Inversion count , invcnt (spoj)

#1

how to solve this problem by modified merge sort in nlogn time ??? as described by instructer at 01;47 min

#2

Suppose we know the number of inversions in the left half and right half of the array (let be inv1 and inv2), what kinds of inversions are not accounted for in Inv1 + Inv2? The answer is – the inversions we have to count during the merge step. Therefore, to get number of inversions, we need to add number of inversions in left subarray, right subarray and merge()

#3

In merge process, let i is used for indexing left sub-array and j for right sub-array. At any step in merge(), if a[i] is greater than a[j], then there are (mid – i) inversions. because left and right subarrays are sorted, so all the remaining elements in left-subarray (a[i+1], a[i+2] … a[mid]) will be greater than a[j]

#4

read my code for implementation…

#5
#6

thanks for the ans , i hope i would ask this before codechef lunctime so that i can able to solve similar question