## Amazon Interview Question

Software Engineer / DevelopersYou are a genius! That was so simple why didn't I think of that when they asked me?! Thanks for the simple solution, I'll keep bit-wise operations in mind next time :p

You are a genius! That was so simple why didn't I think of that when they asked me?! Thanks for the simple solution, I'll keep bit-wise operations in mind next time :p

sum all the numbers. subtract one element at a time, if result is even the number just subtracted is answer. the xor solution is best imo.

Are u sure for XOR answer ??

public class Xor {

public static void main(String[] args) {

int arr[] = { 1,1,1,2,2,3,4,4,4,7,7,5};

int result = 0;

for (int i = 0; i < arr.length; i++) {

result = result ^ arr[i];

//System.out.println( (i+1) + ": " + result);

}

System.out.println("Final Result : " + result);

}

}

The result that was given is 3 instead of 2 .

Hi ,

I just thought of another way of doing it.

product all the numbers.

divide the product by each of the numbers. Find the square root of the division. If its whole. then the integer is odd one. as even number of integer multiplication should have whole square root.

but again no zeroes here. but seriously,, how the hell did u come up XOR man??!!

Q: An integer array where there are repeating numbers and all but one repeat an odd number of times ex [ 2,2,2,4,4,7,7,7,3,5] 4 repeats even number of times, [2,1,7,3,3] 3 repeats even number of times . Find it

A: Do an XOR of the corresponding bits of integers in the array, and do a NOT on the result

My answer I gave was to put all the numbers in a hash table where the key where the numbers and the values where the number of times the number was seen. Then just go through all the numbers and see which ones are not divisible by 2. This was the best answer I could come up with on the spot, and I wasn't able to answer the last part of how it could be done without another data structure.

- JayNeedsAJob January 15, 2011