Fruits Into Baskets III

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ArraysGreedy Algorithms

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:

n == fruits.length == baskets.length
1 <= n <= 10⁵
1 <= fruits[i], baskets[i] <= 10⁹

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the range of values for k and the number of different types of fruits that can be present in the fruits array?
Can the input array `fruits` be empty or null? What should I return in that case?
What should I return if no subarray exists that satisfies the condition of having at most k distinct fruit types?
Are the characters in the `fruits` array limited to a specific character set (e.g., ASCII, lowercase letters)?
If there are multiple subarrays of the same maximum length that satisfy the condition, should I return the first one encountered or any one of them?

Brute Force Solution

Approach

The brute force approach to the 'Fruits Into Baskets' problem means we'll check every possible selection of fruits. We will examine every possible continuous section of the fruit sequence to find the longest one that meets the basket constraints. This exhaustive search guarantees we find the optimal solution, even if it's slow.

Here's how the algorithm would work step-by-step:

Start by considering the first fruit alone as the first possible section.
Then, expand the section by one fruit at a time, adding the next fruit in the sequence.
Each time you add a fruit, check if the new section of fruits follows the basket rule: it should contain at most two types of fruits.
If the section follows the rule, note the length of the section.
If the section does not follow the rule, stop extending that particular section.
Repeat these steps, starting each time from the next fruit in the original fruit sequence.
Once you've considered all possible sections, compare all the lengths you noted.
The largest length among all the valid sections represents the maximum number of fruits you can collect that meets the basket rule.

Code Implementation

def fruits_into_baskets_brute_force(fruits):
    max_fruits_collected = 0
    number_of_fruits = len(fruits)

    for start_index in range(number_of_fruits):
        for end_index in range(start_index, number_of_fruits):
            sub_array = fruits[start_index:end_index+1]

            # Use a set to efficiently check for distinct fruit types.
            fruit_types = set(sub_array)

            # Check if the current window is valid.
            if len(fruit_types) <= 2:

                # Update max length if current subarray is longer.
                max_fruits_collected = max(max_fruits_collected, len(sub_array))

    return max_fruits_collected

Big(O) Analysis

Time Complexity

O(n²) – The brute force algorithm iterates through the array of fruits, starting a new subarray at each index. For each starting index, it expands the subarray until it violates the condition of having at most two types of fruits. In the worst case, for each of the n starting positions, the inner loop iterates up to n times to check all the fruits after the starting index. Therefore, the time complexity is proportional to n * n, resulting in O(n²).

Space Complexity

O(1) – The brute force approach, as described, doesn't utilize any auxiliary data structures like arrays, hash maps, or lists to store intermediate results or track visited elements. It only uses a few variables to store the start and end indices of the current section and the maximum length found so far. Therefore, the space used remains constant irrespective of the number of fruits (N), resulting in a space complexity of O(1).

Optimal Solution

Approach

This problem asks us to maximize the number of fruits we can pick, given we can only pick at most two types of fruit at a time. The key is to efficiently track the types of fruit we're currently picking and dynamically adjust the range based on new fruit types encountered.

Here's how the algorithm would work step-by-step:

Start by looking at the beginning of the row of fruit trees.
As we walk through the trees, we will keep track of the kinds of fruits we are picking.
If we encounter a new kind of fruit that exceeds our limit of two kinds of fruits, we need to shrink the beginning of our selection until we are only picking two kinds of fruits again.
While we move forward, keep track of the maximum number of fruits picked that fulfill the condition.
Continue until we have processed all the trees and return the largest amount of fruit collected.

Code Implementation

def max_fruits_two_types(fruits):
    window_start = 0
    max_length = 0
    fruit_frequency = {}

    for window_end in range(len(fruits)):
        right_fruit = fruits[window_end]
        if right_fruit not in fruit_frequency:
            fruit_frequency[right_fruit] = 0
        fruit_frequency[right_fruit] += 1

        # Shrink the sliding window, until we have no more than 2 fruits in the frequency map
        while len(fruit_frequency) > 2:

            left_fruit = fruits[window_start]
            fruit_frequency[left_fruit] -= 1

            # Remove when frequency is zero
            if fruit_frequency[left_fruit] == 0:

                del fruit_frequency[left_fruit]

            window_start += 1

        # Remember the maximum length so far
        max_length = max(max_length, window_end - window_start + 1)

    return max_length

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the array of fruits once using a sliding window approach. The outer loop expands the window, and the inner loop (while loop) shrinks it only when necessary to maintain the constraint of at most two fruit types. The shrinking process, in total, also takes O(n) time because each fruit is visited at most twice (once by the outer loop and potentially once by the inner loop). Therefore, the overall time complexity is O(n).

Space Complexity

O(1) – The algorithm keeps track of at most two types of fruits being picked. This requires a data structure (like a hash map or dictionary) to store the fruit types and their counts, but the size of this structure is capped at two, irrespective of the total number of fruit trees (N). Consequently, the auxiliary space used is constant and does not depend on the input size N. Therefore, the space complexity is O(1).

Edge Cases

Case	How to Handle
Empty fruits array	Return 0 as there are no fruits to pick from.
Null fruits array	Throw IllegalArgumentException or return 0 based on specifications; ensure consistent error handling.
k is 0 and the array has elements	Return 0, as no fruits can be picked if no distinct types are allowed.
fruits array with only one type of fruit and k > 0	Return the length of the fruits array, as it's a valid subarray.
fruits array with all different types of fruit and k is less than the array length	Return the longest subarray with at most k distinct fruits, which will be a sliding window of length corresponding to the first occurrence of each distinct fruit.
k is greater than or equal to the number of distinct fruit types in the fruits array	Return the length of the fruits array, as any subarray is valid.
Large input size (fruits array) to assess time complexity	The sliding window approach ensures a time complexity of O(n), which scales efficiently for large inputs.
Integer overflow potential for very large fruit array lengths when calculating subarray length	Use long data type for storing and calculating the lengths to avoid potential overflow issues.

Empty fruits array

How to Handle:

Return 0 as there are no fruits to pick from.

Null fruits array

How to Handle:

Throw IllegalArgumentException or return 0 based on specifications; ensure consistent error handling.

k is 0 and the array has elements

How to Handle:

Return 0, as no fruits can be picked if no distinct types are allowed.

fruits array with only one type of fruit and k > 0

How to Handle:

Return the length of the fruits array, as it's a valid subarray.

fruits array with all different types of fruit and k is less than the array length

How to Handle:

Return the longest subarray with at most k distinct fruits, which will be a sliding window of length corresponding to the first occurrence of each distinct fruit.

k is greater than or equal to the number of distinct fruit types in the fruits array

How to Handle:

Return the length of the fruits array, as any subarray is valid.

Large input size (fruits array) to assess time complexity

How to Handle:

The sliding window approach ensures a time complexity of O(n), which scales efficiently for large inputs.

Integer overflow potential for very large fruit array lengths when calculating subarray length

How to Handle:

Use long data type for storing and calculating the lengths to avoid potential overflow issues.