MANG Problem #47: Valid Arrangement of Pairs (Hard)

1. Problem Statement

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. You may assume a valid arrangement exists.

Input: pairs = [[5,1],[4,5],[11,9],[9,4]]
Output: [[11,9],[9,4],[4,5],[5,1]]

2. Approach: Eulerian Path (Hierholzer's Algorithm)

Connecting dominoes like end == start is equivalent to finding a path in a graph that visits every edge exactly once. This is known as an Eulerian Path.

Build Graph & Degrees:
- Treat each pair [u, v] as a directed edge from u to v.
- Track the outDegree and inDegree of every node.
Find Start Node:
- In an Eulerian Path, the start node has outDegree - inDegree == 1.
- If all nodes have outDegree == inDegree, it's an Eulerian Circuit, and any node can be the start.
Hierholzer's Algorithm (DFS):
- Start DFS from the start node.
- Follow an outgoing edge, remove it from the graph, and recurse.
- When a node has no more outgoing edges, push it to a result list.
Reverse: The result list will have the path in reverse order. Reverse it to get the final valid arrangement.

3. Java Implementation

public int[][] validArrangement(int[][] pairs) {
    Map<Integer, Stack<Integer>> adj = new HashMap<>();
    Map<Integer, Integer> outDegree = new HashMap<>();
    Map<Integer, Integer> inDegree = new HashMap<>();

    // 1. Build Graph
    for (int[] pair : pairs) {
        int u = pair[0], v = pair[1];
        adj.computeIfAbsent(u, k -> new Stack<>()).push(v);
        outDegree.put(u, outDegree.getOrDefault(u, 0) + 1);
        inDegree.put(v, inDegree.getOrDefault(v, 0) + 1);
    }

    // 2. Find Start Node
    int startNode = pairs[0][0]; // Default start
    for (int node : outDegree.keySet()) {
        if (outDegree.get(node) - inDegree.getOrDefault(node, 0) == 1) {
            startNode = node;
            break;
        }
    }

    // 3. Hierholzer's DFS
    List<Integer> path = new ArrayList<>();
    dfs(adj, startNode, path);

    // 4. Construct Result (Reverse)
    int[][] res = new int[pairs.length][2];
    Collections.reverse(path);
    for (int i = 0; i < path.size() - 1; i++) {
        res[i][0] = path.get(i);
        res[i][1] = path.get(i + 1);
    }
    
    return res;
}

private void dfs(Map<Integer, Stack<Integer>> adj, int curr, List<Integer> path) {
    Stack<Integer> neighbors = adj.getOrDefault(curr, new Stack<>());
    while (!neighbors.isEmpty()) {
        int next = neighbors.pop();
        dfs(adj, next, path);
    }
    path.add(curr);
}

4. 5-Minute "Video-Style" Walkthrough

The "Aha!" Moment: It looks like Topological Sort, but it's not. Topo Sort visits every node once. We need to use every pair (edge) exactly once. This is the definition of an Eulerian Path.
The Start Node Logic: If a path goes straight through a node, it enters once and leaves once (in == out). The only exception is the very start of the entire chain (it leaves one more time than it enters: out - in = 1) and the very end (in - out = 1).
The Post-Order Trap: Why do we add to path after the while loop? Because we might take a sub-cycle that returns to the current node. By adding nodes after exploring all edges, we guarantee that dead ends are appended first (which become the end of the path when reversed).

5. Interview Discussion