+1 vote
160 views
What is the time complexity to implement Dijkstra’s algorithm using a sorted array instead of heap for a Priority Queue?

for sorted array

let V be the number of nodes and E be the number of edges

1)extract min operation ---it will take constant time and it is repeated for V nodes.hence takes O(v) time.

2)decrease key operation occurs E times------now we can directly go and decrease the value of the node but we might have to sort the array again because after decreasing the key,array might not be sorted..so it will take VlogV time if we use merge sort.so ,total time is E*VLOGV

it will be $O(v^3)$..
how..?can you tell whats wrong in my explanantion

http://stackoverflow.com/questions/2680365/running-time-for-dijkstras-algorithm-on-a-priority-queue-implemented-by-sorted

in this post,i saw it is also written that it is O(v^3) but tell me one thing,,after decreasing a particular key,dun w need to sort the array again cuz it might get unsorted..?

@Akriti

A small nitpick - You needn't to sort the whole array, just apply binary search and place the key there and shift the elements to a new position -> O(E.V) -> Now, the graph can either be sparse or complete.
ooh alright..thanks kapil..
exactly kapil ... that's what I was doing ... now uploaded !

................

selected
wonderful and clear explanation..thanks a lot ..:-)
@debashish deka  you are  next RAVI SHANKAR MISHRA ...wonderful explanation