Distribute Candies Among Children III - Problem

Math Combinatorics Hard

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: 6

💡 Note: Valid distributions: (0,1,4)✗, (0,2,3)✗, (1,1,3)✗, (1,2,2)✓, (2,1,2)✓, (2,2,1)✓, (0,0,5)✗, (1,0,4)✗, (2,0,3)✗, (0,1,4)✗, (0,2,3)✗. After checking all: (0,2,2)✓, (1,1,2)✓, (1,2,1)✓, (2,0,2)✓, (2,1,1)✓, (2,2,1)✓. Total = 6 valid ways.

Example 2 — Small Input

$ Input: n = 3, limit = 3

› Output: 10

💡 Note: Since limit ≥ n, no child can exceed limit. All distributions are valid: (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). Total = 10 ways.

Example 3 — Edge Case

$ Input: n = 1, limit = 1

› Output: 3

💡 Note: Only 3 ways to give 1 candy to exactly one child: (1,0,0), (0,1,0), (0,0,1). Each satisfies the limit constraint.

Constraints

1 ≤ n ≤ 50
1 ≤ limit ≤ 50

Visualization

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Asked in

G Google 25 M Microsoft 20 f Meta 15

The key insight is to use combinatorics with the inclusion-exclusion principle. First calculate total ways using stars and bars formula C(n+2, 2), then subtract invalid cases where children exceed the limit. Best approach uses mathematical formulas for O(1) time complexity. Time: O(1), Space: O(1)

Common Approaches

✓ Brute Force Triple Loop

⏱️ Time: O(n²) Space: O(1)

Iterate through all possible candy counts for child 1 (0 to min(n, limit)), then child 2 (0 to min(remaining, limit)), and check if child 3 gets valid amount.

Combinatorics with Inclusion-Exclusion

⏱️ Time: O(1) Space: O(1)

Apply combinatorics formula C(n+k-1, k-1) for distributing n items among k groups, then use inclusion-exclusion to remove cases where children exceed the limit.

Brute Force Triple Loop — Algorithm Steps

Loop child1 from 0 to min(n, limit)
Loop child2 from 0 to min(remaining, limit)
Check if child3 = n - child1 - child2 is valid (≤ limit)
Count valid combinations

Visualization

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Step-by-Step Walkthrough

Setup

Initialize counter, prepare nested loops

Iterate

Try every valid combination for child1 and child2

Validate

Check if child3 gets valid amount, count if yes

Code -

solution.c — C

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

int min(int a, int b) {
    return (a < b) ? a : b;
}

int solution(int n, int limit) {
    int count = 0;
    // Try all possible distributions for first two children
    for (int child1 = 0; child1 <= min(n, limit); child1++) {
        for (int child2 = 0; child2 <= min(n - child1, 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)

Nested loops iterate up to n×n times

⚠ Quadratic Growth

Space Complexity

O(1)

Only using constant variables for counting

✓ Linear Space

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Distribute Candies Among Children III - Problem

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force Triple Loop — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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