I recently took an online test on codility as part of a recruitment process. I was given two simple problems to solve in 1 hour. For those who don't know codility, its an online coding test site where you can solve ACM style problems in many different languages.
if you have 30 or so mins then check this http://codility.com/demo/run/
My weapon of choice is usually Java.
So on of the problems I have is as follows (I will try to remember, should have taken a screenshot)
Lets say you have array A[0]=1 A[1]=-1 ....A[n]=x
Then what would be the smartest way to find out the number of times when A[i]+A[j] is even where i < j
So if we have {1,2,3,4,5}
we have 1+3 1+5 2+4 3+5 = 4 pairs which are even
The
code I wrote was some thing along the lines
int sum=0;
for(int i=0;i<A.length-1;i++){
for (int j=i+1;j<A.length;j++){
if( ((A[i]+A[j])%2) == 0 && i<j) {
sum++;
}
}
}
There was one more restriction that if the number of pairs is greater than 1e9 then it should retrun -1, but lets forget it.
Can you suggest a better solution for this. The number of elements won't exceed 1e9 in normal cases.
I think I got 27 points deducted for the above
code (ie it's not perfect). Codility gives out a detailed assessment of what went wrong, I don't have that right now.
