3015. Count the Number of Houses at a Certain Distance I | Floyd Warshall

[@yuvanshanmugarajt6362]

waiting for part II bro

[@satwiktatikonda764]

Waiting for higher tc version

[@deepjyotsinghkapoor1955]

Hi Aryan, I'm inclined towards using Dijkstra's algorithm as it boasts a time complexity of O(V + E Log V). With N nodes, the overall time complexity becomes O(V^2 + VE Log V). Considering the typical scenario where E is close to V, we can simplify it further to O(V^2 * Log V). In contrast, the Floyd-Warshall algorithm has a time complexity of O(V^3). I also considered implementing BFS, where each of the V nodes serves as the root. The time complexity of BFS is O(V + E), and when applied to V nodes as roots, it becomes O(V^2 + EV). Since E is equal to V for this problem, the overall time complexity simplifies to O(V^2). Feel free to correct me if there's any flaw in my understanding. Thanks!

[@ashutoshchoubey369]

Just a question, can't I use dijkstra's in this question which may have slightly better TC then Floyd. And the most intuitive also is the dijkstra's algo.

[@himanshuadhikari9440]

Floyd is more intuitive ig

[@anubhavfitnessclub]

can u make a video using some other solution for higher time complexities

[@FA513M]

n=5,x=1,y=5 ans=[10,10,0,0,0] in here why 3=0 we can go from 2 to 5 in which min-distance will be 3. can you help???

[@anasmaher4783]

minimum distance is 2, remember its undirected, so you can go 2 -> 1 1 -> 5

[@craftsworld3237]

birader part II

[@learningmaths786]

❤❤❤

[@anubhavfitnessclub]

👍👍👍👍👍👍👍👍👍👍👍👍