How do you solve Fruits Into Baskets III on LeetCode?

Fruits Into Baskets III (LeetCode #3479) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Sorting helps in efficient allocation.

What is the time complexity of Fruits Into Baskets III?

The optimal solution for Fruits Into Baskets III runs in O(n log n) time and uses O(n) space. Sorting takes O(n log n) and each fruit uses binary search in O(log n).

What is the brute force solution for Fruits Into Baskets III?

The brute force approach for Fruits Into Baskets III is Brute Force, with O(n²) time complexity and O(1) space complexity. Try placing each fruit in every basket until you find a suitable one. This method is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Fruits Into Baskets III?

Fruits Into Baskets III uses the following concepts: Array, Binary Search, Segment Tree, Ordered Set. The recommended patterns to study are: Binary Search, Greedy.

What are the interview tips for Fruits Into Baskets III?

Explain your thought process clearly. Discuss time and space complexities. Practice similar problems to build confidence.

What are common mistakes when solving Fruits Into Baskets III?

Not sorting the baskets first. Using a basket more than once.

#3479

Fruits Into Baskets III

Medium

ArrayBinary SearchSegment TreeOrdered SetBinary SearchGreedy

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n log n)Space O(n)

Sort the baskets and use binary search to efficiently find the first suitable basket for each fruit.

⚙️

Algorithm

3 steps

1Step 1: Sort the baskets in ascending order.
2Step 2: For each fruit, use binary search to find the first basket that can hold it.
3Step 3: Mark the basket as used and continue to the next fruit.

solution.py13 lines

1def unplaced_fruits(fruits, baskets):
2    baskets.sort()
3    used = [False] * len(baskets)
4    unplaced = 0
5    for fruit in fruits:
6        idx = bisect.bisect_left(baskets, fruit)
7        while idx < len(baskets) and used[idx]:
8            idx += 1
9        if idx < len(baskets):
10            used[idx] = True
11        else:
12            unplaced += 1
13    return unplaced

ℹ

Complexity note: Sorting takes O(n log n) and each fruit uses binary search in O(log n).

1Sorting helps in efficient allocation.
2Binary search reduces search time for baskets.

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