Microsoft Interview Question for Software Engineer / Developers


Team: Bing
Country: India
Interview Type: In-Person




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

Assuming that it is literally impossible to compare nuts vs nuts and bolts vs bolts (i.e. we cannot label them arbitrarily as a preprocess to impose a linear order), you could do the following: Choose a pivot nut. Find its corresponding bolt in O(n) time. In O(n) time partition the remaining nuts and bolts by comparing to either the partition nut or the partition bolt depending on their type. Then recurse on the two halves. This is exactly like quicksort except for the additional pass to find the bolt that goes with the pivot nut, which however does not change the asymptotic complexity.

- psykotic February 20, 2012 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

Agreed with Psykotic..it takes O(nlogn) in avg case & O(n^2) in worst case as it is Like a Quick Sort

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

You have to pickup a nut and findout the corresponding bolt.So while comparing you have to ignore all nuts,just pick the bolt.So take the nut(value) as pivot element and start comparing with the bolts using quick sort.During this comparision, ignore the nuts(isNotNut()).

- skylap March 20, 2012 | Flag Reply
Comment hidden because of low score. Click to expand.
-1
of 1 vote

the said question was that the they are pairs , in which case if we sort both of them(arrays) by quick sort which is possible as we know the size already , we can then correspond the corresponding bolts and can say with confidence that they are a perfect pair

- Usman February 25, 2012 | Flag Reply


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