The “Maximum Number of Balls in a Box” problem on Leetcode tests your understanding of hash maps and elementary number theory. This article aims to dissect the problem, explore multiple solutions, and analyze their complexities.

## Problem Statement

The problem can be summarized as follows:

You are given two integers, `lowLimit`

and `highLimit`

. Starting from `lowLimit`

, you have to consider each integer up to `highLimit`

(inclusive) and find the sum of its digits. Then, you have to put each integer into a box whose number is equal to the sum of the digits of the integer.

Your task is to return the number of balls in the box that contains the most balls.

### Input

- Two integers
`lowLimit`

and`highLimit`

where 1≤ lowLimit ≤highLimit ≤ 10^5.

### Output

- An integer, representing the maximum number of balls in any box.

### Example

For `lowLimit = 1`

and `highLimit = 10`

, the maximum number of balls in a box would be 2, since both 1 and 10 would go into the box numbered 1+0=1.

## Brute-Force Solution

The brute-force approach for this problem would involve iterating through each number between `lowLimit`

and `highLimit`

, calculating the sum of its digits, and storing the frequency of each sum in a data structure like a list or a dictionary.

Here’s how you could implement this in Python:

```
def countBalls(lowLimit, highLimit):
box_counter = {}
for i in range(lowLimit, highLimit + 1):
digit_sum = sum(int(digit) for digit in str(i))
if digit_sum in box_counter:
box_counter[digit_sum] += 1
else:
box_counter[digit_sum] = 1
return max(box_counter.values())
```

### Time and Space Complexity

The time complexity is O(n⋅d), where n is the range (`highLimit - lowLimit + 1`

) and d is the maximum number of digits in any number within that range. The space complexity is O(n) due to the storage requirements of the box_counter dictionary.

## Optimized Solution Using a Default Dictionary

While the previous solution is sufficient for solving the problem, we can make our code slightly cleaner by utilizing a default dictionary from Python’s `collections`

module:

Python Code:

```
from collections import defaultdict
def countBalls(lowLimit, highLimit):
box_counter = defaultdict(int)
for i in range(lowLimit, highLimit + 1):
digit_sum = sum(int(digit) for digit in str(i))
box_counter[digit_sum] += 1
return max(box_counter.values())
```

### Time and Space Complexity

The complexities remain the same; however, the code is more readable and eliminates the need for checking if a key exists in the dictionary before incrementing its value.

## Further Optimizations

Since we are dealing with integers and their digit sums, we can use some mathematical operations for a minor performance improvement. Instead of converting the integer to a string and iterating through its digits, we can use modulo and division operations to get the sum of the digits:

Python Code:

```
def countBalls(lowLimit, highLimit):
box_counter = defaultdict(int)
for i in range(lowLimit, highLimit + 1):
digit_sum = 0
num = i
while num:
digit_sum += num % 10
num //= 10
box_counter[digit_sum] += 1
return max(box_counter.values())
```

The time complexity remains O(n⋅d), but the modulo and division operations might be slightly faster than string conversion and iteration, especially for large numbers.

## Testing the Code

It is crucial to test the solution against a variety of cases to ensure its correctness:

```
assert countBalls(1, 10) == 2
assert countBalls(5, 15) == 2
assert countBalls(19, 28) == 2
assert countBalls(1, 100000) >= 2 # Sanity check for large numbers
```

## Conclusion

The “Maximum Number of Balls in a Box” problem serves as a valuable exercise for understanding hash maps and basic number theory. The problem can be solved effectively with a time complexity of O(n⋅d), which is acceptable given the constraints.