Distribute Candies Among Children I - Practice Coding Problems

Distribute Candies Among Children I - Problem

You are given two positive integers n and limit.

Return the total number of ways to distribute n candies among 3 children such that no child gets more than limit candies.

Each child can receive 0 or more candies, and the sum of all candies distributed must equal n.

Input & Output

Example 1 — Basic Case

$ Input: n = 5, limit = 2

› Output: 3

💡 Note: Valid distributions: (0,2,3) is invalid since 3>2, but (1,2,2), (2,1,2), (2,2,1) are valid. Actually there are 3 valid ways total.

Example 2 — Small Values

$ Input: n = 3, limit = 3

› Output: 10

💡 Note: With high limit, all combinations work: (0,0,3), (0,1,2), (0,2,1), (0,3,0), (1,0,2), (1,1,1), (1,2,0), (2,0,1), (2,1,0), (3,0,0)

Example 3 — Tight Constraint

$ Input: n = 4, limit = 1

› Output: 0

💡 Note: Impossible: need to distribute 4 candies but each child can only get max 1, so 3×1 = 3 < 4

Constraints

1 ≤ n ≤ 50
1 ≤ limit ≤ 50

Visualization

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Understanding the Visualization

Input

n=5 candies, limit=2 per child

Constraint

Each child gets 0 to limit candies, sum = n

Count

Find all valid (child1, child2, child3) combinations

Key Takeaway

🎯 Key Insight: This is a constrained stars-and-bars combinatorial problem that can be solved efficiently using mathematical formulas

Asked in

G Google 15 M Microsoft 12 a Amazon 8

This is a combinatorics problem asking to count the number of ways to distribute n candies among 3 children with an upper limit constraint. The brute force approach uses nested loops to try all possibilities in O(n²) time. The optimal mathematical approach uses the inclusion-exclusion principle to count valid distributions in O(1) time. Key insight: transform the constrained stars-and-bars problem using combinatorial formulas.

Common Approaches

Approach	Time	Space	Notes
✓ Triple Nested Loop	O(n²)	O(1)	Try all possible distributions with three nested loops
Inclusion-Exclusion Principle	O(1)	O(1)	Use combinatorial math to count valid distributions efficiently

Triple Nested Loop — Algorithm Steps

Step 1: Loop through candies for child 1 (0 to min(n, limit))
Step 2: Loop through candies for child 2 (0 to min(remaining, limit))
Step 3: Calculate child 3's candies and check if valid (≤ limit)

Visualization

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Step-by-Step Walkthrough

Outer Loop

Try all values for child 1 (0 to limit)

Inner Loop

Try all values for child 2 (0 to remaining limit)

Validation

Calculate child 3 and check if within limit

Code -

solution.c — C

#include <stdio.h>

int solution(int n, int limit) {
    int count = 0;
    int min_n_limit = (n < limit) ? n : limit;
    
    for (int child1 = 0; child1 <= min_n_limit; child1++) {
        int remaining = n - child1;
        int min_remaining_limit = (remaining < limit) ? remaining : limit;
        
        for (int child2 = 0; child2 <= min_remaining_limit; child2++) {
            int child3 = n - child1 - child2;
            if (child3 >= 0 && child3 <= limit) {
                count++;
            }
        }
    }
    return count;
}

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

Time & Space Complexity

Time Complexity

⏱️

O(n²)

Two nested loops, each running up to n times

⚠ Quadratic Growth

Space Complexity

O(1)

Only using a few variables for counting

✓ Linear Space

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Medium Frequency

~15 min Avg. Time

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Triple Nested Loop — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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