Introduction

In this blog post, we will explore how to solve the LeetCode problem "3477" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You are given two arrays of integers, fruits and baskets, each of length n, where fruits[i] represents the quantity of the ith type of fruit, and baskets[j] represents the capacity of the jth 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 <= 100 1 <= fruits[i], baskets[i] <= 1000

Explanation

Here's the breakdown of the solution:

• Greedy Approach: Iterate through the fruits array and for each fruit, find the leftmost available basket with sufficient capacity.
• Marking Baskets: Keep track of which baskets are already occupied to ensure each basket holds only one type of fruit.

• Counting Unplaced: Count the number of fruits that couldn't be placed in any basket.

• Runtime Complexity: O(n^2), where n is the length of the arrays. Storage Complexity: O(n).

Code

    def unplaced_fruits(fruits, baskets):
    """
    Calculates the number of unplaced fruit types after allocation.

    Args:
        fruits (list of int): Quantities of each fruit type.
        baskets (list of int): Capacities of each basket.

    Returns:
        int: The number of fruit types that remain unplaced.
    """
    n = len(fruits)
    basket_occupied = [False] * n  # Keep track of occupied baskets
    unplaced_count = 0

    for fruit_quantity in fruits:
        placed = False
        for i in range(n):
            if not basket_occupied[i] and baskets[i] >= fruit_quantity:
                basket_occupied[i] = True
                placed = True
                break
        if not placed:
            unplaced_count += 1

    return unplaced_count