Is Fruits Into Baskets II easy or hard?

Fruits Into Baskets II is classified as an Easy problem with an acceptance rate around 70%. The base solution relies on straightforward simulation, but the problem can also introduce advanced data structures like segment trees when discussing optimized approaches.

Fruits Into Baskets II Python/Java solution

A Python or Java solution typically simulates the process: iterate through fruits, scan baskets from left to right, and mark the first basket with sufficient capacity as used. The same logic translates directly across Python, Java, C++, Go, and TypeScript implementations.

How to solve Fruits Into Baskets II in O(n log m)?

Build a segment tree over the baskets array where each node stores the maximum basket capacity in its range. For each fruit, query the tree to find the leftmost basket with capacity >= fruit size. After placing the fruit, update that basket's value to mark it as used. Each query and update takes O(log m).

What is the best approach for Fruits Into Baskets II?

The most practical approach is direct simulation: iterate through each fruit and place it into the first unused basket with enough capacity. This runs in O(n*m) time and O(1) space and is sufficient for typical constraints. For larger inputs, a segment tree or ordered set can reduce the search cost to O(log m) per fruit.

Is Fruits Into Baskets II asked at Google/Amazon/Meta?

Problems involving allocation, greedy placement, and segment trees appear frequently in interviews at companies like Google, Amazon, and Meta. While this exact problem may come from coding platforms, the underlying pattern of finding the first valid position using efficient data structures is a common interview theme.

What data structure is used in Fruits Into Baskets II?

The simplest solution uses arrays and simulation. Optimized approaches often rely on segment trees, ordered sets, or balanced binary search trees to efficiently locate the leftmost basket that satisfies the capacity requirement.

What is the time complexity of Fruits Into Baskets II?

The straightforward simulation runs in O(n * m) time because each fruit may scan all baskets to find the first valid placement. Optimized solutions using a segment tree or balanced ordered set reduce this to O(n log m) by quickly locating the leftmost basket with enough capacity.

Description Hints 1 Solution Video 28:00

3477. Fruits Into Baskets II

Easy70.2% acceptance5 companies

You are given two arrays of integers, fruits and baskets, each of length n, where fruits[i] represents the quantity of the i^th type of fruit, and baskets[j] represents the capacity of the j^th basket.

From left to right, place the fruits according to these rules:

Each fruit type must be placed in the leftmost available basket with a capacity greater than or equal to the quantity of that fruit type.
Each basket can hold only one type of fruit.
If a fruit type cannot be placed in any basket, it remains unplaced.

Return the number of fruit types that remain unplaced after all possible allocations are made.

Example 1:

Input: fruits = [4,2,5], baskets = [3,5,4]

Output: 1

Explanation:

fruits[0] = 4 is placed in baskets[1] = 5.
fruits[1] = 2 is placed in baskets[0] = 3.
fruits[2] = 5 cannot be placed in baskets[2] = 4.

Since one fruit type remains unplaced, we return 1.

Example 2:

Input: fruits = [3,6,1], baskets = [6,4,7]

Output: 0

Explanation:

fruits[0] = 3 is placed in baskets[0] = 6.
fruits[1] = 6 cannot be placed in baskets[1] = 4 (insufficient capacity) but can be placed in the next available basket, baskets[2] = 7.
fruits[2] = 1 is placed in baskets[1] = 4.

Since all fruits are successfully placed, we return 0.

Constraints: