Valid Arrangement of Pairs

• 1. Define a recursive helper function permute(list, start_index) to generate all permutations of the input pairs.
• 2. The base case for the recursion is when start_index reaches the end of the list. At this point, a complete permutation has been generated.
• 3. For each generated permutation, check if it's a valid arrangement by iterating from the second pair and ensuring pair[i-1][1] == pair[i][0].
• 4. Since a valid arrangement is guaranteed to exist, the first one found is the answer. Store it and terminate the search.
• 5. The recursive step involves swapping the element at start_index with all subsequent elements to explore different orderings, and backtracking after the recursive call.

The core idea is to explore all N! permutations of the pairs array, where N is the number of pairs. We can implement a recursive function that generates these permutations. The function, say generatePermutations(index, currentPermutation), would place each available pair at the current index and recurse for the next index. Once a full permutation of size N is formed, we validate it. Validation involves iterating from the second pair to the end and checking the condition pairs[i-1][1] == pairs[i][0]. Since the problem guarantees that a valid arrangement exists, the first one we find is a valid answer. However, this approach is computationally infeasible for the given constraints.

1// Conceptual structure for brute-force. This will time out.2class Solution {3    int[][] result;4    public int[][] validArrangement(int[][] pairs) {5        List<int[]> pairList = new ArrayList<>();6        for (int[] p : pairs) {7            pairList.add(p);8        }9        permute(pairList, 0);10        return result;11    }12 13    private void permute(List<int[]> arr, int k) {14        if (result != null) return; // Already found a solution15        if (k == arr.size()) {16            if (isValid(arr)) {17                result = new int[arr.size()][2];18                for (int i = 0; i < arr.size(); i++) {19                    result[i] = arr.get(i);20                }21            }22            return;23        }24        for (int i = k; i < arr.size(); i++) {25            Collections.swap(arr, i, k);26            permute(arr, k + 1);27            Collections.swap(arr, k, i); // backtrack28        }29    }30 31    private boolean isValid(List<int[]> arrangement) {32        for (int i = 1; i < arrangement.size(); i++) {33            if (arrangement.get(i - 1)[1] != arrangement.get(i)[0]) {34                return false;35            }36        }37        return true;38    }39}

• 1. Pre-process the pairs to build a map where keys are starting numbers and values are lists of pairs that begin with that number. This allows for efficient lookups of next possible pairs.
• 2. Use a boolean array used to track which pairs (by their original index) have been included in the current path.
• 3. Define a recursive function backtrack(lastPair, currentPath).
• 4. The base case: if currentPath.size() equals the total number of pairs, a valid arrangement has been found. Store it and return true.
• 5. In the recursive step, get the end value of the lastPair. Find all unused pairs that start with this end value.
• 6. For each such candidate pair, add it to the path, mark it as used, and make a recursive call.
• 7. If the recursive call returns true, it means a solution was found down that branch, so propagate true upwards.
• 8. If the call returns false, backtrack by removing the pair from the path and unmarking it as used.
• 9. To start the process, iterate through all pairs, trying each one as the first pair in the arrangement.

We can model this as finding a path of length N in a state-space graph. We start by trying each pair as the first pair in the arrangement. Then, we recursively search for the next pair. If the current path ends with pair p1, the next pair p2 must have p2.start == p1.end. We use a boolean array or a bitmask to keep track of which pairs have already been included in the current path to avoid reusing pairs. The search backtracks if it hits a dead end (i.e., no unused pair can be appended). The search terminates successfully when a path of length N (using all pairs) is formed.

1// Conceptual structure for backtracking. This is also likely to time out.2class Solution {3    int[][] result;4    int n;5    int[][] pairs;6 7    public int[][] validArrangement(int[][] pairs) {8        this.n = pairs.length;9        this.pairs = pairs;10        boolean[] used = new boolean[n];11        List<int[]> path = new ArrayList<>();12 13        // Try starting with each pair14        for (int i = 0; i < n; i++) {15            path.add(pairs[i]);16            used[i] = true;17            if (backtrack(path, used)) {18                return result;19            }20            used[i] = false;21            path.remove(path.size() - 1);22        }23        return null; // Should not be reached24    }25 26    private boolean backtrack(List<int[]> path, boolean[] used) {27        if (path.size() == n) {28            result = path.toArray(new int[n][]);29            return true;30        }31 32        int[] lastPair = path.get(path.size() - 1);33        int endNode = lastPair[1];34 35        for (int i = 0; i < n; i++) {36            if (!used[i] && pairs[i][0] == endNode) {37                used[i] = true;38                path.add(pairs[i]);39                if (backtrack(path, used)) {40                    return true;41                }42                path.remove(path.size() - 1);43                used[i] = false; // backtrack44            }45        }46 47        return false;48    }49}

• 1. Build Graph and Degrees:
  • Create an adjacency list adj (e.g., Map<Integer, Deque<Integer>>) to store the graph.
  • Create a degree map degree (e.g., Map<Integer, Integer>) to store out-degree - in-degree for each node.
  • Iterate through each pair [u, v]: add an edge u -> v to adj, increment degree[u], and decrement degree[v].
• 2. Find Start Node:
  • The start node of the Eulerian path is the unique node with degree == 1.
  • Iterate through the degree map to find this node.
  • If no such node exists, the graph has an Eulerian circuit. Any node with outgoing edges can be the start node (e.g., pairs[0][0])
• 3. Hierholzer's Algorithm (Iterative DFS):
  • Use a stack for the traversal and a path list to store the result.
  • Push the startNode onto the stack.
  • While the stack is not empty:
    • Peek at the current node u on top of the stack.
    • If u has any unvisited outgoing edges (i.e., adj.get(u) is not empty), pop an edge u -> v, and push v onto the stack.
    • If u has no more outgoing edges, pop u from the stack and add it to the path list.
• 4. Construct Final Result:
  • The path list is generated in reverse post-order. Reverse it to get the correct vertex sequence.
  • Create the final int[][] arrangement by taking consecutive nodes from the path list: [path.get(i), path.get(i+1)].

1. Graph Representation: We first build a graph. An adjacency list, represented by a Map<Integer, Deque<Integer>>, is suitable. The keys are the source vertices, and the values are deques of destination vertices. We also compute the in-degrees and out-degrees of all vertices to find the starting point of the path. A Map<Integer, Integer> degree can store out-degree - in-degree.

2. Finding the Start Node: An Eulerian path starts at a vertex v where out-degree(v) - in-degree(v) = 1. If no such vertex exists, the graph has an Eulerian circuit, and any vertex can be a starting point. We iterate through our degree map to find this special start node. If not found, we can default to pairs[0][0].

3. Path Finding (Hierholzer's Algorithm): We use a post-order DFS traversal. An iterative approach with a stack is common.

• Push the start node onto a stack.
• While the stack is not empty, look at the top vertex u.
• If u has unvisited outgoing edges, pick one (e.g., u -> v), push v onto the stack, and remove the edge u -> v from the graph to mark it as visited.
• If u has no more outgoing edges, pop it from the stack and add it to a result list path.

4. Final Arrangement: The path list we constructed contains the sequence of vertices in reverse order. We reverse it to get the correct path. Then, we construct the final int[][] result by creating pairs [path[i], path[i+1]] for i from 0 to path.length - 2.

1import java.util.*;2 3class Solution {4    public int[][] validArrangement(int[][] pairs) {5        int n = pairs.length;6        Map<Integer, Deque<Integer>> adj = new HashMap<>();7        Map<Integer, Integer> degree = new HashMap<>();8 9        for (int[] pair : pairs) {10            int u = pair[0];11            int v = pair[1];12            adj.computeIfAbsent(u, k -> new ArrayDeque<>()).add(v);13            degree.put(u, degree.getOrDefault(u, 0) + 1);14            degree.put(v, degree.getOrDefault(v, 0) - 1);15        }16 17        int startNode = -1;18        for (Map.Entry<Integer, Integer> entry : degree.entrySet()) {19            if (entry.getValue() == 1) {20                startNode = entry.getKey();21                break;22            }23        }24 25        if (startNode == -1) {26            // It's an Eulerian circuit, start from any node with an edge27            startNode = pairs[0][0];28        }29 30        List<Integer> path = new ArrayList<>();31        Deque<Integer> stack = new ArrayDeque<>();32        stack.push(startNode);33 34        while (!stack.isEmpty()) {35            int u = stack.peek();36            if (adj.containsKey(u) && !adj.get(u).isEmpty()) {37                int v = adj.get(u).pop();38                stack.push(v);39            } else {40                path.add(stack.pop());41            }42        }43 44        Collections.reverse(path);45 46        int[][] result = new int[n][2];47        for (int i = 0; i < n; i++) {48            result[i][0] = path.get(i);49            result[i][1] = path.get(i + 1);50        }51 52        return result;53    }54}