Created
February 5, 2013 23:39
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Algorithm to count the number of inversions in an array
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| let countSplitInv (l : 'a array) (r : 'a array) n = | |
| let res = Array.zeroCreate<'a> n | |
| let ln = l.Length | |
| let mutable i, j, inv = 0, 0, 0 | |
| for k = 0 to n-1 do | |
| if i >= ln then res.[k] <- r.[j]; j <- j + 1 | |
| elif j >= r.Length then res.[k] <- l.[i]; i <- i + 1 | |
| elif l.[i] < r.[j] then res.[k] <- l.[i]; i <- i + 1 | |
| else res.[k] <- r.[j]; j <- j + 1; inv <- inv + ln - i | |
| res, inv | |
| let rec sortAndCount (arr : 'a array) = | |
| match arr.Length with | |
| | 0 | 1 -> arr, 0 | |
| | n -> let (b, x) = sortAndCount(arr.[0..n/2-1]) | |
| let (c, y) = sortAndCount(arr.[n/2..n-1]) | |
| let (d, z) = countSplitInv b c n | |
| d, x + y + z | |
| let countInversion arr = snd <| sortAndCount arr |
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