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Modified merge sort that counts inversions during the merge step:
when an element from the right half is selected, the remaining
elements of the left half (mid - i + 1) all form inversions with
that right element.
count_inv [|8; 4; 2; 1; 3; 5; 7; 6|] = 12
Inversions of [8;4;2;1;3;5;7;6]:
with 8: (8,4)(8,2)(8,1)(8,3)(8,5)(8,7)(8,6) = 7
with 4: (4,2)(4,1)(4,3) = 3
with 2: (2,1) = 1
with 7: (7,6) = 1
total = 12
Tests: let rec ... and ... mutual recursion, while + ref + array
mutation, in-place sort with auxiliary scratch array.
145 baseline programs total.
43 lines
941 B
OCaml
43 lines
941 B
OCaml
let count_inv arr =
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let n = Array.length arr in
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let temp = Array.make n 0 in
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let count = ref 0 in
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let rec merge lo mid hi =
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let i = ref lo and j = ref (mid + 1) and k = ref lo in
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while !i <= mid && !j <= hi do
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if arr.(!i) <= arr.(!j) then begin
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temp.(!k) <- arr.(!i);
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i := !i + 1
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end else begin
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temp.(!k) <- arr.(!j);
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count := !count + (mid - !i + 1);
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j := !j + 1
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end;
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k := !k + 1
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done;
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while !i <= mid do
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temp.(!k) <- arr.(!i);
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i := !i + 1; k := !k + 1
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done;
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while !j <= hi do
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temp.(!k) <- arr.(!j);
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j := !j + 1; k := !k + 1
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done;
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for x = lo to hi do
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arr.(x) <- temp.(x)
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done
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and sort lo hi =
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if lo < hi then begin
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let mid = (lo + hi) / 2 in
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sort lo mid;
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sort (mid + 1) hi;
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merge lo mid hi
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end
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in
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sort 0 (n - 1);
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!count
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;;
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count_inv [|8; 4; 2; 1; 3; 5; 7; 6|]
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