#765

Hard

auto_awesomeGraph traversal with depth-first search

LeetCode Problem Workspace

Couples Holding Hands

This problem requires arranging couples sitting apart in a row with the minimum number of swaps using graph traversal and greedy techniques.

greedy depth first search breadth first search union find graph

Problem Statement

In the Couples Holding Hands problem, n couples are seated in a row of 2n seats. The people and seats are represented by an integer array row, where each person has an ID. Couples are numbered sequentially, with the first couple being (0, 1), the second couple being (2, 3), and so on. Your task is to return the minimum number of swaps required to ensure that each couple is seated side by side.

A swap involves choosing two people, having them stand up, and then swapping their seats. The goal is to determine the fewest swaps needed so that every couple is seated next to each other. For instance, in the case of row = [0, 2, 1, 3], only one swap is needed to align the couples properly.

Examples

Example 1

Input: row = [0,2,1,3]

Output: 1

We only need to swap the second (row[1]) and third (row[2]) person.

Example 2

Input: row = [3,2,0,1]

Output: 0

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.

Solution Approach

Graph Traversal with DFS

The problem can be viewed as a graph, where each seat and person is a node. For each couple, create an edge between the two seats they should occupy. The minimal swaps needed to arrange the couples next to each other can be identified by using depth-first search (DFS) to find connected components. Each component will represent a couple that can be rearranged with one swap.

Greedy Approach for Pairing

A greedy approach can be used where each time you find two people who are not seated next to each other, you swap them. By traversing through the row and making local optimal swaps (minimizing the number of misplacements), you can solve the problem efficiently.

Union-Find Approach

Union-find (also known as disjoint set union, DSU) can be used to manage and track the seating arrangement. By treating each pair as a group and applying union-find, you can identify which people are in the same group and determine how to swap them until all pairs are together.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity depends on the approach chosen. For DFS, it is O(n), where n is the number of people. The space complexity is O(n) due to the need to store the connections between individuals. Union-find has a time complexity of O(α(n)) for each union and find operation, where α is the inverse Ackermann function, and a space complexity of O(n).

What Interviewers Usually Probe

Look for understanding of graph traversal and the ability to apply DFS in a seating arrangement problem.
Assess the candidate's ability to use greedy algorithms in a constrained optimization scenario.
Evaluate knowledge of union-find and its application to solve seating problems efficiently.

Common Pitfalls or Variants

Common pitfalls

Failing to identify the problem as a graph traversal task can lead to inefficient solutions.
Not considering all possible swap configurations, leading to suboptimal results.
Using the wrong data structures, such as simple arrays instead of graph-related structures, can complicate the solution.

Follow-up variants

Instead of using swaps, solve the problem using only adjacent seat moves.
Modify the problem to consider a situation where multiple people can swap at once.
Instead of ensuring only couples are next to each other, generalize to a larger group problem, where people need to be grouped with others of specific criteria.

FAQ

What is the core pattern used in solving the Couples Holding Hands problem?

The problem is primarily solved through graph traversal, especially using depth-first search (DFS) to model the couples as connected components in the graph.

How do swaps work in the Couples Holding Hands problem?

A swap involves choosing two people who are not sitting next to each other and swapping their positions until all couples are seated together.

Can a greedy approach be applied to the Couples Holding Hands problem?

Yes, a greedy approach can work by making local optimal swaps at each step, such as swapping people until each couple is seated next to each other.

Is Union-Find a good approach for the Couples Holding Hands problem?

Yes, Union-Find is effective for tracking and grouping people who should sit together, efficiently handling swaps needed to group each couple.

What is the time complexity of the Couples Holding Hands problem?

The time complexity depends on the approach: DFS is O(n), and Union-Find has a time complexity of O(α(n)) for each union-find operation.

terminal

Solution

Solution 1: Union-Find

We can assign a number to each pair of couples. Person with number $0$ and $1$ corresponds to couple $0$, person with number $2$ and $3$ corresponds to couple $1$, and so on. In other words, the person corresponding to $row[i]$ has a couple number of $\lfloor \frac{row[i]}{2} \rfloor$.

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class Solution:
    def minSwapsCouples(self, row: List[int]) -> int:
        def find(x: int) -> int:
            if p[x] != x:
                p[x] = find(p[x])
            return p[x]

        n = len(row) >> 1
        p = list(range(n))
        for i in range(0, len(row), 2):
            a, b = row[i] >> 1, row[i + 1] >> 1
            p[find(a)] = find(b)
        return n - sum(i == find(i) for i in range(n))

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