Amazon Interview Question for Software Engineer / Developers






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2nlogn + 2n-1

- Anonymous July 21, 2011 | Flag Reply
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of 1 vote

O(n) - Create Hashmap for the first array and check for the keys using the second array. If the key exists, print it or add it to an auxiliary array and display it.

- RiTZ July 25, 2011 | Flag
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of 0 vote

sort the smaller array. now binary search for each element of larger array in the smaller array. complexity=O(mlgm+nlgm). mlgm to sort and nlgm to check if each element of larger array exists in smaller array . doing other way round leads to complexity of O((m+n)lgn). so first way is better.

- anon August 08, 2011 | Flag Reply


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