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sureshmangssureshmangs
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🚀 19-Jul-2020
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‎algorithms/graph/topological_sorting.cpp

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Original file line numberDiff line numberDiff line change
@@ -113,7 +113,7 @@ vector<int> topoSort(int V, vector<int> adj[]) {
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vector<bool> vis(V, false);
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for(int i=0;i<V;i++){
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if(!vis[i])
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dfs(adj, 0, vis);
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dfs(adj, i, vis);
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}
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vector<int>res;
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while(!s.empty()){

‎competitive programming/leetcode/2020-July-Challenge/Day-18-Course Schedule II.cpp

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There are a total of n courses you have to take, labeled from 0 to n-1.
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Some courses may have prerequisites, for example to take course 0 you have to first take course 1, which is expressed as a pair: [0,1]
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Given the total number of courses and a list of prerequisite pairs, return the ordering of courses you should take to finish all courses.
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There may be multiple correct orders, you just need to return one of them. If it is impossible to finish all courses, return an empty array.
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Example 1:
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Input: 2, [[1,0]]
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Output: [0,1]
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Explanation: There are a total of 2 courses to take. To take course 1 you should have finished
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course 0. So the correct course order is [0,1] .
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Example 2:
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Input: 4, [[1,0],[2,0],[3,1],[3,2]]
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Output: [0,1,2,3] or [0,2,1,3]
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Explanation: There are a total of 4 courses to take. To take course 3 you should have finished both
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courses 1 and 2. Both courses 1 and 2 should be taken after you finished course 0.
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So one correct course order is [0,1,2,3]. Another correct ordering is [0,2,1,3] .
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Note:
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The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about how a graph is represented.
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You may assume that there are no duplicate edges in the input prerequisites.
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Hide Hint #1
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This problem is equivalent to finding the topological order in a directed graph. If a cycle exists, no topological ordering exists and therefore it will be impossible to take all courses.
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Hide Hint #2
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Topological Sort via DFS - A great video tutorial (21 minutes) on Coursera explaining the basic concepts of Topological Sort.
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Hide Hint #3
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Topological sort could also be done via BFS.
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class Solution {
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public:
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stack<int> s;
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void dfs(vector<int> adj[],int node, vector<bool> &vis){
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vis[node]=true;
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for(auto x: adj[node]){
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if(!vis[x])
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dfs(adj, x, vis);
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}
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s.push(node);
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}
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vector<int> topoSort(int V, vector<int> adj[]) {
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vector<bool> vis(V, false);
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for(int i=0;i<V;i++){
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if(!vis[i])
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dfs(adj, i, vis);
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}
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vector<int>res;
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while(!s.empty()){
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res.push_back(s.top());
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s.pop();
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}
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return res;
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}
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bool detectCycleUtil(vector<int>adj[], int node, vector<bool> &vis, vector<bool> &rec){
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if(!vis[node]){
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vis[node]=true;
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rec[node]=true;
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}
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for(auto x: adj[node]){
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if(!vis[x] && detectCycleUtil(adj, x, vis, rec)) return true;
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else if(rec[x]==true) return true;
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}
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rec[node]=false;
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return false;
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}
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bool detectCycle(int V, vector<int> adj[]){
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vector<bool> vis(V, false);
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vector<bool> rec(V, false);
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for(int i=0;i<V;i++){
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if(!vis[i]){
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if(detectCycleUtil(adj, i, vis, rec)) return true; // cycle exists
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}
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}
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return false;
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}
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vector<int> findOrder(int numCourses, vector<vector<int>>& prerequisites) {
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vector<int> adj[numCourses];
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int n=prerequisites.size();
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for(int i=0;i<n;i++){
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int u=prerequisites[i][1];
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int v=prerequisites[i][0];
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adj[u].push_back(v);
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}
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vector<int> res;
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if(detectCycle(numCourses, adj)) return res; // not a DAG
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res=topoSort(numCourses, adj);
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return res;
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}
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};

‎competitive programming/leetcode/210. Course Schedule II.cpp

Lines changed: 112 additions & 0 deletions
Original file line numberDiff line numberDiff line change
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There are a total of n courses you have to take, labeled from 0 to n-1.
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Some courses may have prerequisites, for example to take course 0 you have to first take course 1, which is expressed as a pair: [0,1]
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Given the total number of courses and a list of prerequisite pairs, return the ordering of courses you should take to finish all courses.
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There may be multiple correct orders, you just need to return one of them. If it is impossible to finish all courses, return an empty array.
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Example 1:
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Input: 2, [[1,0]]
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Output: [0,1]
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Explanation: There are a total of 2 courses to take. To take course 1 you should have finished
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course 0. So the correct course order is [0,1] .
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Example 2:
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Input: 4, [[1,0],[2,0],[3,1],[3,2]]
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Output: [0,1,2,3] or [0,2,1,3]
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Explanation: There are a total of 4 courses to take. To take course 3 you should have finished both
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courses 1 and 2. Both courses 1 and 2 should be taken after you finished course 0.
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So one correct course order is [0,1,2,3]. Another correct ordering is [0,2,1,3] .
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Note:
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The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about how a graph is represented.
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You may assume that there are no duplicate edges in the input prerequisites.
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class Solution {
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public:
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stack<int> s;
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void dfs(vector<int> adj[],int node, vector<bool> &vis){
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vis[node]=true;
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for(auto x: adj[node]){
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if(!vis[x])
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dfs(adj, x, vis);
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}
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s.push(node);
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}
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vector<int> topoSort(int V, vector<int> adj[]) {
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vector<bool> vis(V, false);
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for(int i=0;i<V;i++){
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if(!vis[i])
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dfs(adj, i, vis);
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}
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vector<int>res;
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while(!s.empty()){
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res.push_back(s.top());
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s.pop();
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}
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return res;
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}
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bool detectCycleUtil(vector<int>adj[], int node, vector<bool> &vis, vector<bool> &rec){
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if(!vis[node]){
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vis[node]=true;
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rec[node]=true;
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}
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for(auto x: adj[node]){
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if(!vis[x] && detectCycleUtil(adj, x, vis, rec)) return true;
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else if(rec[x]==true) return true;
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}
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rec[node]=false;
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return false;
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}
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bool detectCycle(int V, vector<int> adj[]){
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vector<bool> vis(V, false);
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vector<bool> rec(V, false);
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for(int i=0;i<V;i++){
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if(!vis[i]){
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if(detectCycleUtil(adj, i, vis, rec)) return true; // cycle exists
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}
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}
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return false;
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}
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vector<int> findOrder(int numCourses, vector<vector<int>>& prerequisites) {
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vector<int> adj[numCourses];
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int n=prerequisites.size();
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for(int i=0;i<n;i++){
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int u=prerequisites[i][1];
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int v=prerequisites[i][0];
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adj[u].push_back(v);
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}
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vector<int> res;
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if(detectCycle(numCourses, adj)) return res; // not a DAG
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res=topoSort(numCourses, adj);
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return res;
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}
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};

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