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Interview Question for Software Engineer / Developers


Country: India
Interview Type: In-Person



Comment hidden because of low score. Click to expand.
2
of 2 vote

1) First Find the shortest path between two nodes E1->E2->E3 ....
2) For all the edges : Remove one and calculate the shortest path agin store the path in array with it's weight
path1= {E1,E2, ... En, pathWeight}
pathsArray.add(Path1);
3) then sort the pathsArray depending on pathWeight of the path and get the Kth path you want

- Maher Rezeq September 25, 2012 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 votes

seems good...plz clarify this,..let's say u hv two paths from 1 to 6..
1->3->6
1->2->4->3->6
then which edge is gonna be removed?

- anandaroop.acharia September 25, 2012 | Flag
Comment hidden because of low score. Click to expand.
0
of 0 votes

in this case you will not remove the Start and End node the one in between ,in your case the 3 , lets say we have 1->4->6 , store the path and weight and do again and again for all nodes , then sort the weights and get the k'th one

- Maher Rezeq January 02, 2013 | Flag
Comment hidden because of low score. Click to expand.
0
of 0 vote

DIjkstra's algorithm

- Fargushan September 25, 2012 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

Travel till least common ancestor.

that will be the path

- vnvaditya February 10, 2013 | Flag Reply
Comment hidden because of low score. Click to expand.
-1
of 1 vote

Which company?

- Anonymous September 25, 2012 | Flag Reply
Comment hidden because of low score. Click to expand.
-1
of 1 vote

Think of BFS, count the times you reach the goal node and stop after the k-th time. Adapt BFS to work with weighted edges.

- Emil November 12, 2012 | Flag Reply


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