Please explain that mid -(i-1) again with more clarity .
Recursion inversion Count
Hey @vanshikasharma1645, i guess you are having a problem here,
since prateek bhaiya has explained that if there are two arrays that are to be compared:
elements ->x x 5 6 7 8 && x 2 ! ! !
index val->0 1 2 3 4 5 && 0 1 2 3 4
so you can see that a[2] i.e., 5 is greater then a[1] i.e., 2
so all the elements in front of 5 including 5 will make inversion count. But how can we calculate all the elements in front of 5 including 5 too. for that we can see that mid which is index 5 and i which is index 2 is subtracted and added with 1 , we get all the count in front of 5 including 5 i.e.,
cnt=5-2+1 i.e., 4
Therefore 4 elements will make inversion count with element 2 in another array.
Hope this will help you in understanding question 
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