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Maximum Candies Allocated to K Children

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Binary SearchArrays

You are given a 0-indexed integer array candies. Each element in the array denotes a pile of candies of size candies[i]. You can divide each pile into any number of sub piles, but you cannot merge two piles together.

You are also given an integer k. You should allocate piles of candies to k children such that each child gets the same number of candies. Each child can be allocated candies from only one pile of candies, and some piles of candies may go unused.

Return the maximum number of candies each child can get.

For example:

  1. candies = [5,8,6], k = 3 Output: 5 Explanation: We can divide candies[1] into 2 piles of size 5 and 3, and candies[2] into 2 piles of size 5 and 1. We now have five piles of candies of sizes 5, 5, 3, 5, and 1. We can allocate the 3 piles of size 5 to 3 children. It can be proven that each child cannot receive more than 5 candies.

  2. candies = [2,5], k = 11 Output: 0 Explanation: There are 11 children but only 7 candies in total, so it is impossible to ensure each child receives at least one candy. Thus, each child gets no candy, and the answer is 0.

Could you provide an efficient algorithm to solve this problem?

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:

  1. What are the possible ranges for the number of candies in each pile, and what is the maximum value for k (number of children)?
  2. Can the array of candies be empty or contain zero values, and what should I return in those cases?
  3. If it's impossible to allocate candies to all k children (i.e., no possible allocation results in a value > 0 for all children), what value should I return?
  4. Are the candies divisible, or must each child receive a whole number of candies from a given pile?
  5. If multiple maximum amounts of candies can be allocated such that the number of candies each child gets is maximized, is any one of those amounts acceptable?

Brute Force Solution

Approach

The goal is to find the largest amount of candy we can give to each child such that we can give candy to all the children. The brute force strategy involves trying every possible candy amount, from very small to very large, to see if we can distribute the candies that way.

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

  1. Start by guessing that each child can get only one candy.
  2. See if you have enough candy piles so that you can give one candy to each of the children.
  3. If you can, then try guessing that each child can get two candies.
  4. Again, see if you can give two candies to each of the children using the candy piles you have.
  5. Keep increasing the number of candies you try to give each child, one at a time.
  6. For each guess, you need to check if the total candy amount across all piles is sufficient to distribute that amount of candy to all the children.
  7. If at any point you don't have enough candies to give the guessed amount to all children, then the previous successful guess is the most candy you can give to each child.

Code Implementation

def maximum_candies(candies_piles, number_of_children):
    max_candies_possible = 0

    # Iterate through possible candy amounts, starting from 1.
    for current_candy_amount in range(1, sum(candies_piles) + 1):
        candies_distributed = 0
        for pile in candies_piles:
            candies_distributed += pile // current_candy_amount

        # Check if we can distribute the current amount to all children.
        if candies_distributed >= number_of_children:
            max_candies_possible = current_candy_amount

        # If we can't distribute enough, the previous amount was the maximum
        else:
            break

    return max_candies_possible

Big(O) Analysis

Time Complexity
O(m*n)Let n be the number of candy piles and m be the maximum number of candies in any pile. The algorithm iterates through possible candy amounts from 1 up to m. For each candy amount, it iterates through all n piles to check if the total number of candies is sufficient to give the guessed amount to K children. Therefore, the time complexity is driven by the nested loops, resulting in m iterations for the candy amount and n iterations for each pile. Thus, the overall time complexity is O(m*n).
Space Complexity
O(1)The provided plain English solution uses an iterative approach to guess the maximum candies. It doesn't explicitly mention creating any auxiliary data structures like arrays, lists, or hash maps. The algorithm primarily involves incrementing a guess value and checking a condition. Therefore, the space complexity is dominated by a few constant-sized variables, making the auxiliary space usage O(1).

Optimal Solution

Approach

The key is to avoid checking every possible candy distribution. Instead, we can use a clever technique to quickly narrow down the possibilities, focusing on finding the largest possible number of candies we can give to each child while still giving to all the children.

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

  1. Imagine a range of possibilities for the maximum candies each child could receive. The lowest is 0, and the highest is the size of the largest pile of candies.
  2. Pick a number in the middle of that range and try giving that many candies to each child.
  3. To see if that number works, go through each pile of candies and see how many children you can give that amount to. Then see if you can give candies to enough children.
  4. If you can give candies to enough children using that number, that means the maximum candies could be even higher. So, we'll look at numbers in the upper half of our range.
  5. If you can't give candies to enough children, that means that number is too high. So, we'll look at numbers in the lower half of our range.
  6. Repeat this process of picking a number in the middle of the remaining range and checking if it works. Keep narrowing the range until you find the absolute highest number of candies each child can receive.

Code Implementation

def maximum_candies_allocated_to_k_children(candies, number_of_children):
    left_bound = 0
    right_bound = max(candies)

    while left_bound <= right_bound:
        max_candies_per_child = (left_bound + right_bound) // 2
        
        # Avoid division by zero.
        if max_candies_per_child == 0:
            if number_of_children > 0:
                return 0
            else:
                return max(candies)

        number_of_children_given_candies = 0
        for pile_of_candies in candies:
            number_of_children_given_candies += pile_of_candies // max_candies_per_child

        # Check if enough children can receive candies.
        if number_of_children_given_candies >= number_of_children:
            # Increase the lower bound, since we can give more candies.
            left_bound = max_candies_per_child + 1
        else:
            # Decrease the upper bound, since the amount is too high.
            right_bound = max_candies_per_child - 1

    # Return the maximum candies that can be given to each child.
    return right_bound

Big(O) Analysis

Time Complexity
O(n log m)The algorithm uses binary search to find the maximum number of candies each child can receive. The search space ranges from 0 to m, where m is the size of the largest pile of candies. Each iteration of the binary search requires iterating through the array of candy piles of size n to determine how many children can receive the chosen number of candies. The binary search performs log m iterations, and each iteration takes O(n) time. Therefore, the overall time complexity is O(n log m).
Space Complexity
O(1)The algorithm described primarily uses variables for the binary search range (low, high) and intermediate calculations. These variables consume a constant amount of space regardless of the size of the input candy piles. No auxiliary data structures like lists or hash maps are created that scale with the input size N (where N is the number of candy piles). Therefore, the space complexity is constant.

Edge Cases

CaseHow to Handle
candies array is null or emptyReturn 0 since no candies can be allocated to any children.
k is 0Return 0 since no children need candies.
candies array contains a zeroThe binary search will still function correctly, with 0 as a valid allocation size.
candies array contains very large numbers that could lead to integer overflow when dividingUse long data type to prevent integer overflow during calculations.
The candies array contains only 1 valueReturn the candies[0] / k.
The sum of all candies is less than kThe maximum possible candy allocation per child is 0.
k is extremely large, much larger than the number of candies.The result of the division will likely be small, but the binary search still works.
Maximum sized input array and k, close to integer limits.Ensure algorithm is O(n log m) where n is number of piles and m is maximum value of a pile, and avoid unnecessary memory allocation.