We use cookies to ensure you have the best browsing experience on our website. Please read our cookie policy for more information about how we use cookies.
Loading...
  1. Practice
  2. Algorithms
  3. Greedy
  4. Minimum Absolute Difference in an Array
  5. Discussions

Minimum Absolute Difference in an Array

  • LeHarkunwar + 3 comments

    Yeah, it was pretty easy after reading your comment.

    n,a = input(),sorted(map(int, input().split()))
print(min(abs(x-y) for x,y in zip(a,a[1:])))
    • antdro + 4 comments

      Check for duplicates helps to avoid unnecessary computations. 

      if len(arr) != len(list(set(arr))):
    result = 0

      This is relevant for one of the tests.

      • Sourov_Roy + 1 comment
        [deleted]
      • it_henrik + 0 comments

        to avoid even more unnecessary computations:

        len(list(set(arr))) -> len(set(arr))

      • machinekoder + 0 comments

        You can also abort when diff is zero to pass the test.

      • _elf_ + 0 comments

        Not sure how relevant, this passes all tests:

        def minimumAbsoluteDifference(arr):
    diffs = []
    arr.sort()
    for i in range(len(arr)-1):
        diffs.append(abs(arr[i]-arr[i+1]))
    return min(diffs)
    • wangjiehui11235 + 0 comments

      it's really nice!

    • satyavinay456 + 1 comment

      you are not considering all pairs, zip is not useful in this case

      • danc1005 + 1 comment

        Exactly, this just compares 'neighbors' i.e. numbers at consecutive indices (i, i+1)...that's a set of O(n) whereas the set of all possible pairs (given by, for example, set([i, i for i in range(n)]).symmetric_difference(itertools.product(a, a)) is O(n^2)

        • namik + 0 comments

          Note that the array was sorted beforehand, so that considering differences of consecutive elements is all we need to check to find the minimum difference. No point in checking the difference between i and i+2 if we know the difference between i and i+1 is definitely less.