimit + 1):
        box_number = digit_sum(ball)
        box_count[box_number] += 1

    return max(box_count.values())

# Example usage: 
lowLimit = 1
highLimit = 10
# Expected output: 2 (because box 1 has balls 1 and 10)
print(countBalls(lowLimit, highLimit))

Time Complexity

Overall, the time complexity of the solution is O(n * log(highLimit)), where n is the number of balls and log(highLimit) accounts for the digit sum calculation.

Conclusion

This solution efficiently computes the maximum number of balls in any box by leveraging a hash map to count balls and then finds the maximum count. The helper function for digit sum ensures the code remains readable and modular.

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