Other replies mentioned the "prefer ones" algorithm. If you think, instead, of it as a "prefer last" algorithm, it appears to work for your stated problem. Start with T of the first character. Then try to add the last character to the answer, moving from last character to second last character to third last character, etc., as long as a repetition exists.

#!/usr/bin/perl -l

# http://perlmonks.org/?node_id=1188292

use strict;
use warnings;

my $n = shift // 3;  # alphabet size
my $t = shift // $n; # size of tuples

my @alphabet = ('A'..'Z', '0'..'9', 'a'..'z');
my %previous;
@previous{@alphabet[1..$#alphabet]} = @alphabet; # last to 2nd last, e
+tc.
my $need = $n ** $t + $t - 1; # length of the answer

$_ = $alphabet[0] x $t; # start string with first chars

printf "%77s\n", $_ while
  s/^ (?=(.{$t})) (?=.+\1) . /$previous{$&}/x   # prev char if repeat
  or $need > length && s/^/$alphabet[$n - 1]/;  # or add last char

print; # the answer

my $chars = join '', @alphabet[0..$n-1];      # test if valid
/^[$chars]+$/ or die "invalid character";
my %all;
$all{$1}++ while /(?=(.{$t}))/g;
print "\nwant @{[ $n ** $t ]} unique tuples, have @{[ scalar keys %all
+ ]}";
print "solution passes tests";
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I haven't found a case yet where backtracking is required.