Q. Give a divide- and- conquer algorithm that computes the number of inversions in array A in O(n log n) time. Show that your algorithm takes O(n log n).
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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().
How to get number of inversions in
merge()?
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 equal to 2*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]
Since we know that merge sort takes time T(n)=O(n*log(n)). The above algo basically modifies merge sort. So, it also takes O(nlog(n))
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Q. Give a divide- and- conquer algorithm that computes the number of inversions in array A...
2. Using Python, implement the Divide-and-Conquer algorithm to count the number of inversions between two arrays. The algorithm is based on Mergesort and counts the inversions while merging the sorted lists.
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