## VMWare Inc Interview Question Software Engineer / Developers

- -2of 2 votes
Given a sorted array find all possible |ai - aj| where ai,aj belongs to Array A. n^2 is obvious. Find a solution in O(N).

*Sigh* Probably yet another idiot question poster posting an inaccurate statement of the problem. The guy(s) who said you can't do better than O(n^2) are of course correct. The rest of you who gave O(n) algorithms or can't post a proper question need not apply for any software job ... please. My best guess is that the question was asking for the *sum* over all possible |ai-aj|. Now try to do O(n) for this problem.

I doubt if O(n) is possible. It seems that if this problem could be solved in O(n) sorting could be done in O(n) as well.

If I'm allowed to used extra space,

I would construct an array such that S[i] = Sum(A[0..i]). Takes O(n).

1, 2, 4, 6, 7, 9, 20

|S[i]-S[j]| is the Sum(A[i+1..j]).

|S[i-1]-S[j-1]| is the Sum(A[i..j-1])

i.e,

|S[i]-S[j]| = a[i]+a[i+1]+a[i+2]..+a[j]

|S[i-1]-S[j-1]| = a[i-1]+a[i]+a[i+1]+..+a[j-1]

So,

|S[i]-S[j]| + |S[i-1]-S[j-1]| = |a[i]-a[j]|

Correct me if it's wrong...

-Hari.

As given array is sorted( assuming descending order) , lets say size of array is 10, a[0,...9]. and the |ai - aj| = x.

Assuming that all numbers are unique in the array ( if they are not this algo can be easily modified). Now define two indexes such that i = 0 and j=1, find the first aj such that either |a0 -aj| = x or |a0 - aj| < x .

Now we have two indexes such that i=0 and j. if |ai - aj| = x then ++j else if |ai-aj|< x then ++i. Follow this procedure and we will get the pairs.

Complexity: O(n)

Don't need to process a array location n1 if current array location n2 is greater than n1.

void printDiff(int a[], size_t size)

{

for (unsigned int i = 0; i < size; ++i)

{

for (unsigned int j = i+1; j < size; ++j)

{

cout << i << " , " << j << " = " << abs(a[i] - a[j]) << endl;

}

}

}

This is impossible to do in O(n). Minimum is O(n^2).

- Alex on September 11, 2011 Edit | Flag ReplyProof:

Let A be {1, 2, 4, 8, ... 2^n}

We must compute all different |A_i - A_j|. We will show that we have 1 + n(n-1)/2 such differences. First of all, because of the modulo, we can only consider |A_j - A_i| with j > i. We also have a difference of 0, from |A_i - A_i| (this is where the 1+ comes from in the formula above).

Assume that we have i1 and j1 with i1<j1 and i2 and j2 with i2<i2, for which |A_i1 - A_j1| == |A_i2 - A_j2|. We will prove that i1 == i2 and j1 == j2. Because of the ordering between i* and j*, we have A_j1 - A_i1 = A_j2 - A_i2. This is equivalent to 2^j2 - 2^i2 == 2^j1 - 2^i1, turns to 2^i2(2^(j2-i2) + 1) == 2^i1(2^(j1-i1) + 1). From this, 2^i2 has to divide 2^i1, since the parenthesis on the right (2^(j1-i1) + 1) is odd and cannot be divided by 2. In turn, 2^i1 has to divide 2^i2. Therefore i2 == i1 == i. From this, 2^(j1-i) + 1 == 2^(j2-i) + 1, which gives 2^(j1-i) == 2^(j2-i), which gives j1 == j2 == j. QED

This proves that among all pairs (i, j) with i < j, each has a unique difference |A_j - A_i|. They are n(n-1) / 2 pairs, which will take O(n^2) to even print, let alone find.

QED