Find next lexicographic - permutation (permutations as it would occur in a dictionary order).
E.g: nextP('ABCD') = ABDC
If you can generate the next permutation, then you can also generate a 'cycle' of all permutations using a construct like,
cycle = +1;
curr = start;
while ( true )
curr = nextP(curr);
if ( curr == start )
break;
end
cycle = cycle+1;
end
For fun, you could generate all the n! permutations of a, unique n-letter string.
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I think test case 6 is wrong. The lexicographic follower of 'xyzDOG' is 'xyzGDO', not 'xyzGOD'.
@Stefan, if you see carefully, test case 6 asks for the *next to next* lexicographic permutation of 'DOG', which is 'GOD'.