My Favorite Technical Interview Question

I got the idea to write this post from a google engineer, Alex Golec. He wrote a blog post deconstructing an interview question which he often asked and was particularly fond of. Check out his post: Google Interview Questions Deconstructed: The Knight’s Dialer.

Anyways, I am far from having an interview question I ask often since I’m still in school and am on the recieving end of interview questions but nevertheless I have a favorite.

I was asked this question by my friend Julian Moskovits when I requested a mock interview from him. I will first state the question and then explain my thought process as I go about answering it.


You’re given an array of integers, find the greatest value difference between any 2 adjacent indices.

One reason I love this question is because it is very clear as to what the question is asking, yet leaves a bunch of room for the candidate to ask clarifying questions.

My first move when I was asked this question was to clarify a few things. I wanted to know:

  • Are the elements sorted?
  • What do I return? Both indices? The difference?
  • What happens if there are 2 equal answers?
  • Can I have a negative difference or is it the absolute difference?
  • Can the numbers be negative?

I was told that the elements would not necessarily be sorted and that I was required to return the first of the 2 indicies. If there are 2 equal answers return the first of the them and the type of difference is the absolute difference. Lastly, the integers will all be positive.

Alright, we should have all the information we need.

Let’s draw a basic picture:

0 1 2 3 4
1 3 2 8 7

We can loop through this list one time, checking the difference between the current index and the next index (We only loop until length - 2 to avoid and out of bounds error). At each index we check if the difference between the current value and the next value is greater than the previous greatest difference we had seen. If it is, store the current index and the new greatest difference. If it is not, move on.

This will look something like:

	int greatestDifference = -1, currentDifference;
	int index = 0;
	int numbers[] = {1, 3, 2, 8, 7}
	for(int i = 0; i < sizeof(numbers)/ sizeof(int) - 1; i++) { 
		currentDifference = abs(numbers[i] - numbers[i+1]);
		if(currentDifference > greatestDifference) {
			index = i;
			greatestDifference = currentDifference;
	std::cout << index;

Visually we are looking at something like this (The arrow indicates which index we are looking at ):

Index Diff 1 3 2 8 7
0 -1 ⬆︎        
Index Diff 1 3 2 8 7
0 2 ⬆︎        
Index Diff 1 3 2 8 7
0 2   ⬆︎      
Index Diff 1 3 2 8 7
2 6     ⬆︎    
Index Diff 1 3 2 8 7
2 6       ⬆︎  
Index 2

If you got this, GREAT JOB!! If not, don’t worry about it, you know it now!

But the question does not end there.

Julian expanded the question by asking me what happens if you can compare any 2 elements, not just those adjacent to each other.

This expansion means we need to approach the problem differently. We can no longer loop through the array one time since we need to test each index against every other index, not just the one next to it. This would mean we would need a nested for-loop. I asked Julian if this solution would do and he said although it works, he wants the asymptotic time to be O(n) not O(n2).

If you think about this a litte more you’ll hopefully realize that the greatest difference will always be the smallest integer subtracted from the greatest integer. This results in the following discovery: If you loop over the array keeping track of the smallest and greatest integers, you can then return the index of whichever of those two comes first.

The code would look something like this:

	int greatest = 0, smallest = 0;
	int greatestIndex = 0, smallestIndex = 0;
	int numbers[] = {1, 3, 2, 8, 7}
	for(int i = 0; i < sizeof(numbers)/ sizeof(int); i++) { 
		if(numbers[i] > greatest) {
			greatestIndex = i;
			greatest = numbers[i];
		if(numbers[i] < smallest) {
			smallestIndex = i;
			smallest = numbers[i];
	std::cout << greatestIndex < smallestIndex ? greatestIndex : smallestIndex;

Hopefully this makes sense. If you got this, GREAT JOB!! If not, don’t worry, you know it now!

At this point you’ve gotten through 2/3 of the question and hopefully pretty quickly. Usually you would only be asked to program one of these two and just verbally state the other. The second half is clearly a bit more complicated than the first but nothing too tough to deal with.

This is another aspect that I love about this question. Nothing is too bad or too complicated, it just builds on itself and each step requires a bit more critical thinking. The programming itself is pretty straight forward.

Okay so Julian wasn’t going to let me off yet. He had one final part.

Up until now we’ve been using absolute difference. Meaning we substract two numbers and take the absolute value. To answer this next part we must find the greatest difference where difference is defined as A[i] - A[i+j] where j > 0. In otherwords, take the earlier index value and subtract from it the later index value. The key difference here is that we now allow for negative number outcomes.

For example, take the array

0 1 2 3 4
1 3 2 8 7

The difference between index 1 and index 2 is not 2 but -2.

Hopefully you understand. If not, always ask for clarification from your interviewer

This question is hard because in this case the order in which the numbers show up matters. If we tried out the min/max approach that worked for part 2 we would be in trouble whenever we calculated the min/max where the smaller integer came before the larger one. In that scenario we’d get back a negetive which is not (necessarily) the correct answer. (usually not).

I suggest taking a crack at this yourself.

If you came up with a nested for-loop, that works. If you suggested that answer your interviewer would most likley say that works but we want to solve this faster. You’d also probably be asked the asymptotic time complexity (which for a nested for-loop is O(n2)).

If you came up with the nested for-loop solution try to think about it a bit more.

Need a hint? Try starting from the end of the array instead of the beginning.

Okay so what’s the solution?

We can solve this in O(n) or linear time. To do this we start from the end of the array and we use a data structure to hold some key information.

I will first state the algorithm, then show the code and finally provide a visual example.


We will hold currentSmallest variable, and an array of size n.

  1. Set currentSmallest to the last element in the array
  2. Loop from the 2nd to last element until the first element.
    1. If the previous index’s value is less than the currentSmallest
      1. Set currentSmallest to the previous index’s value
    2. Insert the current index’s value - currentSmallest into the equivalent index of our new array
  3. Find the largest value in our new array, the first index which that value sits in the answer.
	int numbers[] = {1, 3, 2, 8, 7};
   	int tracker[5] = {};
   	int length = sizeof(numbers)/ sizeof(int);
   	int currentSmallest = numbers[length - 1];

   	for(int i = length - 2; i >= 0; i--) {
   		tracker[i] = numbers[i] - currentSmallest;
   		if(numbers[i] < currentSmallest) {
   			currentSmallest = numbers[i];
	for(int i : tracker) {
   		std::cout << i << " "; 	// -1 1 -5 1 0
	int index = findFirstMaxIndex(tracker, length);
	std::cout << index;
Smallest 1 3 2 8 7
7       ⬆︎  
Tracker         0

We do 8 (The value at the index we are currently look at) - 7 (the current smallest) and we get 1

Smallest 1 3 2 8 7
7     ⬆︎    
Tracker       1 0

We do 2 (value at current index) - 7 (the current smallest) and we get -5

Smallest 1 3 2 8 7
2   ⬆︎      
Tracker     -5 1 0

We do 3 (value at current index) - 2 (the current smallest) and we get 1

Smallest 1 3 2 8 7
2 ⬆︎        
Tracker -1 1 -5 1 0

We do 1 (value at current index) - 2 (the current smallest) and we get -1

Index 1

That’s it! If you solved this part, GREAT JOB! If not, don’t sweat it, it’s not so simple. Just try and make sure you fully understand it by going back and analyzing the logic in the code.

Thanks goes out to:
Elliot Eisenberg for editing this article.
Yehuda Moskovits for asking me this question.