Count Ways to Distribute Candies

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Dynamic Programming

You are given n identical candies and k children. You want to distribute the candies to the children such that each child receives at least one candy.

You can distribute the candies in two ways:

Distribute the candies to the children such that each child receives at least one candy.
Distribute the candies to the children such that some children receive more than one candy.

Return the number of ways to distribute the candies.

Since the answer may be very large, return it modulo 10⁹ + 7.

Example 1:

Input: n = 5, k = 2
Output: 4
Explanation: We can distribute the candies in the following ways:
- Child 1 receives 1 candy, and child 2 receives 4 candies.
- Child 1 receives 2 candies, and child 2 receives 3 candies.
- Child 1 receives 3 candies, and child 2 receives 2 candies.
- Child 1 receives 4 candies, and child 2 receives 1 candy.

Example 2:

Input: n = 3, k = 3
Output: 1
Explanation: We can only distribute the candies such that each child receives exactly 1 candy.

Example 3:

Input: n = 4, k = 3
Output: 6
Explanation: We can distribute the candies in the following ways:
- Child 1 receives 2 candies, and child 2 and child 3 receive 1 candy each.
- Child 2 receives 2 candies, and child 1 and child 3 receive 1 candy each.
- Child 3 receives 2 candies, and child 1 and child 2 receive 1 candy each.
- Child 1, child 2, and child 3 receive 1, 2, and 1 candies respectively.
- Child 1, child 2, and child 3 receive 1, 1, and 2 candies respectively.
- Child 1, child 2, and child 3 receive 2, 1, and 1 candies respectively.

Constraints:

1 <= n, k <= 1000

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the constraints on the number of candies and the number of children?
Can the number of candies or children be zero?
Are the candies identical, or are they distinguishable?
Do I need to return the number of ways modulo some value to prevent integer overflow?
Is there any constraint related to each child receiving at least one candy?

Brute Force Solution

Approach

The problem asks us to find how many different ways we can distribute a certain number of candies to a certain number of kids. The brute force way involves trying every single possible distribution of candies.

Here's how the algorithm would work step-by-step:

Imagine giving zero candies to the first kid and giving the rest to the other kids, then try giving one candy to the first kid and giving the rest to the others, then two, and so on.
For each of those possibilities for the first kid, try every possible distribution of the remaining candies to the second kid, and then to the third kid, and so on for all the kids.
Keep track of each way you find to give out all the candies such that the sum of candies given to each kid equals the total number of candies you started with.
Count all the valid ways you distributed the candies.

Code Implementation

def count_ways_distribute_candies_brute_force(
    number_candies,
    number_kids
):
    number_ways = 0

    def distribute_recursive(
        remaining_candies,
        kid_index
    ):
        nonlocal number_ways

        # Base case: all kids have received candies
        if kid_index == number_kids - 1:
            if remaining_candies >= 0:
                number_ways += 1
            return

        # Iterate through the possible number of candies to give to the current kid
        for candies_for_current_kid in range(remaining_candies + 1):

            # Recursively call this function for each possible number of candies
            distribute_recursive(
                remaining_candies - candies_for_current_kid,
                kid_index + 1
            )

    # Initiate the recursive process
    distribute_recursive(
        number_candies,
        0
    )

    return number_ways

Big(O) Analysis

Time Complexity

O(n^k) – The described brute force solution explores every possible distribution of candies. For each candy, we have k choices of which kid to give it to, where n is the number of candies and k is the number of kids. This branching creates a tree of possibilities where each level represents a candy being distributed, and each branch at that level represents assigning that candy to a specific kid. Since we have n candies and k kids, the number of possible distributions grows exponentially. Therefore, the time complexity is O(n^k) reflecting the recursive branching where the depth depends on 'n' and number of branches depends on 'k'.

Space Complexity

O(K) – The described brute force approach uses recursion. The maximum depth of the recursion depends on the number of kids, K. Each recursive call creates a new stack frame to store local variables. Thus, in the worst-case scenario, the recursion depth will be proportional to K, leading to a space complexity of O(K) due to the call stack.

Optimal Solution

Approach

The problem asks us to find how many ways we can distribute candies among people. The key idea is to use a mathematical concept called combinations to avoid checking every single possibility. We calculate the number of ways directly using a formula instead of brute-force checking.

Here's how the algorithm would work step-by-step:

First, understand that the order we give out the candies doesn't matter; it's only about how many candies each person receives.
Imagine we have the candies in a row and we want to divide them among the people. We can do this by placing dividers between the candies.
Think of the problem as arranging candies and dividers in a line. The number of dividers is always one less than the number of people.
The number of ways to distribute the candies is equivalent to the number of ways to choose where to place the dividers among the candies and the dividers themselves.
Use the combinations formula (also known as 'n choose k') to calculate the number of ways to place the dividers. This formula directly tells us how many combinations are possible.
Specifically, you calculate 'n choose k' where 'n' is the total number of candies plus the number of people minus one, and 'k' is the number of people minus one.
The result of this 'n choose k' calculation is the answer to the problem: the number of ways to distribute the candies.
By using combinations, you avoid checking every single arrangement and directly arrive at the answer efficiently.

Code Implementation

def count_ways_to_distribute_candies(number_of_candies,
                                            number_of_people):
    def combinations(total_items, items_to_choose):
        if items_to_choose < 0 or items_to_choose > total_items:
            return 0
        if items_to_choose == 0 or items_to_choose == total_items:
            return 1
        if items_to_choose > total_items // 2:
            items_to_choose = total_items - items_to_choose

        result = 1
        for i in range(items_to_choose):
            result = result * (total_items - i) // (i + 1)
        return result

    # Calculate total items and items to choose for combinations.
    total_items_to_arrange = number_of_candies + number_of_people - 1

    number_of_dividers = number_of_people - 1

    # Use combinations formula to find the number of ways.
    number_of_ways = combinations(total_items_to_arrange,
                                   number_of_dividers)

    return number_of_ways

Big(O) Analysis

Time Complexity

O(n) – The dominant operation in this solution is the calculation of combinations, which typically involves computing factorials or using a precomputed table of factorials. If we assume factorials are precomputed or calculated iteratively, the calculation of 'n choose k' involves a loop that iterates up to k (or n-k, whichever is smaller). In this problem, k is the number of people minus one, and n is the number of candies plus the number of people minus one. Therefore the time complexity is dominated by a loop that runs proportionally to the size of the number of candies plus the number of people, but we are interested in understanding the number of candies which can be represented as n. Therefore, the time complexity is approximately O(n) due to the iterative calculations of factorials for the combination formula.

Space Complexity

O(1) – The plain English explanation focuses on calculating combinations using the formula 'n choose k'. It doesn't explicitly mention any auxiliary data structures like arrays, lists, or hash maps being used for intermediate storage or calculations. The calculation of combinations typically involves arithmetic operations and storing a few variables (like n, k, and intermediate results during factorial calculations, if implemented naively), but these occupy a constant amount of space. Therefore, the auxiliary space complexity is considered constant, independent of the input size (number of candies or people).

Edge Cases

Case	How to Handle
Zero candies to distribute (n = 0)	Return 1, as there's one way to distribute zero candies (give nothing to each person).
One person (k = 1)	Return 1, as there's only one way to give all candies to that one person.
More people than candies (k > n)	Return 0, as it's impossible to distribute the candies such that everyone gets at least zero.
Large number of candies or people leading to integer overflow	Use a data type capable of storing large numbers (e.g., long long in C++, or appropriate Python integer type) to prevent overflow.
Very large n and k exceeding recursion depth limitations if using a recursive approach	Implement the solution using dynamic programming to avoid excessive recursion depth.
n is a very large number	The solution should be efficient and not iterate n times if possible; dynamic programming provides a relatively constant calculation given n and k.
k is a very large number (while still less or equal to n)	If using combinations, optimize combination calculation, possibly precomputing factorials to avoid redundant calculations.
Negative input for n or k	Throw an IllegalArgumentException or similar, as the number of candies and people cannot be negative.

Zero candies to distribute (n = 0)

How to Handle:

Return 1, as there's one way to distribute zero candies (give nothing to each person).

One person (k = 1)

How to Handle:

Return 1, as there's only one way to give all candies to that one person.

More people than candies (k > n)

How to Handle:

Return 0, as it's impossible to distribute the candies such that everyone gets at least zero.

Large number of candies or people leading to integer overflow

How to Handle:

Use a data type capable of storing large numbers (e.g., long long in C++, or appropriate Python integer type) to prevent overflow.

Very large n and k exceeding recursion depth limitations if using a recursive approach

How to Handle:

Implement the solution using dynamic programming to avoid excessive recursion depth.

n is a very large number

How to Handle:

The solution should be efficient and not iterate n times if possible; dynamic programming provides a relatively constant calculation given n and k.

k is a very large number (while still less or equal to n)

How to Handle:

If using combinations, optimize combination calculation, possibly precomputing factorials to avoid redundant calculations.

Negative input for n or k

How to Handle:

Throw an IllegalArgumentException or similar, as the number of candies and people cannot be negative.