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Author: Kirito
Subtask 1
For the
subtask, it is possible to iterate over all
pairs of numbers and compare the absolute value of their differences, and print the minimum.
Time Complexity: 
Subtask 2
For the
subtask, a greedy approach can be used. Sort all the elements, and iterate through each pair of adjacent numbers in the sorted array, and print the minimum absolute value of their difference.
Proof of greedy algorithm:
Let the array
be the sorted array. The minimum difference between
and any number is
. For the sake of contradiction, assume there exists some
,
, such that
Thus we have two cases:
Case 1: 
This implies that
.
is sorted,
.
, for
.
Substituting, this gives
, which gives:

, this is a contradiction.
Case 2: 
This implies that
.
is sorted,
.
, for
.
Substituting, this gives
, which gives:

, this is a contradiction.
we have proven that
does not exist.
Time Complexity: 
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