Introduction

In this blog post, we will explore how to solve the LeetCode problem "2097" 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

You are given a 0-indexed 2D integer array pairs where pairs[i] = [starti, endi]. An arrangement of pairs is valid if for every index i where 1 <= i < pairs.length, we have endi-1 == starti. 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 endi-1 always equals starti. 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 endi-1 always equals starti. 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 endi-1 always equals starti. end0 = 2 == 2 = start1 end1 = 1 == 1 = start2 Constraints: 1 <= pairs.length <= 105 pairs[i].length == 2 0 <= starti, endi <= 109 starti != endi No two pairs are exactly the same. There exists a valid arrangement of pairs.

Explanation

Here's the breakdown of the solution:

• Eulerian Path/Cycle: The problem can be modeled as finding an Eulerian path in a directed graph. Each pair [start, end] represents a directed edge from start to end. A valid arrangement corresponds to an Eulerian path. Since a valid arrangement is guaranteed to exist, we are certain to find an Eulerian path.

• Hierholzer's Algorithm: We use Hierholzer's algorithm, a classic algorithm for finding Eulerian paths/cycles efficiently. The algorithm avoids explicit graph representation and uses a stack to keep track of the path.

• Iterative Depth-First Search: The core of Hierholzer's algorithm is a depth-first search (DFS) that iteratively explores the graph, removing edges as they are visited. This ensures that we only visit each edge once, leading to an efficient solution.

• Time & Space Complexity:

  • Time Complexity: O(n), where n is the number of pairs.
  • Space Complexity: O(n), due to the stack and the adjacency list (implemented as a dictionary).

Code

    def find_arrangement(pairs):
    """
    Finds a valid arrangement of pairs such that endi-1 == starti for all i.

    Args:
        pairs: A list of pairs (lists of two integers) where pairs[i] = [starti, endi].

    Returns:
        A list of pairs representing a valid arrangement.
    """

    graph = {}
    for start, end in pairs:
        if start not in graph:
            graph[start] = []
        graph[start].append(end)

    start_node = pairs[0][0]  # Start with the first start node

    stack = [start_node]
    path = []

    while stack:
        node = stack[-1]
        if node in graph and graph[node]:
            next_node = graph[node].pop()
            stack.append(next_node)
        else:
            path.append(stack.pop())

    # Reconstruct the pairs from the path
    result = []
    for i in range(len(path) - 1):
        start = path[i+1]
        end = path[i]
        for pair in pairs:
            if pair[0] == start and pair[1] == end:
                result.append(pair)
                pairs.remove(pair)  # avoid duplicates if any
                break

    return result