Introduction

In this blog post, we will explore how to solve the LeetCode problem "1029" 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

A company is planning to interview 2n people. Given the array costs where costs[i] = [aCosti, bCosti], the cost of flying the ith person to city a is aCosti, and the cost of flying the ith person to city b is bCosti. Return the minimum cost to fly every person to a city such that exactly n people arrive in each city. Example 1: Input: costs = [[10,20],[30,200],[400,50],[30,20]] Output: 110 Explanation: The first person goes to city A for a cost of 10. The second person goes to city A for a cost of 30. The third person goes to city B for a cost of 50. The fourth person goes to city B for a cost of 20. The total minimum cost is 10 + 30 + 50 + 20 = 110 to have half the people interviewing in each city. Example 2: Input: costs = [[259,770],[448,54],[926,667],[184,139],[840,118],[577,469]] Output: 1859 Example 3: Input: costs = [[515,563],[451,713],[537,709],[343,819],[855,779],[457,60],[650,359],[631,42]] Output: 3086 Constraints: 2 * n == costs.length 2 <= costs.length <= 100 costs.length is even. 1 <= aCosti, bCosti <= 1000

Explanation

Here's a breakdown of the solution:

• Calculate Cost Differences: Determine the cost difference between sending each person to city A versus city B.
• Prioritize Savings: Sort people based on these cost differences in ascending order. This prioritizes sending people to the city where they offer the most significant cost savings.

• Balance City Allocation: Iterate through the sorted list. Send the first n people to city A and the rest to city B. This ensures exactly n people are sent to each city.

• Runtime Complexity: O(n log n), where n is the number of people (costs.length). This is due to the sorting step.

• Storage Complexity: O(n) in the worst case due to the sorting.

Code

    def two_city_scheduling(costs):
    """
    Calculates the minimum cost to fly 2n people to cities A and B with n people in each city.

    Args:
        costs: A list of lists where costs[i] = [aCosti, bCosti].

    Returns:
        The minimum cost to fly every person to a city such that exactly n people arrive in each city.
    """

    n = len(costs) // 2

    # Calculate the cost difference between city A and city B for each person
    costs.sort(key=lambda x: x[0] - x[1])  # Sort by A_cost - B_cost

    min_cost = 0
    for i in range(n):
        min_cost += costs[i][0]  # Send first n people to city A
    for i in range(n, 2 * n):
        min_cost += costs[i][1]  # Send remaining n people to city B

    return min_cost