sorted array in dijkstra

Question

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

 

please verify this.

Comments

@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.

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ooh alright..thanks kapil..

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exactly kapil ... that's what I was doing ... now uploaded !

Answer 1

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Comments

wonderful and clear explanation..thanks a lot ..:-)

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@debashish deka  you are  next RAVI SHANKAR MISHRA ...wonderful explanation

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hey just need to ask you  that if we use (unsorted) array instead of sorted array.Then Analysis of Dijkstra would be-

1. The while loop will run for O(V)  times untill its empty

2. For extraction the minimum it will take O(V) time rather that O(1) since we have to search the entire array

3. we can do the decrease key operation in O(1) time since no overhead for maintaing the array sorted. This decrease key operation will  run for O(E) times (for every adjacent v for the vertex u. Sum of all length of all the adjacents)

Hence total Time Complexity would be (V.V+E.1)

=>O(V^2)

If this analysis is right can we say that dijkstra algo for unsorted array take lesse time that sorted array