Valid Arrangement of Pairs

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You are given a 0-indexed 2D integer array pairs where pairs[i] = [start_i, end_i]. An arrangement of pairs is valid if for every index i where 1 <= i < pairs.length, we have end_i-1 == start_i.

Return any valid arrangement of pairs.

Note: The inputs will be generated such that there exists a valid arrangement of pairs.

Example 1:

Input: pairs = [[5,1],[4,5],[11,9],[9,4]]
Output: [[11,9],[9,4],[4,5],[5,1]]
Explanation:
This is a valid arrangement since end_i-1 always equals start_i.
end₀ = 9 == 9 = start₁ 
end₁ = 4 == 4 = start₂
end₂ = 5 == 5 = start₃

Example 2:

Input: pairs = [[1,3],[3,2],[2,1]]
Output: [[1,3],[3,2],[2,1]]
Explanation:
This is a valid arrangement since end_i-1 always equals start_i.
end₀ = 3 == 3 = start₁
end₁ = 2 == 2 = start₂
The arrangements [[2,1],[1,3],[3,2]] and [[3,2],[2,1],[1,3]] are also valid.

Example 3:

Input: pairs = [[1,2],[1,3],[2,1]]
Output: [[1,2],[2,1],[1,3]]
Explanation:
This is a valid arrangement since end_i-1 always equals start_i.
end₀ = 2 == 2 = start₁
end₁ = 1 == 1 = start₂

Constraints:

1 <= pairs.length <= 10⁵
pairs[i].length == 2
0 <= start_i, end_i <= 10⁹
start_i != end_i
No two pairs are exactly the same.
There exists a valid arrangement of pairs.

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 expected behavior if no valid arrangement exists? Should I return an empty list, null, or throw an exception?
Are the input pairs represented as a list of tuples/lists or a different data structure, and what are the constraints on the values within each pair (e.g., non-negative integers)?
If multiple valid arrangements exist, is any valid arrangement acceptable, or is there a specific criterion for selecting one (e.g., lexicographically smallest)?
Is it guaranteed that all numbers will appear in at least one pair, or might there be isolated numbers not belonging to any pair?
Is the graph formed by the pairs guaranteed to be connected, or could there be multiple disconnected components?

Brute Force Solution

Approach

The brute force approach involves trying out all possible arrangements of the given pairs. It systematically explores every combination to find one that satisfies the arrangement rules, without using clever tricks or shortcuts.

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

Consider all possible orderings of the pairs.
For each possible ordering, check if the arrangement is valid based on the pairs' relationship: the second element of one pair must match the first element of the next pair in the sequence.
If a sequence is valid, then remember it.
Continue checking every single possible ordering until we've gone through all of them.
Finally, if we found at least one valid arrangement, return one of them. If we found no valid arrangement, then there is no solution.

Code Implementation

import itertools

def find_valid_arrangement_brute_force(pairs):
    # Generate all possible permutations of the input pairs.
    possible_arrangements = list(itertools.permutations(pairs))

    for arrangement in possible_arrangements:
        is_valid = True

        # Check if the current arrangement is valid.
        for i in range(len(arrangement) - 1):
            if arrangement[i][1] != arrangement[i+1][0]:
                is_valid = False
                break

        # If arrangement is valid, return it.
        if is_valid:
            return list(arrangement)

    # No valid arrangement found.
    return []

Big(O) Analysis

Time Complexity

O(n! * n) – The brute force approach considers all possible orderings (permutations) of the n pairs. Generating all permutations of n elements takes O(n!) time. For each of these n! permutations, we need to check if the arrangement is valid. Checking the validity of a single permutation involves iterating through the n pairs and comparing adjacent elements. Therefore, checking validity for a single permutation takes O(n) time. The overall time complexity is thus O(n! * n), as we generate n! permutations and perform O(n) operations for each.

Space Complexity

O(N!) – The brute force approach explores all possible orderings of the N pairs. Storing a single permutation requires O(N) space. Since it needs to check and potentially store a valid arrangement out of all N! possible orderings in the worst-case scenario before finding a solution, the space complexity is dominated by the number of permutations to check. Thus, the auxiliary space complexity is O(N!).

Optimal Solution

Approach

The key is to visualize the pairs as edges in a graph. By cleverly traversing this graph, we can efficiently construct the valid arrangement without needing to explore all possible permutations.

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

First, imagine each number in the pairs as a point, and each pair as a connection between two points.
Figure out if there is a starting point where you can travel all the connections without lifting your pen, and without revisiting a connection.
If there are points with an odd number of connections, pick one of them as your start and another as your end.
If all the points have an even number of connections, pick any point as a start.
Now, begin from that starting point, and follow the connections, but remove the connection you're currently using.
Keep going until you can't go any further from the current point.
Once you are stuck, go back and look for other connections that haven't been used up yet, and repeat the process.
Link the journeys that you created into a final journey, making sure that the start and end points are connected correctly.
The sequence of numbers in your journey represents a valid arrangement of the pairs.

Code Implementation

def find_valid_arrangement(pairs):
    graph = {}
    in_degree = {}
    out_degree = {}

    for start_node, end_node in pairs:
        graph.setdefault(start_node, []).append(end_node)
        in_degree[end_node] = in_degree.get(end_node, 0) + 1
        out_degree[start_node] = out_degree.get(start_node, 0) + 1
        if start_node not in in_degree:
            in_degree[start_node] = 0
        if end_node not in out_degree:
            out_degree[end_node] = 0

    start_node_to_use = None
    end_node_to_use = None

    for node in in_degree.keys() | out_degree.keys():
        if out_degree.get(node, 0) - in_degree.get(node, 0) == 1:
            start_node_to_use = node
        elif in_degree.get(node, 0) - out_degree.get(node, 0) == 1:
            end_node_to_use = node

    # Determine a starting node for traversal
    if start_node_to_use is None:
        start_node_to_use = next(iter(graph.keys()), None)

    result = []

    def traverse(current_node):
        while graph.get(current_node):
            next_node = graph[current_node].pop()
            traverse(next_node)
        result.append(current_node)

    if start_node_to_use is not None:
        traverse(start_node_to_use)

    # Reverse the result to get the correct order
    result.reverse()

    final_arrangement = []
    for i in range(len(result) - 1):

        # Store the ordered pair correctly
        final_arrangement.append([result[i], result[i+1]])

    return final_arrangement

Big(O) Analysis

Time Complexity

O(n + m) – The algorithm constructs a graph from the input pairs, which takes O(m) time where m is the number of pairs (edges). Finding the start node involves iterating through all nodes (numbers in the pairs), taking O(n) time in the worst case, where n is the number of distinct numbers. The Euler path traversal visits each edge exactly once, effectively processing all pairs, which is O(m). Combining these operations, the overall time complexity is O(n + m), where n is the number of unique numbers and m is the number of pairs.

Space Complexity

O(N) – The space complexity is dominated by the adjacency list representation of the graph and the resulting path. The adjacency list, implicitly used for storing connections, can potentially store each pair's elements. Therefore, the space used by the adjacency list scales with the number of pairs, N. The stack space used during the depth-first traversal can also reach a depth proportional to N in the worst-case scenario where the graph is a single long chain. Consequently, the auxiliary space complexity is O(N).

Edge Cases

Case	How to Handle
Empty input array	Return an empty array or null to indicate no valid arrangement possible.
Input array with only one pair	The algorithm should correctly identify the single pair as the valid arrangement.
Input array representing a disconnected graph	The algorithm must find a valid starting node in one of the connected components and correctly traverse only that component.
Input array representing a graph with multiple possible starting nodes (nodes with outdegree > indegree)	The algorithm needs a deterministic way (e.g. smallest value) to choose a starting node among all possible candidates.
Input array where some numbers appear multiple times, creating multiple edges between nodes	The Euler path algorithm must handle multi-edges in the graph formed by the pairs.
Input array where a valid Eulerian path exists, but no node has outdegree exceeding indegree	The algorithm must correctly identify an appropriate starting node, which will be a node with (outdegree - indegree == 0)
Input array representing a graph with cycles	The algorithm should still be able to find the Eulerian path despite the presence of cycles.
Large input array exceeding memory limits	Optimize the algorithm to minimize memory usage, possibly using iterative approaches instead of recursive ones to avoid stack overflow.

Empty input array

How to Handle:

Return an empty array or null to indicate no valid arrangement possible.

Input array with only one pair

How to Handle:

The algorithm should correctly identify the single pair as the valid arrangement.

Input array representing a disconnected graph

How to Handle:

The algorithm must find a valid starting node in one of the connected components and correctly traverse only that component.

Input array representing a graph with multiple possible starting nodes (nodes with outdegree > indegree)

How to Handle:

The algorithm needs a deterministic way (e.g. smallest value) to choose a starting node among all possible candidates.

Input array where some numbers appear multiple times, creating multiple edges between nodes

How to Handle:

The Euler path algorithm must handle multi-edges in the graph formed by the pairs.

Input array where a valid Eulerian path exists, but no node has outdegree exceeding indegree

How to Handle:

The algorithm must correctly identify an appropriate starting node, which will be a node with (outdegree - indegree == 0)

Input array representing a graph with cycles

How to Handle:

The algorithm should still be able to find the Eulerian path despite the presence of cycles.

Large input array exceeding memory limits

How to Handle:

Optimize the algorithm to minimize memory usage, possibly using iterative approaches instead of recursive ones to avoid stack overflow.