this is just brute force:
count = 0
for i in range(0, len(A)):
for j in range(i+1, len(A)):
if A[i] > A[j]:
count += 1
Time complexity = O(n**2)
Anil
On Sun, Jun 27, 2010 at 9:59 AM, sharad kumar <[email protected]>wrote:
> Give an unsorted array find count of pairs of numbers[a,b] where a > b and
> b comes after a in the array.
>
> Eg. {8,3,6,10,5}
>
> the count of such numbers is 5. i.e. (8,3), (8,6), (8,5), (6,5) and (10,5)
>
>
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