Clone Graph

Clone an undirected graph. Each node in the graph contains a label and a list of its neighbors.

OJ's undirected graph serialization:

Nodes are labeled uniquely.

We use # as a separator for each node, and , as a separator for node label and each neighbor of the node.

As an example, consider the serialized graph {0,1,2#1,2#2,2}.

The graph has a total of three nodes, and therefore contains three parts as separated by #.

1. First node is labeled as 0. Connect node 0 to both nodes 1 and 2.
2. Second node is labeled as 1. Connect node 1 to node 2.
3. Third node is labeled as 2. Connect node 2 to node 2 (itself), thus forming a self-cycle.

Visually, the graph looks like the following:

       1
      / \
     /   \
    0 --- 2
         / \
         \_/

/**
 * Definition for undirected graph.
 * class UndirectedGraphNode {
 *     int label;
 *     List<UndirectedGraphNode> neighbors;
 *     UndirectedGraphNode(int x) { 
 *        label = x; 
 *        neighbors = new ArrayList<UndirectedGraphNode>(); 
 *     }
 * };
 */

/*
Use Queue to implement BFS and use HashMap to deep clone every graph node and connect their neighbors.

time:O(V+E), space:O(V), E is the number of neighbors, V is the number of nodes.
*/

public class Solution {
    public UndirectedGraphNode cloneGraph(UndirectedGraphNode node) {
        if (node == null) {
            return null;
        }
        Queue<UndirectedGraphNode> queue = new LinkedList<>();
        Map<UndirectedGraphNode, UndirectedGraphNode> map = new HashMap<>();
        queue.offer(node);
        map.put(node, new UndirectedGraphNode(node.label));
        while (!queue.isEmpty()) {
            UndirectedGraphNode curr = queue.poll();
            for (UndirectedGraphNode neighbor : curr.neighbors) {
                if (!map.containsKey(neighbor)) {
                    queue.offer(neighbor);
                    map.put(neighbor, new UndirectedGraphNode(neighbor.label));
                }
                map.get(curr).neighbors.add(map.get(neighbor));
            }
        }
        return map.get(node);
    }
}


//DFS, time:O(V+E), space:O(V)
public class Solution {
    public UndirectedGraphNode cloneGraph(UndirectedGraphNode node) {
        Map<UndirectedGraphNode, UndirectedGraphNode> map = new HashMap<>();
        return deepClone(map, node);
    }
    public UndirectedGraphNode deepClone(Map<UndirectedGraphNode,                                              UndirectedGraphNode> map, UndirectedGraphNode node) {
        if (node == null) {
            return null;
        }
        if (map.containsKey(node)) {
            return map.get(node);
        }
        map.put(node, new UndirectedGraphNode(node.label)); 
        for (UndirectedGraphNode neighbor : node.neighbors) {
            map.get(node).neighbors.add(deepClone(map, neighbor));
        }
        return map.get(node);
    }

}

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.
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 the correct course order is [0,1]
4, [[1,0],[2,0],[3,1],[3,2]]
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].


/*
The problem reduces to find a topological sort order of the courses. If a node has incoming edges, it has prerequisites. Therefore, the first few in the order must be those with no prerequisites, i.e. no incoming edges. Notice, if there is a cycle in the graph, it means it is impossible to finish all the courses.

time:O(n), space:O(n)
*/

public class Solution {
    public int[] findOrder(int numCourses, int[][] prerequisites) {
        Map<Integer, Integer> inMap = countInlinks(prerequisites);
        Map<Integer, ArrayList<Integer>> outMap =                                                 initNeighbors(prerequisites);
        return findOrderByBFS(numCourses, inMap, outMap);
    }
    public Map<Integer, Integer> countInlinks(int[][] pre) {
        Map<Integer, Integer> map = new HashMap<>();
        for (int[] pair : pre) {
            if (!map.containsKey(pair[0])) {
                map.put(pair[0], 1);
            } else {
                map.put(pair[0], map.get(pair[0]) + 1);
            }
        }
        return map;
    }
    public Map<Integer, ArrayList<Integer>> initNeighbors(int[][]                                                             pre) {
        Map<Integer, ArrayList<Integer>> map = new HashMap<>();
        for (int[] pair : pre) {
            if (!map.containsKey(pair[1])) {
                ArrayList<Integer> list = new ArrayList<>();
                list.add(pair[0]);
                map.put(pair[1], list);
            } else {
                map.get(pair[1]).add(pair[0]);
            }
        }
        return map;
    }
    public int[] findOrderByBFS(int numCourses, Map<Integer,                   Integer> inMap, Map<Integer, ArrayList<Integer>> outMap) {
        int[] ans = new int[numCourses];
        Queue<Integer> queue = new LinkedList<>();
        int index = 0;
        for (int course = 0; course < numCourses; course++) {
            if (!inMap.containsKey(course)) {
                ans[index++] = course;
                queue.offer(course);
            }
        }
        while (!queue.isEmpty()) {
            int course = queue.poll();
            if (!outMap.containsKey(course)) {
                continue;
            }
            for (Integer i : outMap.get(course)) {
             inMap.put(i, inMap.get(i) - 1);
                if (inMap.get(i) == 0) {
                    ans[index++] = i;
                    queue.offer(i);
                }
            }
        }
        return index == numCourses ? ans : new int[0];
    }
}