Introduction

In this blog post, we will explore how to solve the LeetCode problem "1217" 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

We have n chips, where the position of the ith chip is position[i]. We need to move all the chips to the same position. In one step, we can change the position of the ith chip from position[i] to: position[i] + 2 or position[i] - 2 with cost = 0. position[i] + 1 or position[i] - 1 with cost = 1. Return the minimum cost needed to move all the chips to the same position. Example 1: Input: position = [1,2,3] Output: 1 Explanation: First step: Move the chip at position 3 to position 1 with cost = 0. Second step: Move the chip at position 2 to position 1 with cost = 1. Total cost is 1. Example 2: Input: position = [2,2,2,3,3] Output: 2 Explanation: We can move the two chips at position 3 to position 2. Each move has cost = 1. The total cost = 2. Example 3: Input: position = [1,1000000000] Output: 1 Constraints: 1 <= position.length <= 100 1 <= position[i] <= 10^9

Explanation

Here's a breakdown of the solution strategy:

• Parity Focus: The cost of moving between positions with the same parity (both even or both odd) is 0. The cost of moving between positions with different parity is 1.
• Counting Even/Odd: Count the number of chips at even positions and the number of chips at odd positions.

• Minimum Cost: The minimum cost will be the smaller of the two counts (the cost of moving all chips to the more frequent parity).

• Runtime Complexity: O(n), where n is the number of chips. Storage Complexity: O(1).

Code

    def minCostToMoveChips(position):
    """
    Calculates the minimum cost to move all chips to the same position.

    Args:
        position: A list of integers representing the positions of the chips.

    Returns:
        The minimum cost needed to move all the chips to the same position.
    """

    even_count = 0
    odd_count = 0

    for pos in position:
        if pos % 2 == 0:
            even_count += 1
        else:
            odd_count += 1

    return min(even_count, odd_count)