#2226

Medium

auto_awesomeBinary search over the valid answer space

LeetCode Problem Workspace

Maximum Candies Allocated to K Children

Maximize candies allocation for k children by dividing piles into sub-piles using binary search.

Problem Statement

You are given a 0-indexed integer array, candies, where each element denotes the number of candies in a pile. You can divide each pile into as many sub-piles as needed but cannot merge piles together. Your task is to allocate candies to k children such that each child receives the same number of candies, and each child can be allocated candies from only one pile.

Return the maximum number of candies each child can get. Some piles of candies may remain unused. For a fixed number of candies c, determine if it's possible to allocate c candies to each child using binary search over the valid answer space.

Examples

Example 1

Input: candies = [5,8,6], k = 3

Output: 5

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.

Example 2

Input: candies = [2,5], k = 11

Output: 0

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.

Constraints

1 <= candies.length <= 105
1 <= candies[i] <= 107
1 <= k <= 1012

Solution Approach

Binary Search on the Answer Space

Perform binary search to find the maximum number of candies each child can receive. For a given number of candies, check if it is possible to allocate that amount to all k children. This check is done by calculating how many full allocations of candies can be made from the available piles.

Checking Feasibility for Each Candidate

For each candidate number of candies, iterate through the piles and calculate how many children can be satisfied by dividing the candies into smaller sub-piles. If the total number of children that can be satisfied is greater than or equal to k, the allocation is feasible.

Optimizing the Search Range

Adjust the binary search range based on whether the current number of candies can be allocated to all k children. Narrow the search space by focusing on the higher or lower half of the range depending on the feasibility check.

Complexity Analysis

Metric	Value
Time	O(n \log m)
Space	O(1)

The time complexity of this solution is O(n log m), where n is the number of piles and m is the maximum number of candies in any pile. The binary search reduces the problem space, while iterating through the piles takes linear time. The space complexity is O(1) since only a few variables are needed to track the search range and allocation count.

What Interviewers Usually Probe

Candidate should be able to explain binary search principles and how they apply to this problem.
Candidate must show an understanding of how to validate each candidate number of candies during the binary search.
Candidate should demonstrate clarity in adjusting the binary search range based on feasibility checks.

Common Pitfalls or Variants

Common pitfalls

Confusing the binary search logic with linear search or brute force approaches.
Failing to correctly check if a certain number of candies can be allocated to all k children.
Not properly handling edge cases like not having enough candies to distribute to all children.

Follow-up variants

What if there are fewer children than piles? In this case, the algorithm will still work but some piles will remain unused.
What if some piles have fewer candies than the required amount? The binary search will help find the maximum possible value for candy allocation.
Consider an optimized solution that minimizes redundant calculations during the binary search process.

FAQ

What is the maximum number of candies each child can get in the 'Maximum Candies Allocated to K Children' problem?

The goal is to find the maximum number of candies that can be allocated to each child while ensuring that each child receives an equal number, using binary search to optimize the allocation.

How does binary search help in solving the 'Maximum Candies Allocated to K Children' problem?

Binary search allows us to efficiently narrow down the possible maximum number of candies by testing feasible solutions within a valid answer space.

What happens if there are fewer candies than children in the 'Maximum Candies Allocated to K Children' problem?

If there are fewer candies than children, the result will be 0 since it's impossible to allocate at least one candy to each child.

How do I determine if a certain number of candies can be allocated to k children?

For a candidate number of candies, iterate through the piles and count how many full allocations can be made. If the total number is greater than or equal to k, it's feasible.

What is the time complexity of the solution for the 'Maximum Candies Allocated to K Children' problem?

The time complexity is O(n log m), where n is the number of piles and m is the maximum number of candies in any pile.

terminal

Solution

Solution 1: Binary Search

We notice that if each child can receive $v$ candies, then for any $v' \lt v$, each child can also receive $v'$ candies. Therefore, we can use binary search to find the maximum $v$ such that each child can receive $v$ candies.

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class Solution:
    def maximumCandies(self, candies: List[int], k: int) -> int:
        l, r = 0, max(candies)
        while l < r:
            mid = (l + r + 1) >> 1
            if sum(x // mid for x in candies) >= k:
                l = mid
            else:
                r = mid - 1
        return l

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