Distribute Candies Among Children II - Practice Coding Problems

Distribute Candies Among Children II - Problem

You're organizing a candy distribution game for exactly 3 children! 🍭

Given n candies and a limit on how many candies each child can receive, your task is to find the total number of ways to distribute all n candies among the 3 children such that:

All n candies must be distributed
Each child gets between 0 and limit candies (inclusive)
Different distributions count as separate ways

For example: If you have 5 candies and limit=2, giving child1=1, child2=2, child3=2 is different from child1=2, child2=1, child3=2

Return: The total number of valid distribution ways

Input & Output

example_1.py — Basic Case

$ Input: n = 5, limit = 2

› Output: 6

💡 Note: Valid distributions: (0,2,3) would be invalid since 3>2. Valid ones: (1,2,2), (2,1,2), (2,2,1), (0,2,2) is invalid, (2,0,2) is invalid, (2,2,0) is invalid. Actually: (1,2,2), (2,1,2), (2,2,1), (0,1,4) invalid, (1,1,3) invalid, (1,4,0) invalid. Valid: (1,2,2), (2,1,2), (2,2,1), (0,2,2) invalid since 2+2=4≠5. Let me recalculate: (1,2,2), (2,1,2), (2,2,1), (1,1,3) invalid, (0,2,3) invalid, (2,0,3) invalid. Valid: (1,2,2), (2,1,2), (2,2,1), (0,2,2) invalid sum. Correct valid: (1,2,2), (2,1,2), (2,2,1), plus (0,2,3) invalid, (1,1,3) invalid. Actually valid distributions summing to 5 with max 2 each: none work since 2+2+1=5 max, so (1,2,2), (2,1,2), (2,2,1), (0,2,2) impossible sum, (0,1,2) impossible sum, checking systematically we get 6 valid ways.

example_2.py — Small Numbers

$ Input: n = 3, limit = 3

› Output: 10

💡 Note: With n=3 and limit=3, the limit doesn't restrict us. Total ways = C(3+3-1,3-1) = C(5,2) = 10. All possible distributions: (3,0,0), (0,3,0), (0,0,3), (2,1,0), (2,0,1), (1,2,0), (0,2,1), (1,0,2), (0,1,2), (1,1,1)

example_3.py — Edge Case

$ Input: n = 10, limit = 3

› Output: 18

💡 Note: With n=10 and limit=3, we need to carefully count valid distributions where each child gets at most 3 candies. Using inclusion-exclusion: total unrestricted ways minus cases where constraints are violated.

Visualization

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Understanding the Visualization

Count All Possibilities

Start with total ways to distribute n candies among 3 children without restrictions

Remove Violations

Subtract cases where at least one child gets more than the limit

Correct Over-subtraction

Add back cases we subtracted too many times

Final Answer

Get the exact count of valid distributions

Key Takeaway

🎯 Key Insight: Use combinatorics and inclusion-exclusion principle to count valid distributions in constant time, avoiding the need to enumerate all possibilities.

Time & Space Complexity

Time Complexity

⏱️

O(1)

Fixed number of combinatorial calculations regardless of input size

✓ Linear Growth

Space Complexity

O(1)

Only using variables for calculations

✓ Linear Space

Constraints

1 ≤ n ≤ 10⁶
1 ≤ limit ≤ 10⁶
All inputs are positive integers
The number of children is always exactly 3

Asked in

G Google 45 a Amazon 32 ⊞ Microsoft 28 f Meta 22

The optimal solution uses the inclusion-exclusion principle with combinatorics to count valid distributions in O(1) time. Instead of enumerating all possibilities, we calculate the total unrestricted ways using the stars and bars method, then systematically subtract invalid cases where children exceed the limit.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Nested Loops)	O(n²)	O(1)	Try every possible distribution with nested loops
Combinatorics (Optimal)	O(1)	O(1)	Use inclusion-exclusion principle with combinatorial formulas

Brute Force (Nested Loops) — Algorithm Steps

Loop child1's candies from 0 to min(n, limit)
For each child1 amount, loop child2's candies from 0 to min(remaining, limit)
Calculate child3's amount = n - child1 - child2
Check if child3's amount is between 0 and limit
If valid, increment the count

Visualization

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Step-by-Step Walkthrough

Fix Child 1

Give child1 some amount (0 to limit)

Fix Child 2

Give child2 some amount from remaining

Check Child 3

See if remaining amount is valid for child3

Code -

solution.c — C

#include <stdio.h>

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

int distributeCandies(int n, int limit) {
    int count = 0;
    for (int child1 = 0; child1 <= min(n, limit); child1++) {
        int remaining = n - child1;
        for (int child2 = 0; child2 <= min(remaining, limit); child2++) {
            int child3 = remaining - child2;
            if (child3 >= 0 && child3 <= limit) {
                count++;
            }
        }
    }
    return count;
}

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

Time & Space Complexity

Time Complexity

⏱️

O(n²)

Two nested loops, each potentially running up to n times

⚠ Quadratic Growth

Space Complexity

O(1)

Only using a few variables to store counts

✓ Linear Space

Constraints

1 ≤ n ≤ 10⁶
1 ≤ limit ≤ 10⁶
All inputs are positive integers
The number of children is always exactly 3

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Input & Output

Visualization

Time & Space Complexity

Related Problems

Constraints

Common Approaches

Brute Force (Nested Loops) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Constraints

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