Couples Holding Hands

“Couples Holding Hands” could involve graph theory, specifically permutations, as well as concepts related to union-find data structures or greedy algorithms. Here are some problems to prepare:

  1. 200. Number of Islands: This problem helps to understand the basic usage of union-find data structure.

  2. 547. Friend Circles: This problem requires an understanding of how to use union-find for groupings, which might be beneficial for understanding the couples grouping.

  3. 684. Redundant Connection: This problem helps to understand union-find structure in the context of dealing with connections.

  4. 733. Flood Fill: This problem provides practice with problems related to modifying an arrangement, similar to rearranging the couples.

  5. 765. Couples Holding Hands: This is actually the problem in question but it would be good to attempt once some basic understanding of union-find and permutations is established.

  6. 767. Reorganize String: This problem provides practice with problems related to reordering or reorganizing elements.

  7. 785. Is Graph Bipartite?: This problem is useful for understanding graph theory and how to check conditions on graph structure.

  8. 886. Possible Bipartition: This problem is a good step up from “Is Graph Bipartite?” It is a more complex graph problem, but less complex than “Couples Holding Hands.”

  9. 990. Satisfiability of Equality Equations: This problem involves handling and reasoning about equality relationships, which may be helpful for reasoning about couples.

  10. 1202. Smallest String With Swaps: This problem involves performing permutations to reach a desired state, which could be useful for reasoning about swapping couples.

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from typing import List

class Solution:
    def minSwapsCouples(self, row: List[int]) -> int:
        pos = {person: i for i, person in enumerate(row)}
        num_swaps = 0
        for i in range(0, len(row), 2):
            if (row[i]^1) != row[i+1]:
                swap_pos = pos[row[i]^1]
                row[i+1], row[swap_pos] = row[swap_pos], row[i+1]
                pos[row[swap_pos]] = pos[row[i+1]]
                num_swaps += 1
        return num_swaps

Problem Classification

This problem is a combination of several categories:

  1. Array Manipulation: The problem revolves around manipulating a given array that represents people sitting in a row.

  2. Graph Theory / Union Find: The problem can be seen as finding a way to pair all nodes (people) in a graph with the minimum number of operations (swaps). Each couple can be considered as an undirected edge connecting two nodes, and we need to rearrange the nodes in such a way that each edge is contiguous.

  3. Greedy Algorithm: The problem asks for the minimum number of swaps, which hints at the possibility of a greedy approach, where we try to solve the problem step-by-step always making the choice that seems the best at that moment.

The ‘What’ components of the problem are:

  1. Input: An array ‘row’ representing the seating arrangement of people, with each element representing the ID of a person.

  2. Output: The minimum number of swaps needed to ensure that every couple is sitting side by side.

  3. Constraints: The constraints provided guide the solution towards specific algorithmic paradigms (like those that operate within polynomial time complexity).

  4. Difficulty Level: The problem requires understanding of several concepts like array manipulation, graph theory, and greedy algorithms.

  5. Problem Type: This is an optimization problem as we are required to find the minimum number of operations (swaps) to reach the goal state (couples sitting side by side).

Identifying Problem Isomorphism

“Minimum Swaps to Make Sequences Increasing” is an approximate isomorphism to “Couples Holding Hands”.

In “Minimum Swaps to Make Sequences Increasing”, you are given two integer arrays of the same length, and the task is to make both arrays strictly increasing, i.e., array[i] < array[i+1] for i from 0 to the length of the array minus 2. The minimum number of swaps you need to apply to the arrays in order to achieve this goal is required.

Similarly, in “Couples Holding Hands”, you are given an array of integers where the integers in the pairs are adjacent to each other and the task is to arrange the pairs such that every pair is adjacent to each other, again with the goal of minimizing the number of swaps.

Both problems involve the manipulation of array elements through swapping to meet a certain condition. While the conditions are different - one seeks strictly increasing order and the other requires adjacency of pairs - the core computational concept of minimizing the number of swaps to satisfy these conditions is shared.

“Minimum Swaps to Make Sequences Increasing” is a simpler problem as it involves only one condition (strictly increasing order), while “Couples Holding Hands” is more complex due to the additional condition of adjacency of pairs.

Thought Process

To solve this problem, we can use the following approach:

  1. We first need to realize that each couple should be together. Therefore, our goal is to make all couples sit together with the minimum number of swaps.

  2. We can iterate through the array, two indices at a time, representing a pair of seats. If the pair of seats is not occupied by a couple, we can swap one of the people with his/her couple.

  3. The trick here is to find the couple’s location, which could be anywhere in the array. We can use a HashMap to store each person’s location for constant time lookup.

  4. For each pair of seats, we find the other half of the couple for the first person, and swap it with the person not belonging to this couple in this pair of seats.

  5. Count the number of swaps during the process.

Let’s put this into Python code:

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def minSwapsCouples(row):
    """
    :type row: List[int]
    :rtype: int
    """
    # Mapping from person to their current position
    pos = {person: i for i, person in enumerate(row)}
    num_swaps = 0
    for i in range(0, len(row), 2):
        # If two people are a couple, do nothing
        if (row[i]^1) == row[i+1]:
            continue
        else:
            # Swap
            row[pos[row[i]^1]], row[i+1] = row[i+1], row[pos[row[i]^1]]
            pos[row[i+1]] = pos[row[i]^1]
            num_swaps += 1
    return num_swaps

In the code, row[i]^1 will give the couple for person row[i]. For example, if row[i] is even, row[i]^1 will give the next odd number, which is the couple of row[i] and vice versa. This makes use of the property that couples are numbered in order, and the bitwise XOR operation (^) with 1 flips the least significant bit, effectively mapping an even number to the next odd and an odd number to the previous even.

This solution operates in O(n), as it only performs a single pass through the array, where n is the length of the array row.

Language Agnostic Coding Drills

  1. Identifying the distinct concepts

    The given code snippet leverages a few distinct concepts:

    a. Creating a dictionary: In Python, a dictionary is used to store (key, value) pairs. Here, we are creating a dictionary to store the index position of each person as its value and the person’s identifier as the key.

    b. Bit manipulation (XOR operation): The XOR operation is being used to find the partner of a person. In the problem, couples are numbered in a way that two partners have sequential IDs and one of them is an even number. Hence, XORing with 1 can give us the partner’s ID.

    c. Looping through a list in steps: The code is iterating over the ‘row’ list in steps of 2 to process each pair of people sitting together.

    d. Swapping elements in a list: The code swaps elements in the list when necessary, which is a common operation in many algorithms.

    e. Updating dictionary entries: Whenever a swap is made, the dictionary that stores the positions of people is updated accordingly.

  2. Difficulty levels of the concepts

    a. Creating a dictionary: Easy - This is one of the fundamental concepts of Python programming.

    b. Bit manipulation (XOR operation): Intermediate - While not complex itself, understanding when and how to use XOR operation might be tricky for beginners.

    c. Looping through a list in steps: Easy - This is a very basic concept, though slightly more advanced than regular looping as it involves step size.

    d. Swapping elements in a list: Easy - This is a simple and frequently used operation in Python.

    e. Updating dictionary entries: Intermediate - Although not difficult, it requires a good understanding of dictionaries and how they work.

  3. Problem-solving approach

    The given problem is an application of the Greedy algorithm. The approach is to iterate over the people in the row by pairs. For each pair, we check if they are a couple. If not, we need to make a swap.

    We use the dictionary to quickly find the position of the partner of the person, perform the swap, update the swapped person’s position in the dictionary, and increment our swap counter. We continue this until we have processed all the pairs.

    We first need to understand how to create and update dictionaries (Concept a & e). We also need to understand how to iterate over a list in steps and swap elements (Concept c & d). Finally, we need to understand the use of the XOR operation to find the partner of a person (Concept b). Each of these drills plays a vital role in solving the problem.

Targeted Drills in Python

  1. Python-based coding drills for each concept

    a. Creating a dictionary:

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    def create_dictionary():
    dict_example = {1: "a", 2: "b", 3: "c"}
    print(dict_example)
create_dictionary()

    b. Bit manipulation (XOR operation):

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    def find_partner(id):
    partner_id = id ^ 1
    print(partner_id)
find_partner(2)  # Output will be 3

    c. Looping through a list in steps:

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    def step_iteration():
    for i in range(0, 10, 2):
        print(i, end=' ')
step_iteration()  # Outputs: 0 2 4 6 8

    d. Swapping elements in a list:

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    def swap_elements():
    list_example = [1, 2, 3, 4, 5]
    list_example[0], list_example[4] = list_example[4], list_example[0]
    print(list_example)
swap_elements()  # Outputs: [5, 2, 3, 4, 1]

    e. Updating dictionary entries:

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    def update_dictionary():
    dict_example = {1: "a", 2: "b", 3: "c"}
    dict_example[1] = "z"
    print(dict_example)
update_dictionary()  # Outputs: {1: "z", 2: "b", 3: "c"}

  2. Problem-specific concepts

    The problem-specific concept is the Greedy algorithm. This strategy involves making the most optimal (greedy) choice at each stage with the hope that these local optimums will lead to a global optimum. In this problem, the greedy choice is to always swap the current person with their partner if they are not already together.

    This strategy is important because we need to minimize the number of swaps to arrange couples side by side.

  3. Integrating the drills

    To solve the initial problem, we first need to create a dictionary that stores the positions of each person (Concept a). We then start iterating over the row in steps of 2, processing each pair at a time (Concept c).

    For each pair, we find the partner of the first person using the XOR operation (Concept b). If this partner is not the current second person, we make a swap. The swapping involves updating the list (Concept d) and the dictionary (Concept e).

    We continue this process until all pairs have been processed. The count of swaps gives us the minimum number of swaps required to arrange the couples side by side. The overall strategy aligns with the Greedy algorithm concept.