Nearest absolute difference

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(Example from Hackerrank problem 'Minimum Absolute Difference in an Array'.)

The built-in Python array sorting algorithm invoked by `sorted(-)` uses the timsort algorithm which has worst case time complexity `O(n log(n))`. This means that if we have a programme involving arrays which has complexity worse than `O(n log(n))` it might be appropriate to sort the arrays first. (Note that finding a way to use a hash table cache as in Nearest repeated entries is even better. But that might not be possible.)

Consider the following example from the Hackerrank problem 'Minimum Absolute Difference in an Array':-

The brute-force algorithm that comes immediately to mind is to iterate twice through the array thereby comparing each pair of numbers:

def nearest_dist(arr): # O(n^2)
    minimum = float('inf')
    for i in range(len(arr)):
        for j in range(i+1, len(arr)):
            if abs(arr[i] - arr[j]) < minimum:
                minimum = abs(arr[i] - arr[j])
    return minimum

which has time complexity `O(n^2)`.

However this solution takes too long to solve the Hackerrank exercise (one gets a `timed out`-style error). Reasoning as above: we see that `O(n^2)` is worse than `O(n log(n))` and furthermore a moment's thought tells us that solving the problem would be much easier with a sorted array. So we can try:

def nearest_dist(arr): # O(n log(n))
    minimum = float('inf')
    a = sorted(arr)
    for i, _ in enumerate(a):
        if abs(a[i] - a[i+1]) < minimum:
            minimum = abs(a[i] - a[i+1])
    return minimum

which has time complexity `O(n log(n))` and passes as a Hackerrank submission.