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Directi Interview Question for Software Engineer / Developers






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Think the matrix as a graph, we visit graph with BFS. For each node, it has state {non-visited, visit-drop, visit-target}. We add all neighbour into queue, then check & mark status until we finished the queue.

- Anonymous August 24, 2011 | Flag Reply
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So funny... even for this type of problems people can't think out of graph!

O(n^2) time is sufficient to do this in straight forward manner. No solution CAN'T exist that has lower complexity to find ALL local maxima!

- anon August 27, 2011 | Flag
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n^2 is too much...

- -=tisko=- October 07, 2011 | Flag
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@tisko any algo to solve it in less than O(n^2) ?

- Nitish May 01, 2012 | Flag


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