Introduction

In this blog post, we will explore how to solve the LeetCode problem "765" 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

There are n couples sitting in 2n seats arranged in a row and want to hold hands. The people and seats are represented by an integer array row where row[i] is the ID of the person sitting in the ith seat. The couples are numbered in order, the first couple being (0, 1), the second couple being (2, 3), and so on with the last couple being (2n - 2, 2n - 1). Return the minimum number of swaps so that every couple is sitting side by side. A swap consists of choosing any two people, then they stand up and switch seats. Example 1: Input: row = [0,2,1,3] Output: 1 Explanation: We only need to swap the second (row[1]) and third (row[2]) person. Example 2: Input: row = [3,2,0,1] Output: 0 Explanation: All couples are already seated side by side. Constraints: 2n == row.length 2 <= n <= 30 n is even. 0 <= row[i] < 2n All the elements of row are unique.

Explanation

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

• High-Level Approach:

  • We can model the problem as finding cycles in a graph where an edge exists between two couples seated incorrectly.
  • The minimum number of swaps needed to fix a cycle of length k is k - 1.
  • Iterate through the row, find misplaced couples, and trace the cycles they form.

• Complexity:

  • Runtime Complexity: O(n)
  • Storage Complexity: O(n)

Code

    def minSwapsCouples(row):
    """
    Calculates the minimum number of swaps required to seat couples side by side.

    Args:
        row: A list of integers representing the seating arrangement.

    Returns:
        The minimum number of swaps required.
    """
    n = len(row) // 2
    adj = [0] * n

    for i in range(n):
        u = row[2 * i] // 2
        v = row[2 * i + 1] // 2
        if u != v:
            adj[u] = v

    visited = [False] * n
    cycles = 0
    for i in range(n):
        if not visited[i]:
            cycles += 1
            j = i
            while not visited[j]:
                visited[j] = True
                j = adj[j]

    return n - cycles