How do you solve Fruits Into Baskets II on LeetCode?

Fruits Into Baskets II (LeetCode #3477) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(1) space complexity. Key insight: Sorting allows for efficient matching of fruits to baskets.

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

The brute force approach for Fruits Into Baskets II is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach tries to place each type of fruit in the leftmost basket that can hold it. If no suitable basket is found, the fruit remains unplaced. This method is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n log n).

What are the interview tips for Fruits Into Baskets II?

Always consider edge cases, such as when all fruits can fit or none can fit. Explain your thought process clearly while coding; it helps the interviewer understand your approach. Practice similar problems to recognize patterns and improve your problem-solving speed.

What are common mistakes when solving Fruits Into Baskets II?

Failing to sort the arrays before attempting to match fruits to baskets. Not properly managing the used baskets, leading to incorrect counts of unplaced fruits.

#3477

Fruits Into Baskets II

Easy

ArrayBinary SearchSegment TreeSimulationOrdered SetTwo PointersSorting

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

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

The optimal solution sorts both arrays first, allowing us to efficiently match fruits to baskets using a two-pointer technique. This reduces unnecessary checks and speeds up the placement process.

⚙️

Algorithm

5 steps

1Step 1: Sort both the fruits and baskets arrays.
2Step 2: Initialize two pointers, one for fruits and one for baskets.
3Step 3: Iterate through the fruits, and for each fruit, move the basket pointer until a suitable basket is found or all baskets are checked.
4Step 4: If a suitable basket is found, mark it as used and move to the next fruit. If not, increment the unplaced counter.
5Step 5: Return the count of unplaced fruits.

solution.py19 lines

1# Full working Python code
2fruits = [4, 2, 5]
3baskets = [3, 5, 4]
4
5fruits.sort()
6baskets.sort()
7
8unplaced = 0
9j = 0
10
11for fruit in fruits:
12    while j < len(baskets) and baskets[j] < fruit:
13        j += 1
14    if j < len(baskets):
15        j += 1  # Use this basket
16    else:
17        unplaced += 1
18
19print(unplaced)

ℹ

Complexity note: The time complexity is O(n log n) due to sorting both arrays. The subsequent placement of fruits is O(n), making the overall complexity dominated by the sorting step. The space complexity is O(1) since we only use a few extra variables.

1Sorting allows for efficient matching of fruits to baskets.
2Using two pointers minimizes unnecessary checks and speeds up the process.

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