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Microsoft Interview Question for Software Engineer / Developers






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Use a Hash table and fill the list of all the customers who paid money. Next scan through the list of all the customers and see if he is there in the hash table. If yes that means he is the one who paid, if not, then he is the one who didn't pay and hence print the name of the customer. Repeat it for the entire list of customers. This is linear in both time and space complexity.

- gauravk.18 February 23, 2008 | Flag Reply
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Hi Gaurav,

Your solution gives the obvious answer... But have you given a thought of how you can improve it?

With some realistic assumptions that the list of customers is ordered... we could use Bit vector ...maybe?? just a thought.

- chinni September 16, 2008 | Flag
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Using array list may not make sense. It'll lead to O(nm) in time complexity and O(n+m) in space complexity.

But using hashmap approach, time complexity: O(n+1)=>O(n) and space complexity: O(n)

- sathishp123 April 25, 2012 | Flag


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