Find the missing permutation: Difference between revisions
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Alternate method : if we had all permutations, each letter would appear an even number of times at each position.
Since there is only one permutation missing, we can find where each letter goes by looking at the parity of
the number of occurences of each letter. The following program works with permutations of at least 3 letters.
<lang ocaml>let array_of_perm s =
let n = String.length s in
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let rec aux = function
[ ] -> () | s::w ->
let u =
for i=0 to n-1 do a.(i).(u.(i)) <- a.(i).(u.(i)) + 1 done;
aux w
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