Thursday, July 13th

Courses with pre-requisites.  Given prerequisite requirements, can all courses be taken?  We are given a directed edgelist where a node from u to v means to take course u, you must first take node u.

A collection of courses cannot be completed when there's cyclic dependencies.  This problem is the same as yesterday's.  We will use topological sort to find cycles.

First we reverse the graph.  For topological sort, we want to start with nodes that have no incoming edges.  This means we reverse the graph so an edge v to u means v must be taken before u.  We will create an adjacency list to represent the reversed graph

       vector<vector<int> > adjlist(numCourses);
int u,v;
for(int i=0; i<prerequisites.size(); i++) {
u=prerequisites[i][0];
v=prerequisites[i][1];
adjlist[v].push_back(u);
}

Next we calculate the indegree for each vertex.  This is accomplished by iterating over the adjacency list, and incrementing a node's indegree whenever the node appears in the adjacency list

        vector<int> indegree(numCourses, 0);
vector<int> adj;
for(int i=0; i<numCourses; i++) {
adj=adjlist[i];
for(int j=0; j<adj.size(); j++) {
v=adj[j];
indegree[v]+=1;
}
}

We want to begin topological sort using nodes with indegree of zero, so iterate through the indegree vector and push nodes with 0 indegree

        queue<int> q;      
for(int i=0; i<numCourses; i++) {
if(indegree[i]==0) {
q.push(i);
}
}

Finally, we perform breadth first search using the queue of nodes with 0 indegree.  We count each time a node is popped from the queue.  Every time a node is popped, we iterate over its neighbors and decrement each neighbor's indegree by 1.  If the new indegree is 0, it's added to the back of the queue.  Like breadth-first search, this process continues until the queue is empty.

        int node, numdone=0, qsize, neighbor;
vector<int> neighbors;
while(!q.empty()) {
qsize=q.size();
for(int i=0; i<qsize; i++) {
node = q.front();
q.pop();
numdone++;
neighbors = adjlist[node];
for(int j=0; j<neighbors.size(); j++) {
neighbor=neighbors[j];
indegree[neighbor]-=1;
if(indegree[neighbor]==0) {
q.push(neighbor);
}
}
}
}

Counting the number of processed ("popped") nodes, if the number processed equals the total number of nodes in the original graph, then we can finish all the courses.  If the number of processed nodes is less than the total number of nodes, then we have a cycle, and the courses cannot be finished

return numdone == numCourses ? true : false;

Thanks to the cycle problem yesterday for a primer on topological sort

Comments

Post a Comment

Popular posts from this blog

Wednesday, September 20th

Image
 Reduce X to Zero Given an integer array and value x, use elements at the right and left of the array to reduce the x value to zero A collection of windows, by Daderot .  Shared under the Creative Commons Attribution-Share Alike 3.0 Unported License .  No changes made A key observation is that if elements at the ends of the array are used to sum up to a target value x , then elements in a contiguous subarray will equal the total_sum - x .  So we will look for the largest window equal to  total_sum - x Since all elements in the array are positive, then we can solve the problem using 2 pointers that form a sliding window. The general idea is to begin 2 pointers right and left at index 0.  Continue incrementing the right index by 1, adding up the elements it passes into the current_window  sum.  If the sum equals the target value, update the length.  Otherwise, if the sum is greater than the target sum, increment the left index until the window...
Read more

Monday, September 11th

Image
 Groups There are a given number of people, each of whom is assigned to a group.  The number identifier of the group is also its size.  The size of a group may be a multiple of its size, e.g. a group of 3 may have 6 people.  Return the  collection of all groups . Team USA Softball (a group).  Shared by USASoftball , via Wikipedia , distributed under the Creative Commons Attribution-Share Alike 3.0 Unported license.  No changes made. This problem is pretty straightforward.  The idea is to create a map (or a vector) keyed by the group, storing an array of individuals belonging to the group. First we create the vector-of-vectors to store our solution, then we initialize a vector of vectors that will store individuals according to their group numbers.  The size of the solution vector is n+1, because contrary to what the beginning of the problem statement said, the indices/group sizes go from 1 to n, according to the stated constraints. vect...
Read more

Tuesday, September 19th

Image
 Find the duplicate number Given an array of integers of length n+1, which contains numbers within the range [1,n], and contains on element that is duplicated at least once, find the duplicate element , considering use of only constant space and no changes to the array "How They Met Themselves" by Dante Rossetti, depicting doppelgängers The problem of finding duplicates is made difficult by the constraints of no modifications to the original array, and use of only constant extra space.  Still, despite these constraints there are still several ways to solve the problem.   Since yesterday's solution involved binary search, we will hone our BS skills by applying it to today's problem The general idea is based on the observation that without duplicates, the number of elements less than or equal an element is equal to the element itself.  E.g., in the collection {1,2,3,4}, the number of elements less than or equal to 2 is 2, less than or equal to 3 is 3, etc.   Howe...
Read more