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Distribute Candies Among Children III

Number: 3216

Difficulty: Hard

Paid? Yes

Companies: Amazon, Rubrik

Problem Description

Given n candies, count the total number of distributions of these candies among 3 children such that every child gets no more than limit candies. A distribution is defined by an ordered triple (a, b, c) where a + b + c = n and 0 ≤ a, b, c ≤ limit.

Key Insights

  • Notice that without the upper bound, the number of solutions (non-negative integer triples) is given by C(n+2, 2).
  • The constraint that each child can receive at most limit candies can be handled via the Inclusion-Exclusion Principle.
  • Specifically, subtract the distributions where at least one child exceeds limit, add back those where at least two exceed, and so on.
  • The formula becomes:   ans = Σ from k=0 to 3 [(-1)^k * C(3, k) * C(n - k*(limit+1) + 2, 2)] for terms where n - k*(limit+1) ≥ 0.
  • If n > 3*limit then no distribution is possible.

Space and Time Complexity

Time Complexity: O(1) since the solution uses a closed-form mathematical expression. Space Complexity: O(1) as only a constant amount of extra space is used.

Solution

We leverage combinatorics to count the distributions. The number of ways to distribute n candies among 3 children without restrictions is C(n+2, 2). However, since each child can receive at most limit candies, we need to subtract those distributions where one or more children exceed this limit. Using the Inclusion-Exclusion Principle, for every k (from 0 to 3), we subtract or add back the count of distributions where exactly k children get more than limit candies. We perform a variable substitution for a child that exceeds the limit (by subtracting limit+1 from its candy count), and then count the number of solutions for the adjusted total, ensuring we only count valid cases where the remaining candies is non-negative. This approach gives us a formula that calculates the result in constant time without iterating over all possible distributions.

Code Solutions

Below are implementations in Python, JavaScript, C++, and Java with detailed inline comments.

# Function to compute number of distributions
def count_distributions(n, limit):
    # If n is more than the maximum achievable candies, return 0
    if n > 3 * limit:
        return 0
    
    total = 0

    # Loop over k children who might have exceeded the limit
    for k in range(4):  # k = 0, 1, 2, 3
        # Calculate the remaining candies after subtracting (limit+1) from k children
        remaining = n - k * (limit + 1)
        
        # If remaining is negative, break out since further k will only decrease the remaining candies
        if remaining < 0:
            break
        
        # Calculate number of ways to distribute the remaining candies without restrictions.
        # The formula for combinations with repetition is C(remaining + 2, 2)
        ways = (remaining + 2) * (remaining + 1) // 2
        
        # Add or subtract the current term based on Inclusion-exclusion
        # Multiply by C(3,k) as there are C(3,k) ways to choose k children
        # (-1)^k manages addition and subtraction.
        total += (-1)**k * (comb := (1 if k == 0 else (3 if k == 1 else (3 if k == 2 else 1))) ) * ways
        # Note: For k=0, C(3,0)=1; k=1, C(3,1)=3; k=2, C(3,2)=3; k=3, C(3,3)=1.
    return total

# Example usage:
print(count_distributions(5, 2))  # Expected output: 3
print(count_distributions(3, 3))  # Expected output: 10
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