Jobs

Interviews

Discussions

Couples Holding Hands

Hard

Asked by:

44 views

Topics:

ArraysGreedy Algorithms

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 i^th 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.

Try coding it in LeetCode

Couples Holding Hands

Hard

Asked by:

44 views

Topics:

ArraysGreedy Algorithms

There are n couples sitting in 2n seats arranged in a row and want to hold hands.

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.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the size of the input array `row`? Are there any limits or constraints on the number of couples or the total number of seats?
Does each number from 0 to N-1 (where N is the number of couples) appear exactly twice in the input array?
If no swaps are needed (the couples are already paired correctly), should I return 0?
What data type are the elements in the `row` array? Can I assume they are non-negative integers?
If there are multiple valid sets of swaps that achieve the desired pairing, is any valid set acceptable?

Brute Force Solution

Approach

The brute force strategy for this problem involves trying every possible swap of people until all couples are holding hands. We essentially try every single configuration to see if it works.

Here's how the algorithm would work step-by-step:

Look at the first person. Check if they are holding hands with their partner.
If they are not, find their partner somewhere else in the line.
Try swapping the person next to the first person with the partner of the first person.
Check if this swap makes the first couple hold hands. If it does not, put the original configuration back.
Repeat the swapping process with every other person in the line until you find the partner or you reach the end of the line.
Keep track of how many swaps you made to get the first couple to hold hands.
Now move on to the next couple and repeat the same process.
Once you have all couples holding hands, record the total number of swaps you made.
Repeat this entire process, starting with different initial swaps, to explore all possible swap combinations.
After exploring all possible combinations, choose the arrangement with the fewest swaps.

Code Implementation

def couples_holding_hands_brute_force(row):
    number_of_people = len(row)
    minimum_swaps = float('inf')

    def calculate_swaps(current_row):
        swaps = 0
        for i in range(0, number_of_people, 2):
            person_one = current_row[i]
            person_two = current_row[i + 1]

            if person_one // 2 != person_two // 2:
                # If not a couple, find the partner.
                for j in range(i + 2, number_of_people):
                    if current_row[j] == person_one ^ 1:
                        # Swap current and found person.
                        current_row[i + 1], current_row[j] = current_row[j], current_row[i + 1]
                        swaps += 1
                        break
        return swaps

    import itertools
    # Trying all possible permutations is the core of brute force
    for permutation in itertools.permutations(row):
        current_swaps = calculate_swaps(list(permutation))
        minimum_swaps = min(minimum_swaps, current_swaps)

    # Return the minimum swaps from all the permutations
    return minimum_swaps

Big(O) Analysis

Time Complexity

O(n!) – The provided algorithm explores every possible swap combination. In the worst-case scenario, to check all possible arrangements of n people (n/2 couples) in a line, the algorithm considers all permutations. The number of permutations of n items is n factorial (n!). Therefore, the time complexity is dominated by the exploration of all possible permutations. Each swap operation takes constant time O(1), but the number of swaps explored grows factorially.

Space Complexity

O(1) – The brute force algorithm described primarily involves swapping elements in place. The explanation doesn't explicitly create any auxiliary data structures like lists or hash maps that scale with the input size N (where N is the number of people, and N is even). The algorithm keeps track of how many swaps are made, but this is a single integer. Therefore, the extra space required remains constant regardless of the input size, resulting in O(1) space complexity.

Optimal Solution

Approach

The core idea is to fix mismatched couples one at a time using swaps. We aim to perform the minimum number of swaps to get all couples sitting together correctly.

Here's how the algorithm would work step-by-step:

Imagine each person has a partner number assigned based on their position; for example, people at positions 0 and 1 are a couple, 2 and 3 are a couple, and so on.
Go through the seats one pair at a time. If the two people sitting in the current pair of seats are a couple already, great, move on to the next pair.
If the two people are not a couple, figure out who the partner of the person in the first seat should be.
Find where that partner is currently sitting.
Swap the person in the second seat of our current pair with the partner we just found.
Count that swap. By swapping the correct partner into position, we've fixed one couple.
Repeat the process for all pairs of seats. The total number of swaps is the answer.

Code Implementation

def minSwapsCouples(row):
    number_of_people = len(row)
    number_of_couples = number_of_people // 2
    swaps = 0

    for couple_index in range(number_of_couples):
        first_person_index = 2 * couple_index
        second_person_index = 2 * couple_index + 1

        first_person = row[first_person_index]
        second_person = row[second_person_index]

        # If they are not a couple, we need to perform a swap.
        if not are_couple(first_person, second_person):
            partner = find_partner(first_person)

            # Find the index of the correct partner.
            partner_index = row.index(partner)

            # Swap the person with the correct partner.
            row[second_person_index], row[partner_index] = row[partner_index], row[second_person_index]
            swaps += 1

    return swaps

def are_couple(person_a, person_b):
    return (person_a ^ 1) == person_b

def find_partner(person):
    if person % 2 == 0:
        return person + 1
    else:
        return person - 1

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the input array of size n (representing the seating arrangement) one pair at a time, where n is the number of people (and therefore even). For each pair, it potentially needs to find the correct partner's position. Finding the partner involves searching the array, but in the described algorithm, the partner is not found by a nested loop over the remaining part of the array. The number of swaps is directly dependent on how many wrong couples were sitting next to each other, and swaps are performed until no wrong couples are there. The search for the partner takes, on average, a constant time (or is precomputed with a hashmap which can be done in linear time). Overall, since we are iterating through the n elements once, the time complexity is O(n).

Space Complexity

O(1) – The provided algorithm operates in place by swapping elements within the input array. It uses a few constant space variables for iteration and temporary storage during the swap operation. The algorithm doesn't create any auxiliary data structures that scale with the input size N (the number of people). Therefore, the space complexity is constant.

Edge Cases

Case	How to Handle
Input row is null or empty	Return 0 swaps, indicating no swaps are needed because no couples are present.
Row has an odd number of people	Throw an IllegalArgumentException, because each person must have a partner
Maximum row size exceeds memory capacity (very large N)	The solution must be designed to avoid excessive memory usage, potentially by processing the row in chunks or using an alternative data structure if N is excessively large.
A person is partnered with themselves (row[i] == i)	This should be handled correctly as no swaps are needed for that pair, and the algorithm should proceed to the next pair.
All couples are initially seated correctly (no swaps needed)	The algorithm correctly identifies that no swaps are necessary and returns 0.
The row contains non-integer values or integers outside the valid range (0 to N-1, inclusive).	Throw an IllegalArgumentException since the input violates problem constraints
Integer overflow when calculating row size if the input is read from a file	Use long data type when calculating row size and check for overflows, throwing exception when size exceeds maximum allowed.
The same person ID (number) appears more than twice in the row	Throw an IllegalArgumentException, as each person should appear exactly twice or not at all.

Input row is null or empty

How to Handle:

Return 0 swaps, indicating no swaps are needed because no couples are present.

Row has an odd number of people

How to Handle:

Throw an IllegalArgumentException, because each person must have a partner

Maximum row size exceeds memory capacity (very large N)

How to Handle:

The solution must be designed to avoid excessive memory usage, potentially by processing the row in chunks or using an alternative data structure if N is excessively large.

A person is partnered with themselves (row[i] == i)

How to Handle:

This should be handled correctly as no swaps are needed for that pair, and the algorithm should proceed to the next pair.

All couples are initially seated correctly (no swaps needed)

How to Handle:

The algorithm correctly identifies that no swaps are necessary and returns 0.

The row contains non-integer values or integers outside the valid range (0 to N-1, inclusive).

How to Handle:

Throw an IllegalArgumentException since the input violates problem constraints

Integer overflow when calculating row size if the input is read from a file

How to Handle:

Use long data type when calculating row size and check for overflows, throwing exception when size exceeds maximum allowed.

The same person ID (number) appears more than twice in the row

How to Handle:

Throw an IllegalArgumentException, as each person should appear exactly twice or not at all.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the size of the input array `row`? Are there any limits or constraints on the number of couples or the total number of seats?
Does each number from 0 to N-1 (where N is the number of couples) appear exactly twice in the input array?
If no swaps are needed (the couples are already paired correctly), should I return 0?
What data type are the elements in the `row` array? Can I assume they are non-negative integers?
If there are multiple valid sets of swaps that achieve the desired pairing, is any valid set acceptable?

Brute Force Solution

Approach

The brute force strategy for this problem involves trying every possible swap of people until all couples are holding hands. We essentially try every single configuration to see if it works.

Here's how the algorithm would work step-by-step:

Look at the first person. Check if they are holding hands with their partner.
If they are not, find their partner somewhere else in the line.
Try swapping the person next to the first person with the partner of the first person.
Check if this swap makes the first couple hold hands. If it does not, put the original configuration back.
Repeat the swapping process with every other person in the line until you find the partner or you reach the end of the line.
Keep track of how many swaps you made to get the first couple to hold hands.
Now move on to the next couple and repeat the same process.
Once you have all couples holding hands, record the total number of swaps you made.
Repeat this entire process, starting with different initial swaps, to explore all possible swap combinations.
After exploring all possible combinations, choose the arrangement with the fewest swaps.

Code Implementation

def couples_holding_hands_brute_force(row):
    number_of_people = len(row)
    minimum_swaps = float('inf')

    def calculate_swaps(current_row):
        swaps = 0
        for i in range(0, number_of_people, 2):
            person_one = current_row[i]
            person_two = current_row[i + 1]

            if person_one // 2 != person_two // 2:
                # If not a couple, find the partner.
                for j in range(i + 2, number_of_people):
                    if current_row[j] == person_one ^ 1:
                        # Swap current and found person.
                        current_row[i + 1], current_row[j] = current_row[j], current_row[i + 1]
                        swaps += 1
                        break
        return swaps

    import itertools
    # Trying all possible permutations is the core of brute force
    for permutation in itertools.permutations(row):
        current_swaps = calculate_swaps(list(permutation))
        minimum_swaps = min(minimum_swaps, current_swaps)

    # Return the minimum swaps from all the permutations
    return minimum_swaps

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

The core idea is to fix mismatched couples one at a time using swaps. We aim to perform the minimum number of swaps to get all couples sitting together correctly.

Here's how the algorithm would work step-by-step:

Imagine each person has a partner number assigned based on their position; for example, people at positions 0 and 1 are a couple, 2 and 3 are a couple, and so on.
Go through the seats one pair at a time. If the two people sitting in the current pair of seats are a couple already, great, move on to the next pair.
If the two people are not a couple, figure out who the partner of the person in the first seat should be.
Find where that partner is currently sitting.
Swap the person in the second seat of our current pair with the partner we just found.
Count that swap. By swapping the correct partner into position, we've fixed one couple.
Repeat the process for all pairs of seats. The total number of swaps is the answer.

Code Implementation

def minSwapsCouples(row):
    number_of_people = len(row)
    number_of_couples = number_of_people // 2
    swaps = 0

    for couple_index in range(number_of_couples):
        first_person_index = 2 * couple_index
        second_person_index = 2 * couple_index + 1

        first_person = row[first_person_index]
        second_person = row[second_person_index]

        # If they are not a couple, we need to perform a swap.
        if not are_couple(first_person, second_person):
            partner = find_partner(first_person)

            # Find the index of the correct partner.
            partner_index = row.index(partner)

            # Swap the person with the correct partner.
            row[second_person_index], row[partner_index] = row[partner_index], row[second_person_index]
            swaps += 1

    return swaps

def are_couple(person_a, person_b):
    return (person_a ^ 1) == person_b

def find_partner(person):
    if person % 2 == 0:
        return person + 1
    else:
        return person - 1

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Input row is null or empty	Return 0 swaps, indicating no swaps are needed because no couples are present.
Row has an odd number of people	Throw an IllegalArgumentException, because each person must have a partner
Maximum row size exceeds memory capacity (very large N)	The solution must be designed to avoid excessive memory usage, potentially by processing the row in chunks or using an alternative data structure if N is excessively large.
A person is partnered with themselves (row[i] == i)	This should be handled correctly as no swaps are needed for that pair, and the algorithm should proceed to the next pair.
All couples are initially seated correctly (no swaps needed)	The algorithm correctly identifies that no swaps are necessary and returns 0.
The row contains non-integer values or integers outside the valid range (0 to N-1, inclusive).	Throw an IllegalArgumentException since the input violates problem constraints
Integer overflow when calculating row size if the input is read from a file	Use long data type when calculating row size and check for overflows, throwing exception when size exceeds maximum allowed.
The same person ID (number) appears more than twice in the row	Throw an IllegalArgumentException, as each person should appear exactly twice or not at all.

Input row is null or empty

How to Handle:

Return 0 swaps, indicating no swaps are needed because no couples are present.

Row has an odd number of people

How to Handle:

Throw an IllegalArgumentException, because each person must have a partner

Maximum row size exceeds memory capacity (very large N)

How to Handle:

The solution must be designed to avoid excessive memory usage, potentially by processing the row in chunks or using an alternative data structure if N is excessively large.

A person is partnered with themselves (row[i] == i)

How to Handle:

This should be handled correctly as no swaps are needed for that pair, and the algorithm should proceed to the next pair.

All couples are initially seated correctly (no swaps needed)

How to Handle:

The algorithm correctly identifies that no swaps are necessary and returns 0.

The row contains non-integer values or integers outside the valid range (0 to N-1, inclusive).

How to Handle:

Throw an IllegalArgumentException since the input violates problem constraints

Integer overflow when calculating row size if the input is read from a file

How to Handle:

Use long data type when calculating row size and check for overflows, throwing exception when size exceeds maximum allowed.

The same person ID (number) appears more than twice in the row

How to Handle:

Throw an IllegalArgumentException, as each person should appear exactly twice or not at all.