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feat(leetcode): 2097 Valid Arrangement of Pairs
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‎leetcode/2097-valid-arrangement-of-pairs/README.md‎

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## [2097. Valid Arrangement of Pairs](https://leetcode.com/problems/valid-arrangement-of-pairs/)
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<p>You are given a <strong>0-indexed</strong> 2D integer array <code>pairs</code> where <code>pairs[i] = [start<sub>i</sub>, end<sub>i</sub>]</code>. An arrangement of <code>pairs</code> is <strong>valid</strong> if for every index <code>i</code> where <code>1 &lt;= i &lt; pairs.length</code>, we have <code>end<sub>i-1</sub> == start<sub>i</sub></code>.</p>
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<p>Return <em><strong>any</strong> valid arrangement of </em><code>pairs</code>.</p>
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<p><strong>Note:</strong> The inputs will be generated such that there exists a valid arrangement of <code>pairs</code>.</p>
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<p>&nbsp;</p>
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<p><strong class="example">Example 1:</strong></p>
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<pre>
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<strong>Input:</strong> pairs = [[5,1],[4,5],[11,9],[9,4]]
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<strong>Output:</strong> [[11,9],[9,4],[4,5],[5,1]]
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<strong>Explanation:
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</strong>This is a valid arrangement since end<sub>i-1</sub> always equals start<sub>i</sub>.
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end<sub>0</sub> = 9 == 9 = start<sub>1</sub>
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end<sub>1</sub> = 4 == 4 = start<sub>2</sub>
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end<sub>2</sub> = 5 == 5 = start<sub>3</sub>
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</pre>
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<p><strong class="example">Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> pairs = [[1,3],[3,2],[2,1]]
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<strong>Output:</strong> [[1,3],[3,2],[2,1]]
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<strong>Explanation:</strong>
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This is a valid arrangement since end<sub>i-1</sub> always equals start<sub>i</sub>.
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end<sub>0</sub> = 3 == 3 = start<sub>1</sub>
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end<sub>1</sub> = 2 == 2 = start<sub>2</sub>
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The arrangements [[2,1],[1,3],[3,2]] and [[3,2],[2,1],[1,3]] are also valid.
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</pre>
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<p><strong class="example">Example 3:</strong></p>
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<pre>
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<strong>Input:</strong> pairs = [[1,2],[1,3],[2,1]]
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<strong>Output:</strong> [[1,2],[2,1],[1,3]]
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<strong>Explanation:</strong>
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This is a valid arrangement since end<sub>i-1</sub> always equals start<sub>i</sub>.
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end<sub>0</sub> = 2 == 2 = start<sub>1</sub>
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end<sub>1</sub> = 1 == 1 = start<sub>2</sub>
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</pre>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>1 &lt;= pairs.length &lt;= 10<sup>5</sup></code></li>
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<li><code>pairs[i].length == 2</code></li>
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<li><code>0 &lt;= start<sub>i</sub>, end<sub>i</sub> &lt;= 10<sup>9</sup></code></li>
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<li><code>start<sub>i</sub> != end<sub>i</sub></code></li>
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<li>No two pairs are exactly the same.</li>
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<li>There <strong>exists</strong> a valid arrangement of <code>pairs</code>.</li>
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</ul>
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## Hints
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1. Could you convert this into a graph problem?
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2. Consider the pairs as edges and each number as a node.
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3. We have to find an Eulerian path of this graph. Hierholzer’s algorithm can be used.

‎leetcode/2097-valid-arrangement-of-pairs/solution.cpp‎

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/**
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* Author: 1chooo<hugo970217@gmail.com>
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* Problem: https://leetcode.com/problems/valid-arrangement-of-pairs/
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* Runtime: 221ms (99.05%)
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*/
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const static auto _ = []() {
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cin.tie(nullptr)->sync_with_stdio(false);
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return nullptr;
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}();
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class Solution {
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public:
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vector<vector<int>> validArrangement(vector<vector<int>> &pairs) {
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unordered_map<int, vector<int>> adjacencyList;
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unordered_map<int, int> inOutDegree;
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for (const auto &pair : pairs) {
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adjacencyList[pair[0]].push_back(pair[1]);
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inOutDegree[pair[0]]++;
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inOutDegree[pair[1]]--;
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}
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// Find starting node
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int startNode = pairs[0][0];
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for (const auto &[node, degree] : inOutDegree) {
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if (degree == 1) {
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startNode = node;
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break;
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}
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}
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vector<int> path;
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stack<int> nodeStack;
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nodeStack.push(startNode);
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while (!nodeStack.empty()) {
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auto &neighbors = adjacencyList[nodeStack.top()];
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if (neighbors.empty()) {
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path.push_back(nodeStack.top());
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nodeStack.pop();
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} else {
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int nextNode = neighbors.back();
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nodeStack.push(nextNode);
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neighbors.pop_back();
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}
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}
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vector<vector<int>> arrangement;
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int pathSize = path.size();
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arrangement.reserve(pathSize - 1);
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for (int i = pathSize - 1; i > 0; --i) {
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arrangement.push_back({path[i], path[i - 1]});
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}
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return arrangement;
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}
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};
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/**
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* Author: 1chooo<hugo970217@gmail.com>
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* Problem: https://leetcode.com/problems/valid-arrangement-of-pairs/
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* Runtime: 1430ms (25.24%)
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*/
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const static auto _ = []() {
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cin.tie(nullptr)->sync_with_stdio(false);
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return nullptr;
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}();
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class Solution {
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private:
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void dfs(vector<int> &path, unordered_map<int, stack<int>> &graph, int start) {
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auto &neighbors = graph[start];
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while (!neighbors.empty()) {
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int next = neighbors.top();
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neighbors.pop();
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dfs(path, graph, next);
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}
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path.push_back(start);
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}
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public:
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vector<vector<int>> validArrangement(vector<vector<int>> &pairs) {
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unordered_map<int, stack<int>> graph;
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unordered_map<int, int> inDegree, outDegree;
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for (const auto &pair : pairs) {
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int u = pair[0], v = pair[1];
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graph[u].push(v);
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outDegree[u]++;
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inDegree[v]++;
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}
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int start = pairs[0][0];
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for (const auto &[node, _] : graph) {
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if (outDegree[node] > inDegree[node]) {
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start = node;
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break;
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}
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}
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vector<int> path;
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dfs(path, graph, start);
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reverse(path.begin(), path.end());
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vector<vector<int>> result;
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for (size_t i = 0; i < path.size() - 1; ++i) {
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result.push_back({path[i], path[i + 1]});
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}
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return result;
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}
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};

‎leetcode/2097-valid-arrangement-of-pairs/solution.go‎

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/**
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* Author: 1chooo<hugo970217@gmail.com>
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* Problem: https://leetcode.com/problems/valid-arrangement-of-pairs/
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* Runtime: ms (%)
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*/

‎leetcode/2097-valid-arrangement-of-pairs/solution.java‎

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/**
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* Author: 1chooo<hugo970217@gmail.com>
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* Problem: https://leetcode.com/problems/valid-arrangement-of-pairs/
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* Runtime: 221ms (99.05%)
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*/
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class Solution {
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public void dfs(List<Integer> pa, Map<Integer, Stack<Integer>> gr, int st) {
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Stack<Integer> nei = gr.getOrDefault(st, new Stack<>());
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while (!nei.isEmpty()) {
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int ne = nei.pop();
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dfs(pa, gr, ne);
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}
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pa.add(st);
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}
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public int[][] validArrangement(int[][] pairs) {
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Map<Integer, Stack<Integer>> gr = new HashMap<>();
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Map<Integer, Integer> in = new HashMap<>();
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Map<Integer, Integer> out = new HashMap<>();
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for (int[] a : pairs) {
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int u = a[0], v = a[1];
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gr.computeIfAbsent(u, k -> new Stack<>()).push(v);
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out.put(u, out.getOrDefault(u, 0) + 1);
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in.put(v, in.getOrDefault(v, 0) + 1);
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}
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int st = pairs[0][0];
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for (int k : gr.keySet()) {
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if (out.getOrDefault(k, 0) > in.getOrDefault(k, 0)) {
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st = k;
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break;
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}
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}
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List<Integer> path = new ArrayList<>();
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dfs(path, gr, st);
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Collections.reverse(path);
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int[][] res = new int[pairs.length][2];
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for (int i = 0; i < path.size() - 1; i++) {
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res[i][0] = path.get(i);
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res[i][1] = path.get(i + 1);
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}
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return res;
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}
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}

‎leetcode/2097-valid-arrangement-of-pairs/solution.py‎

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"""
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* Author: 1chooo<hugo970217@gmail.com>
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* Problem: https://leetcode.com/problems/valid-arrangement-of-pairs/
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* Runtime: ms (%)
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"""
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