The total time complexity of Dijkstra’s Algorithm is given by : E.T(update key / relaxing edges) + V.T(to delete min) .
Where E is the edge and V denotes the vertices. T represents the time to implement the operation given in parenthesis.
The minimum in a sorted array is present at the first position itself (I am assuming an ascending sorted array) hence the time is O(1) . Whereas the time required to update the key will be O(V). [The number of vertices in the array.]
Therefore we get the time as V+EV which can be taken as EV.
Please correct me If I am wrong.
