Distribute Candies Among Children II - 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 distribution must assign a non-negative number of candies to each child, and the sum of all candies distributed must equal n.

Input & Output

Example 1 — Basic Case

$ Input: n = 5, limit = 2

› Output: 3

💡 Note: We need to find all (x,y,z) where x+y+z=5 and 0≤x,y,z≤2. The valid distributions are: (1,2,2), (2,1,2), (2,2,1). Total: 3 ways.

Example 2 — Small Input

$ Input: n = 3, limit = 3

› Output: 10

💡 Note: All possible distributions are valid since limit=3 ≥ n=3. Total ways = C(3+2,2) = C(5,2) = 10.

Example 3 — Tight Constraint

$ Input: n = 2, limit = 1

› Output: 3

💡 Note: Valid ways: (0,1,1), (1,0,1), (1,1,0). Each child gets at most 1 candy and total is 2.

Constraints

1 ≤ n ≤ 50
1 ≤ limit ≤ 50

Visualization

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Asked in

G Google 15 a Amazon 12 M Microsoft 8

Brief HTML summary: The key insight is to use combinatorics with inclusion-exclusion principle. Count total ways using stars and bars, then subtract invalid distributions. Best approach is mathematical formula with Time: O(1), Space: O(1).

Common Approaches

✓ Brute Force Enumeration

⏱️ Time: O(limit²) Space: O(1)

Iterate through all possible ways to give candies to the first two children (0 to limit each), then check if the remaining candies for the third child is within the limit.

Mathematical Formula (Stars and Bars)

⏱️ Time: O(1) Space: O(1)

Apply the stars and bars theorem to count ways to distribute n items into 3 groups, then use inclusion-exclusion to subtract invalid cases where any child gets more than limit candies.

Brute Force Enumeration — Algorithm Steps

Iterate child1 candies from 0 to min(n, limit)
Iterate child2 candies from 0 to min(n-child1, limit)
Calculate child3 = n - child1 - child2
Check if child3 <= limit and count valid cases

Visualization

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Step-by-Step Walkthrough

Try Child1

Give 0 to limit candies to first child

Try Child2

For each Child1, try 0 to limit for second child

Check Child3

Calculate remaining candies and check if valid

Code -

solution.c — C

#include <stdio.h>

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

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

Time & Space Complexity

Time Complexity

⏱️

O(limit²)

Two nested loops, each running up to limit times

✓ Linear Growth

Space Complexity

O(1)

Only using a few variables for counting

✓ Linear Space

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Distribute Candies Among Children II - Problem

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force Enumeration — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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