Maximum Number of Balls in a Box - Problem

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.

Input & Output

Example 1 — Small Range

$ Input: lowLimit = 1, highLimit = 10

› Output: 2

💡 Note: Balls 1,10 both go to box 1 (digit sum 1). Balls 2-9 each go to different boxes. Box 1 has maximum 2 balls.

Example 2 — Sequential Range

$ Input: lowLimit = 5, highLimit = 15

› Output: 2

💡 Note: Balls 5,14 both go to box 5 (5+0=5, 1+4=5). Ball 6→box6, 7→box7, etc. Box 5 has maximum 2 balls.

Example 3 — Same Digit Sum

$ Input: lowLimit = 19, highLimit = 28

› Output: 2

💡 Note: Balls 19,28 both go to box 10 (1+9=10, 2+8=10). Most other balls go to unique boxes.

Constraints

1 ≤ lowLimit ≤ highLimit ≤ 10⁵

Visualization

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Asked in

a Amazon 15 G Google 12

The key insight is to count balls per box using digit sums as keys. Calculate digit sum for each ball number and use hash map or array to count. Best approach is array counting since digit sums have limited range. Time: O(n×d), Space: O(1)

Common Approaches

✓ Hash Map Counting

⏱️ Time: O(n × d) Space: O(k)

Iterate through all ball numbers from lowLimit to highLimit, calculate the sum of digits for each number, use a hash map to count how many balls go in each box, then find the box with maximum count.

Single Pass with Array

⏱️ Time: O(n × d) Space: O(1)

Since the maximum possible digit sum is limited (for 5-digit numbers, max sum is 9×5=45), we can use a simple array instead of hash map for counting, which provides better cache performance and eliminates hash collisions.

Hash Map Counting — Algorithm Steps

Iterate through all ball numbers from lowLimit to highLimit
For each ball number, calculate sum of its digits
Use hash map to count balls in each box
Return the maximum count from all boxes

Visualization

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Step-by-Step Walkthrough

Calculate Digit Sums

For each ball number, sum its digits

Count in Hash Map

Increment count for each box number

Find Maximum

Return the highest count

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int digitSum(int num) {
    int total = 0;
    while (num > 0) {
        total += num % 10;
        num /= 10;
    }
    return total;
}

int solution(int lowLimit, int highLimit) {
    int boxCount[200] = {0};  // Max digit sum for numbers up to 99999 is ~45
    
    for (int ball = lowLimit; ball <= highLimit; ball++) {
        int boxNum = digitSum(ball);
        boxCount[boxNum]++;
    }
    
    int maxCount = 0;
    for (int i = 0; i < 200; i++) {
        if (boxCount[i] > maxCount) {
            maxCount = boxCount[i];
        }
    }
    
    return maxCount;
}

int main() {
    int lowLimit, highLimit;
    scanf("%d", &lowLimit);
    scanf("%d", &highLimit);
    
    int result = solution(lowLimit, highLimit);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n × d)

n iterations where n = highLimit - lowLimit + 1, and d digits per number to sum

✓ Linear Growth

Space Complexity

O(k)

Hash map stores k unique digit sums, where k ≤ 9×log₁₀(highLimit)

✓ Linear Space

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Medium Frequency

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Maximum Number of Balls in a Box - Problem

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Hash Map Counting — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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