207. Course Schedule && 210. Course Schedule II && 310. Minimum Height Trees

207. Course Schedule

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, is it possible for you to finish all courses?

For example:

2, [[1,0]]

There are a total of 2 courses to take. To take course 1 you should have finished course 0. So it is possible.

2, [[1,0],[0,1]]

There are a total of 2 courses to take. To take course 1 you should have finished course 0, and to take course 0 you should also have finished course 1. So it is impossible.

Note:
The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about how a graph is represented.

Hints:

1. This problem is equivalent to finding if a cycle exists in a directed graph. If a cycle exists, no topological ordering exists and therefore it will be impossible to take all courses.
2. Topological Sort via DFS - A great video tutorial (21 minutes) on Coursera explaining the basic concepts of Topological Sort.
3. Topological sort could also be done via BFS.

 Depth-first Search Breadth-first Search Graph Topological Sort

 

Solution:

Easy BFS. Starting points are those who have indegree 0. Iteratively reduce the indegree of connected vertices by 1.

 

public class Solution {
    public boolean canFinish(int numCourses, int[][] prerequisites) {
        if(numCourses<=1)
            return true;
        
        //from -> List(to)
        //out -> List(in)
        ArrayList<Integer>[] adjacencyList = new ArrayList[numCourses];
        for(int i = 0; i<numCourses; ++i)
            adjacencyList[i] = new ArrayList<Integer>();
        
        int[] indegree = new int[numCourses];
        for(int[] prerequesite: prerequisites) {
            adjacencyList[prerequesite[1]].add(prerequesite[0]);
            ++indegree[prerequesite[0]];
        }
        
        //outmostLayer is the layer that we can start with
        ArrayList<Integer> outmostLayer = new ArrayList<Integer>();
        for(int i = 0; i<numCourses; ++i)
            if(indegree[i]==0)
                outmostLayer.add(i);
        
        while(outmostLayer.size()>0) {
            ArrayList<Integer> nextLayer = new ArrayList<Integer>();
            for(int from: outmostLayer) {
                for(int to: adjacencyList[from]) {
                    --indegree[to];
                    if(indegree[to]==0)
                        nextLayer.add(to);
                }
            }
            outmostLayer = nextLayer;
        }
        
        for(int i = 0; i<numCourses; ++i)
            if(indegree[i]!=0)
                return false;
        return true;
    }
}

 

 

DFS solution:

import java.lang.reflect.Array;

public class Solution {
    public boolean canFinish(int numCourses, int[][] prerequisites) {
        HashSet<Integer>[] adjacencyList = (HashSet<Integer>[])Array.newInstance(HashSet.class, numCourses);
        for(int i = 0; i<numCourses; ++i)
        {
            adjacencyList[i] = new HashSet<Integer>();
        }
        for(int[] prerequesite: prerequisites)
        {
            adjacencyList[prerequesite[1]].add(prerequesite[0]);
        }
            
        int[] visit = new int[numCourses]; 
        //0 means unvisited
        //1 means visited/good/no cycles
        //-1 means under current searching path, no determined yet.
        for(int i = 0; i<numCourses; ++i)
        {
            if(!DFS(adjacencyList, i, visit))
                return false;
        }
        return true;
    }
    
    private boolean DFS(HashSet<Integer>[] adjacencyList, int node, int[] visit)
    {
        if(visit[node] == 1)
            return true; //This node is good. There is no cycles.
        visit[node] = -1;
        
        for(int out: adjacencyList[node])
        {
            if(visit[out]==-1  // equals -1 mean a cycle is made from this node(out).
            || 
            !DFS(adjacencyList, out, visit)) // means children nodes have cycles.
                return false;     //Found cycles.
        }
        visit[node] = 1; //Set visit to 1 when you backtrack, meanning that this node and all children nodes are good.
        return true;
    }
}

 

210. Course Schedule II

Given the total number of courses and a list of prerequisite pairs, return the ordering of courses you should take to finish all courses.

Depth-first Search Breadth-first Search Graph Topological Sort

 

 

Almost the same as last problem.

public class Solution {
    public int[] findOrder(int numCourses, int[][] prerequisites) {
        if(numCourses == 0)
            return new int[0];
            
        int[] inbounds = new int[numCourses];
        List<List<Integer>> outBounds = new ArrayList<List<Integer>>();
        for(int i = 0; i<numCourses; ++i) {
            outBounds.add(new ArrayList<Integer>());
        }
        
        for(int[] pair: prerequisites) {
            int from  = pair[1];
            int to  = pair[0];
            ++inbounds[to];
            List<Integer> outs = outBounds.get(from);
            outs.add(to);
        }
        
        List<Integer> starts = new ArrayList<Integer>();
        for(int i = 0; i<numCourses; ++i)
            if(inbounds[i] == 0)
                starts.add(i);
        
        int[] ret = new int[numCourses];
        int i=0;
        while(starts.size()>0) {
            List<Integer> newStarts = new ArrayList<Integer>();
            for(Integer start: starts) {
                ret[i++] = start;
                for(Integer out: outBounds.get(start)) {
                    --inbounds[out];
                    if(inbounds[out] == 0)
                        newStarts.add(out);
                }
            }
            starts = newStarts;
        }
        
        for(int n = 0; n<numCourses; ++n) {
            if(inbounds[n]>0)
                return new int[0];
        }
        return ret;
    }
}

 

310. Minimum Height Trees

For a undirected graph with tree characteristics, we can choose any node as the root. The result graph is then a rooted tree. Among all possible rooted trees, those with minimum height are called minimum height trees (MHTs). Given such a graph, write a function to find all the MHTs and return a list of their root labels.

Format
The graph contains n nodes which are labeled from 0 to n - 1. You will be given the number n and a list of undirected edges (each edge is a pair of labels).

You can assume that no duplicate edges will appear in edges. Since all edges are undirected, [0, 1] is the same as [1, 0] and thus will not appear together in edges.

Example 1:

Given n = 4, edges = [[1, 0], [1, 2], [1, 3]]

        0
        |
        1
       / \
      2   3

return [1]

Example 2:

Given n = 6, edges = [[0, 3], [1, 3], [2, 3], [4, 3], [5, 4]]

     0  1  2
      \ | /
        3
        |
        4
        |
        5

return [3, 4]

Hint: How many MHTs can a graph have at most?

Note:

(1) According to the definition of tree on Wikipedia: “a tree is an undirected graph in which any two vertices are connected by exactly one path. In other words, any connected graph without simple cycles is a tree.”

(2) The height of a rooted tree is the number of edges on the longest downward path between the root and a leaf.

Breadth-first Search Graph

 

public class Solution {
    public List<Integer> findMinHeightTrees(int n, int[][] edges) {
        List<Integer> ret = new ArrayList<Integer>();
        if(n == 0)
            return ret;
        if(n == 1) {
            ret.add(0);
            return ret;
        }
        
        int[] count = new int[n];
        List<LinkedList<Integer>> neighbors = new ArrayList<LinkedList<Integer>>(n);
        for(int i = 0; i<n; ++i)
            neighbors.add(new LinkedList());
        
        for(int[] edge: edges) {
            int node1 = edge[0];
            int node2 = edge[1];
            ++count[node1];
            ++count[node2];
            neighbors.get(node1).add(node2);
            neighbors.get(node2).add(node1);
        }
        
        List<Integer> starts = new ArrayList<Integer>();
        for(int i = 0; i<n; ++i)
            if(count[i] == 1)
                starts.add(i);
        
        while(n>2) {
            List<Integer> newStarts = new ArrayList<Integer>();
            for(Integer surfaceNode: starts) {
                --n;
                for(Integer innerNode: neighbors.get(surfaceNode)) {
                    --count[innerNode];
                    if(count[innerNode] == 1)
                        newStarts.add(innerNode);
                }
            }
            starts = newStarts;
        }
        
        return starts;
    }
}