How do you solve Maximum Candies Allocated to K Children on LeetCode?

Maximum Candies Allocated to K Children (LeetCode #2226) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log m) time complexity and O(1) space complexity. Key insight: Binary search helps efficiently narrow down the maximum candies each child can receive.

What is the brute force solution for Maximum Candies Allocated to K Children?

The brute force approach for Maximum Candies Allocated to K Children is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute-force approach, we can try every possible number of candies each child could receive, starting from the maximum pile size down to zero. For each potential number of candies, we check if we can distribute them to the children. The optimal Optimal Solution approach improves this to O(n log m).

What algorithm or data structure is used to solve Maximum Candies Allocated to K Children?

Maximum Candies Allocated to K Children uses the following concepts: Array, Binary Search. The recommended patterns to study are: Binary Search, Greedy.

What are the interview tips for Maximum Candies Allocated to K Children?

Always clarify the problem constraints and edge cases. Explain your thought process clearly while coding. Practice binary search problems to get comfortable with the pattern.

What are common mistakes when solving Maximum Candies Allocated to K Children?

Not considering edge cases where k is larger than the total number of candies. Failing to optimize the search space using binary search.

#2226

Maximum Candies Allocated to K Children

Medium

ArrayBinary SearchBinary SearchGreedy

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log m)
Space	O(1)	O(1)

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Intuition

Time O(n log m)Space O(1)

The optimal approach uses binary search to efficiently find the maximum number of candies each child can receive. Instead of checking every possible number of candies, we narrow down the search space based on whether a certain number of candies can be allocated to k children.

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Algorithm

5 steps

1Step 1: Set low to 1 and high to the maximum pile size.
2Step 2: While low is less than or equal to high, calculate mid as the average of low and high.
3Step 3: Count how many children can be served with mid candies using a helper function.
4Step 4: If the count is at least k, update the result and move low up; otherwise, move high down.
5Step 5: Return the result.

solution.py17 lines

1# Full working Python code
2
3def canAllocate(candies, mid, k):
4    return sum(pile // mid for pile in candies) >= k
5
6
7def maxCandies(candies, k):
8    low, high = 1, max(candies)
9    result = 0
10    while low <= high:
11        mid = (low + high) // 2
12        if canAllocate(candies, mid, k):
13            result = mid
14            low = mid + 1
15        else:
16            high = mid - 1
17    return result

ℹ

Complexity note: The complexity is O(n log m) where n is the number of piles and m is the maximum number of candies in a pile. The log m comes from the binary search over the possible number of candies.

1Binary search helps efficiently narrow down the maximum candies each child can receive.
2Understanding how to check if a certain number of candies can be allocated is crucial.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.