Introduction

In this blog post, we will explore how to solve the LeetCode problem "3479" 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 <= 105 1 <= fruits[i], baskets[i] <= 109

Explanation

Here's the breakdown of the solution and the code:

• High-Level Approach:

  • Iterate through the fruits array.
  • For each fruit, iterate through the baskets array to find the leftmost available basket with sufficient capacity.
  • If a suitable basket is found, mark it as occupied and move to the next fruit.

• Complexity:

  • Runtime Complexity: O(n^2) in the worst case (when no suitable basket is found until the last one)
  • Storage Complexity: O(n) due to the occupied array

Code

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

    Args:
        fruits (list): A list of integers representing the quantity of each fruit type.
        baskets (list): A list of integers representing the capacity of each basket.

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

    for fruit_qty in fruits:
        placed = False
        for i in range(n):
            if not occupied[i] and baskets[i] >= fruit_qty:
                occupied[i] = True
                placed = True
                break  # Move to the next fruit

        if not placed:
            unplaced_count += 1

    return unplaced_count