Introduction

In this blog post, we will explore how to solve the LeetCode problem "3502" 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 an integer array cost of size n. You are currently at position n (at the end of the line) in a line of n + 1 people (numbered from 0 to n). You wish to move forward in the line, but each person in front of you charges a specific amount to swap places. The cost to swap with person i is given by cost[i]. You are allowed to swap places with people as follows: If they are in front of you, you must pay them cost[i] to swap with them. If they are behind you, they can swap with you for free. Return an array answer of size n, where answer[i] is the minimum total cost to reach each position i in the line. Example 1: Input: cost = [5,3,4,1,3,2] Output: [5,3,3,1,1,1] Explanation: We can get to each position in the following way: i = 0. We can swap with person 0 for a cost of 5. i = 1. We can swap with person 1 for a cost of 3. i = 2. We can swap with person 1 for a cost of 3, then swap with person 2 for free. i = 3. We can swap with person 3 for a cost of 1. i = 4. We can swap with person 3 for a cost of 1, then swap with person 4 for free. i = 5. We can swap with person 3 for a cost of 1, then swap with person 5 for free. Example 2: Input: cost = [1,2,4,6,7] Output: [1,1,1,1,1] Explanation: We can swap with person 0 for a cost of 1, then we will be able to reach any position i for free. Constraints: 1 <= n == cost.length <= 100 1 <= cost[i] <= 100

Explanation

Here's a breakdown of the solution:

• Greedy Approach: The core idea is to realize that the minimum cost to reach position i is the minimum of cost[0] to cost[i].
• Iterative Calculation: We iterate through the cost array and maintain a running minimum. At each index i, the minimum cost to reach that position is the smaller value between the current running minimum and cost[i].

• Building the Answer Array: We store the running minimum at each position in the answer array, effectively creating the desired output.

• Runtime Complexity: O(n), where n is the length of the cost array. Storage Complexity: O(n) to store the result.

Code

    def min_cost_to_reach(cost):
    """
    Calculates the minimum cost to reach each position in a line.

    Args:
      cost: An integer array representing the cost to swap with each person.

    Returns:
      An array of size n, where answer[i] is the minimum cost to reach position i.
    """

    n = len(cost)
    answer = [0] * n
    min_cost = float('inf')

    for i in range(n):
        min_cost = min(min_cost, cost[i])
        answer[i] = min_cost

    return answer