LeetCode_210_Course_Schedule_II.cpp

/*
  210. Course Schedule II
  There are a total of n courses you have to take, labeled from 0 to n-1.
  
  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]
  Given the total number of courses and a list of prerequisite pairs, return the
  ordering of courses you should take to finish all courses.

  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.

  Example 1:
  Input: 2, [[1,0]] 
  Output: [0,1]
  Explanation: There are a total of 2 courses to take. To take course 1 you should have finished   
  course 0. So the correct course order is [0,1] .

  Example 2:
  Input: 4, [[1,0],[2,0],[3,1],[3,2]]
  Output: [0,1,2,3] or [0,2,1,3]
  Explanation: There are a total of 4 courses to take. To take course 3 you should have finished both     
  courses 1 and 2. Both courses 1 and 2 should be taken after you finished course 0. 
  So one correct course order is [0,1,2,3]. Another correct ordering is [0,2,1,3] .

  Note:
  The input prerequisites is a graph represented by a list of edges, not adjacency matrices.
*/

vector<int> findOrder( int numCourses, vector<vector<int>>& prerequisites ) {
  /*Build adjacenecy list and indegree vector */
  vector<int> indegrees( numCourses, 0 );
  unordered_map< int, set<int> > adjList;
  for( auto pre : prerequisites ) {
    adjList[ pre[1] ].insert( pre[0] );
    ++indegrees[ pre[0] ];
  }

  /*Push all nodes with indegree = 0 to queue*/
  queue<int> sources;
  for( int i=0; i < numCourses; i++ )
    if( indegrees[i] == 0 ) sources.push( i );

  /*For each node in queue, process all its neighbours and 
     reduce their indegree by 1, if indegree of nbr becomes 0
     then add them to queue. */
  vector<int> courseOrder;
  while( !sources.empty() ) {
    int curr = sources.front(); sources.pop();
    courseOrder.push_back( curr );
    for( auto nbr : adjList[curr] ) {
      --indegrees[nbr];
      if( indegrees[nbr] == 0 ) 
	sources.push( nbr );
    }
  }  
  return courseOrder.size() == numCourses ? courseOrder : vector<int>();
}