Distribute Candies Among Children III

Hard

Rubrik

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Topics:

ArraysGreedy Algorithms

You are given two positive integers n and k, meaning that there are n children and k types of candies.

You are given a 0-indexed integer array candies of length k where candies[i] denotes the number of candies of the i^th type you have. You need to distribute all the candies fairly among the n children.

A distribution of candies is considered fair if each child receives an equal number of candies. More formally, each child must receive a total of totalCandies / n candies, where totalCandies is the sum of all candies you have.

Return true if it is possible to distribute all the candies fairly among the n children. Otherwise, return false.

Example 1:

Input: n = 3, candies = [5,8]
Output: true
Explanation: You have 5 candies of type 1 and 8 candies of type 2. In total, you have 5 + 8 = 13 candies. You can give 4 candies of type 2 to the first child so that they have 4 candies in total. You can give 4 candies of type 2 to the second child so that they have 4 candies in total. Finally, you can give 5 candies of type 1 to the third child so that they have 5 candies in total.

Example 2:

Input: n = 4, candies = [1,2,3,4]
Output: false
Explanation: You have 1 candy of type 1, 2 candies of type 2, 3 candies of type 3, and 4 candies of type 4. In total, you have 1 + 2 + 3 + 4 = 10 candies. It is impossible to give each of the 4 children exactly 10 / 4 = 2.5 candies each since you can only give a child a whole number of candies.

Constraints:

1 <= n <= 10⁷
1 <= candies.length <= 10⁵
1 <= candies[i] <= 10⁹

Try coding it in LeetCode

Distribute Candies Among Children III

Hard

Rubrik

7 views

Topics:

ArraysGreedy Algorithms

You are given two positive integers n and k, meaning that there are n children and k types of candies.

You are given a 0-indexed integer array candies of length k where candies[i] denotes the number of candies of the i^th type you have. You need to distribute all the candies fairly among the n children.

A distribution of candies is considered fair if each child receives an equal number of candies. More formally, each child must receive a total of totalCandies / n candies, where totalCandies is the sum of all candies you have.

Return true if it is possible to distribute all the candies fairly among the n children. Otherwise, return false.

Example 1:

Input: n = 3, candies = [5,8]
Output: true
Explanation: You have 5 candies of type 1 and 8 candies of type 2. In total, you have 5 + 8 = 13 candies. You can give 4 candies of type 2 to the first child so that they have 4 candies in total. You can give 4 candies of type 2 to the second child so that they have 4 candies in total. Finally, you can give 5 candies of type 1 to the third child so that they have 5 candies in total.

Example 2:

Input: n = 4, candies = [1,2,3,4]
Output: false
Explanation: You have 1 candy of type 1, 2 candies of type 2, 3 candies of type 3, and 4 candies of type 4. In total, you have 1 + 2 + 3 + 4 = 10 candies. It is impossible to give each of the 4 children exactly 10 / 4 = 2.5 candies each since you can only give a child a whole number of candies.

Constraints:

1 <= n <= 10⁷
1 <= candies.length <= 10⁵
1 <= candies[i] <= 10⁹

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? What are the data types for these inputs?
Can the number of candies be zero? What about the number of children?
Are there any constraints on how many candies each child must receive (e.g., can a child receive zero candies)?
If it's impossible to distribute the candies according to the (unspecified) rules, what should I return?
Could you please clarify the objective function - what criteria determines the 'best' distribution if there are multiple valid ways to distribute the candies?

Brute Force Solution

Approach

The brute force approach to distributing candies is like trying every single way to give candies to the children. We explore every possible combination, checking if each one meets the rules of the candy distribution.

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

Start by considering the first child and try giving them zero candies.
Then try giving them one candy, then two candies, and so on, up to the maximum number of candies available.
For each possible number of candies given to the first child, consider the second child and repeat the process of giving them zero, one, two candies, and so on, with the remaining candies.
Continue this process for each child, always making sure you don't give out more candies than you have.
Once you have assigned a number of candies to each child, check if it satisfies the problem's condition. Does this distribution meet all the requirements?
Keep track of all the distributions that satisfy the requirements.
Finally, from all these valid distributions, select the 'best' one according to the specific criteria the problem asks for. For example, you might be looking for the distribution with the most candies given to the last child.

Code Implementation

def distribute_candies_brute_force(number_of_candies, number_of_children):
    best_distribution = None

    def find_distributions(current_child_index, remaining_candies, current_distribution):
        nonlocal best_distribution

        # Base case: All children have been considered
        if current_child_index == number_of_children:
            # Check if all candies have been distributed
            if remaining_candies == 0:
                # Update the best distribution if necessary
                if best_distribution is None or current_distribution[-1] > best_distribution[-1]:
                    best_distribution = current_distribution.copy()
            return

        # Iterate through possible candy amounts for the current child
        for candies_for_current_child in range(remaining_candies + 1):

            # Recursive call to handle the next child
            find_distributions(
                current_child_index + 1,
                remaining_candies - candies_for_current_child,
                current_distribution + [candies_for_current_child],
            )

    # Start the recursive process
    find_distributions(0, number_of_candies, [])

    return best_distribution

Big(O) Analysis

Time Complexity

O(C^K) – The brute force approach explores every possible combination of distributing C candies among K children. For each child, we iterate through all possible numbers of candies they can receive (from 0 to C). This creates a nested loop structure where the depth of nesting is equal to the number of children (K). Therefore, in the worst case, we are considering approximately C possibilities for each of the K children, leading to C multiplied by itself K times, or C^K possibilities. Consequently, the time complexity is O(C^K).

Space Complexity

O(N) – The brute force approach described relies heavily on recursion. Each child being considered adds a new level to the recursion. In the worst-case scenario, where we have N children, the recursion depth can reach N, requiring space on the call stack to store function call information (parameters, return address, local variables) for each level of recursion. Therefore, the auxiliary space used by the recursion stack is proportional to N, leading to O(N) space complexity.

Optimal Solution

Approach

The most efficient way to distribute candies fairly involves giving each child the minimum possible, then sharing the extras in a way that limits the biggest difference between any two children. This avoids unnecessary calculations by focusing on fairness and minimizing the spread.

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

First, figure out how many candies each child would get if you divided them evenly, ignoring any leftovers.
Give each child that initial number of candies.
If you have any candies left over, distribute them one by one to the children. Give them out in a round-robin fashion, making sure no two adjacent children get more than one extra candy between them.
This ensures the candies are as evenly distributed as possible with the fewest possible disparities.

Code Implementation

def distribute_candies(number_of_candies, number_of_children):
    base_candies = number_of_candies // number_of_children
    candies_distributed = [base_candies] * number_of_children
    remaining_candies = number_of_candies - base_candies * number_of_children

    # Distribute remaining candies to minimize difference
    child_index = 0
    while remaining_candies > 0:
        candies_distributed[child_index] += 1
        remaining_candies -= 1

        # Avoid giving adjacent children extra candies
        child_index += 1
        if child_index >= number_of_children:
            child_index = 0

    return candies_distributed

Big(O) Analysis

Time Complexity

O(n) – The algorithm involves an initial division to determine the base number of candies each child receives, which takes constant time. Then, a loop iterates at most 'candies % children' times, where 'candies' is the total number of candies and 'children' is the number of children. In the worst case, this is distributing one candy at a time to each of the 'n' children in a round-robin fashion until all remaining candies are distributed. Therefore, the time complexity is proportional to the number of children 'n', resulting in O(n).

Space Complexity

O(1) – The provided algorithm primarily uses a fixed number of variables to track the initial number of candies per child and distribute the remaining candies in a round-robin fashion. It does not create any data structures (such as lists or dictionaries) whose size depends on the number of children. Therefore, the auxiliary space complexity is constant and independent of the input size N (number of children) or C (number of candies).

Edge Cases

Case	How to Handle
Number of children is zero or negative.	Return 0 since candies cannot be distributed to no or negative children.
Number of candies is zero.	Return 0 because no candies can be distributed.
Number of candies is less than the number of children.	Return 0 as not every child can receive a candy.
Number of candies is a very large number, potentially leading to integer overflow if not handled correctly.	Use a data type that can accommodate large numbers (e.g., long in Java) to prevent overflow during calculations.
Number of candies is equal to the maximum integer value.	Ensure calculations involving the number of candies do not cause an integer overflow.
The problem specifies that candies are distributed equally, but the number of candies is not divisible by the number of children.	The prompt did not specify equal distribution, so return the floor of candies/children to find how many candies each child gets.
Extremely large number of children and candies, possibly exceeding memory limitations if intermediate calculations store large data structures.	Optimize memory usage, perhaps by performing calculations iteratively instead of storing large intermediate results, or considering a formula to calculate the final result without extensive memory.
If there are constraints or conditions on how many maximum candies each child can receive	If there is a constraint that each child cannot receive more than X candies then return minimum of (floor(candies/children) , X).