#1742

Easy

LeetCode Problem Workspace

Maximum Number of Balls in a Box

The problem asks you to find the box with the maximum number of balls based on the sum of digits of ball numbers.

Problem Statement

You are working in a ball factory with balls numbered from lowLimit to highLimit (inclusive). Each ball is placed in a box where the box number equals the sum of the digits of the ball's number. For example, a ball numbered 321 will go into box number 6 (because 3 + 2 + 1 = 6).

Your task is to determine which box contains the most balls and return that number. The range from lowLimit to highLimit will contain at least one ball, and each ball will be assigned to a box based on the sum of its digits.

Examples

Example 1

Input: lowLimit = 1, highLimit = 10

Output: 2

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

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

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 Approach

Use Hash Table to Track Ball Counts

Iterate through all ball numbers from lowLimit to highLimit. For each number, calculate the sum of its digits and use a hash table to count how many balls are placed in each box. The key in the hash table will be the box number, and the value will be the ball count for that box.

Optimize for Time Complexity

Given the constraints (lowLimit and highLimit up to 105), a direct iteration approach is feasible. Each ball number is processed by summing its digits, and this operation is fast due to the small number size (at most 6 digits). Using a hash table ensures that you can quickly update and retrieve the count for each box.

Find Maximum Ball Count

After populating the hash table with ball counts, find the box with the highest ball count by iterating through the table. This step involves scanning the hash table, which is efficient since the number of boxes is limited by the sum of digits of the numbers, not the total count of balls.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity is O(n), where n is the difference between highLimit and lowLimit, because we iterate through each ball once and compute the sum of digits for each. The space complexity is O(m), where m is the number of unique box numbers (sum of digits). Since the sum of digits is bounded (the maximum sum is 45 for numbers up to 99999), the space complexity remains small and manageable.

What Interviewers Usually Probe

Check if the candidate can identify the importance of using a hash table to store the count of balls in each box.
Ensure the candidate understands the sum of digits concept and its relevance to the problem.
Assess the candidate's ability to optimize the solution given the constraints, especially time complexity.

Common Pitfalls or Variants

Common pitfalls

Failing to efficiently compute the sum of digits for each number.
Not handling the edge cases, such as the smallest range (lowLimit = highLimit).
Using inefficient data structures that could increase time complexity unnecessarily.

Follow-up variants

Modify the problem to consider larger ranges, forcing candidates to optimize the approach further.
Introduce additional constraints on the ball or box numbers, requiring adjustments to the sum of digits logic.
Allow multiple queries for different ranges and require a solution that can handle batch processing.

FAQ

What is the main pattern in the 'Maximum Number of Balls in a Box' problem?

The problem primarily uses a hash table to store counts for each box, while the sum of digits of each ball's number determines the box number.

How do I calculate the sum of digits for a ball number?

To calculate the sum of digits, repeatedly extract the last digit of the number by performing a modulus operation and add it to a running total. Then, divide the number by 10 and repeat until the number is reduced to zero.

What is the time complexity of the solution?

The time complexity is O(n), where n is the number of balls from lowLimit to highLimit, because we process each ball exactly once.

Can I solve this problem with a brute force approach?

Yes, but it may be inefficient for larger ranges. A more efficient solution uses a hash table to track ball counts and compute the sum of digits for each ball.

What are some potential optimizations for this problem?

You can optimize the solution by ensuring the sum of digits is computed efficiently and by using a hash table for constant time lookups and updates.

terminal

Solution

Solution 1: Array + Simulation

Observing the problem's data range, the maximum number of balls does not exceed $10^5$, so the maximum sum of the digits of each number is less than $50$. Therefore, we can directly create an array $\textit{cnt}$ of length $50$ to count the number of occurrences of each digit sum.

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class Solution:
    def countBalls(self, lowLimit: int, highLimit: int) -> int:
        cnt = [0] * 50
        for x in range(lowLimit, highLimit + 1):
            y = 0
            while x:
                y += x % 10
                x //= 10
            cnt[y] += 1
        return max(cnt)

Continue Practicing

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