# Solving Leetcode Interviews in Seconds with AI: Minimum Cost to Reach Every Position


	# 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](https://www.chatmagic.app), 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
	```python
	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
	```