# Minimum Absolute Difference in an Array

See the original problem on HackerRank.

## Solutions

The brute force solution consists in calculating the absolute difference of each pair and reducing to the minimum value. This takes $$O(N^2)$$.

We can find better solution, instead, by making a simple observation: the closest two numbers, the lowest their difference. It's just like calculating the distance between points on the same axis. Indeed, the formula to calculate the distance between two points $$x1, x2$$ on the same axis is simply $$|x1-x2|$$.

Thus, we need to “fill the gaps” between numbers. This means, we need to sort the array. Then, we do the difference between each ajacent pair and calculate the minimum value along the way.

Here is a possible implementation in Python 2:

 1 2 3 4 5 6  def minimumAbsoluteDifference(arr, n): arr.sort() minVal = sys.maxint for i in range(n - 1): minVal = min(minVal, abs(arr[i] - arr[i+1])) return minVal 

A pattern brings out from the code above: zip | map | reduce:

 1 2 3  def minimumAbsoluteDifference(arr, n): arr.sort() return min(map(lambda x: abs(x-x), zip(arr[1:], arr))) 

The same approach in C#, by Claudio Bacchelli:

 1 2 3 4 5  static int minimumAbsoluteDifference(int[] arr) { var lista = arr.ToList.Sort(); return lista.Zip(lista.Skip(1), (a, b) => Math.Abs(a - b)).Min(); } 

An alternative way of writing the solution above in Python 3 by Eduard “Eddie” Rubio Cholbi and Stefano Ligabue:

 1 2 3  def minimumAbsoluteDifference(arr): arr = sorted(arr) return min(map(lambda a, b: b - a, arr[:-1], arr[1:])) 

Eddie and Stefano wrote generic way of to zip things in Python and applied it to the problem:

 1 2 3 4 5 6 7  def zipper(arr, n): return (arr[i:len(arr)-n+i+1] for i in range(n)) # Complete the minimumAbsoluteDifference function below. def minimumAbsoluteDifference(arr): arr = sorted(arr) return min(map(lambda a, b: b - a, *zipper(arr, 2))) 

zipper zips a sequence with itself shifted n times. For example, on arr = [1,2,3,4,5]:

With n=1 (does nothing):

 1 2  unzipper(arr, 1) => [1, 2, 3, 4, 5] 

With n=2:

 1 2 3  unzipper(arr, 2) => [1, 2, 3, 4] [2, 3, 4, 5] 

With n=3:

 1 2 3 4  unzipper(arr, 2) => [1, 2, 3] [2, 3, 4] [3, 4, 5] 

Etc.

This is an example of “freely expressing oneself” we encourage at Coding Gym. Eddie and Stefano solved the challenge very quickly and started experimenting with their favourite language, getting out of their “state of the art”. Since Coding Gym is not time nor prize driven, anyone sets her own targets.

Just for fun, in C++ zip|map|reduce can be written in terms of inner_product:

 1 2 3 4 5  sort(begin(v), end(v)); cout << inner_product(next(begin(v)), end(v), begin(v), numeric_limits::max(), [](int a, int b) { return min(a,b); }, [](int a, int b) { return abs(a-b); } ); 

In C++20 we can use ranges, left as an exercise.

We've worked on this challenge in these gyms: modena