# 210. Course Schedule II

Difficulty: Medium

Frequency: N/A

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]`.

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 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.
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.

Test Cases:

Solution 1:

Data structure:
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Code:
```class Solution {
private:
vector<unordered_set<int>> make_graph(int numCourses, vector<pair<int, int>>& prerequisites) {
vector<unordered_set<int>> graph(numCourses);
for (auto pre : prerequisites) {
graph[pre.second].insert(pre.first);
}
return graph;
}
bool dfs(vector<unordered_set<int>>& graph, int node, vector<bool>& onpath, vector<bool>& visited, vector<int>& result) {
if (visited[node]) return false;
visited[node] = onpath[node] = true;
for (int neighbor : graph[node]) {
if (onpath[neighbor] || dfs(graph, neighbor, onpath, visited, result)) {
return true;
}
}
result.push_back(node);
return onpath[node] = false;
}
public:
vector<int> findOrder(int numCourses, vector<pair<int, int>>& prerequisites) {
vector<unordered_set<int>> graph = make_graph(numCourses, prerequisites);
vector<bool> onpath(numCourses, false), visited(numCourses, false);
vector<int> result;
for (int i = 0; i < numCourses; i++) {
if (!visited[i] && dfs(graph, i, onpath, visited, result)) {
return {};
}
}
reverse(result.begin(), result.end());
return result;
}
};
```

Solution 2:

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steps:
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Submission errors:

Things to learn:

# 207. Course Schedule

Difficulty: Medium

Frequency: N/A

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.

Test Cases:

Solution 1:

Data structure:
Steps:
Complexity:
Runtime:
Space:
Code:
```class Solution {
private:
vector<unordered_set<int>> make_graph(int numCourses, vector<pair<int, int>>& prerequisites) {
vector<unordered_set<int>> graph(numCourses);
for (auto pre : prerequisites) {
graph[pre.second].insert(pre.first);
}
return graph;
}
bool dfs_cycle(vector<unordered_set<int>>& graph, int node, vector<bool>& onpath, vector<bool>& visited) {
if (visited[node]) return false;
onpath[node] = visited[node] = true;
for (int neighbor : graph[node]) {
if (onpath[neighbor] || dfs_cycle(graph, neighbor, onpath, visited)) {
return true;
}
}
return onpath[node] = false;
}

public:
bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {
vector<unordered_set<int>> graph = make_graph(numCourses, prerequisites);
vector<bool> onpath(numCourses, false), visited(numCourses, false);
for (int i = 0; i < numCourses; i++) {
if (!visited[i] && dfs_cycle(graph, i, onpath, visited)) {
return false;
}
}
return true;
}
};
```

Solution 2:

Data structure:
steps:
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Submission errors:

Things to learn:

# 133. Clone Graph

Difficulty: Medium

Frequency: N/A

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
/ \
\_/```

Test Cases:

Solution 1:

Data structure:
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Code:
```/**
* Definition for undirected graph.
* struct UndirectedGraphNode {
*     int label;
*     vector<UndirectedGraphNode *> neighbors;
*     UndirectedGraphNode(int x) : label(x) {};
* };
*/
class Solution {
public:
UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) {
if (!node) return NULL;
UndirectedGraphNode *p1 = node;
UndirectedGraphNode *p2 = new UndirectedGraphNode(node->label);
unordered_map<UndirectedGraphNode*, UndirectedGraphNode*> m;
queue<UndirectedGraphNode*> q;
q.push(node);
m[node] = p2;
while (!q.empty()) {
p1 = q.front();
p2 = m[p1];
q.pop();
for (int i = 0; i < p1->neighbors.size(); i++) {
UndirectedGraphNode *nb = p1->neighbors[i];
if (m.count(nb)) {
p2->neighbors.push_back(m[nb]);
} else {
UndirectedGraphNode *tmp = new UndirectedGraphNode(nb->label);
p2->neighbors.push_back(tmp);
m[nb] = tmp;
q.push(nb);
}
}
}
return m[node];
}
};
```

Solution 2:

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