Okay, that complicates things a little, but this seems to work:

#! perl -slw
use strict;
use Data::Dump qw[ pp ];

sub dhm2int {
    my( $d, $h, $m ) = unpack '(A2)*', $_[0];
    return ( ( $d - 1 ) * 24 + $h ) * 60 + $m;
}

sub int2dhm {
    sprintf "%02d%02d%02d", int($_[0]/1440)+1, int($_[0]/60)%24, $_[0]
+%60;
}

sub adjacentPairs (&@) {
    my $code = shift;
    map {
        $code->( shift(), $_[0] );
    } 1 .. @_;
}

#my @listone = 
#    (['010000','010010',2],['010200','010210',5],['012359','020001',3
+]);
#my @listtwo = 
#    (['010005','010015',1],['010207','010211',4]);
my @listone = 
    (['010000','010110',6],['010200','010210',5],['012350','020012',3]
+);
my @listtwo = 
    (['010005','010015',1],['010207','010211',4],['012355','020003',1]
+);

my @res = adjacentPairs{
    defined( $_[1] ) ? do {
        $_[1][3] > $_[0][3]
            ? [ $_[0][0], int2dhm( $_[1][3] - 1 ), $_[0][2] ]
            : [ int2dhm( $_[1][4] + 1 ), $_[0][1], $_[0][2] ]
    }
    : [ @{ $_[0] }[ 0 .. 2 ] ]
} sort {
    $a->[ 3 ] <=> $b->[ 3 ]
    ||
    $b->[ 4 ] <=> $a->[ 4 ]
} map {
    my $in = $_;
    my $s = dhm2int( $in->[ 0 ] );
    my $e = dhm2int( $in->[ 1 ] );
    my @out;
    while( int( $s / 1440 ) != int( $e / 1440 ) ) {
        my $newe = ( int( $s / 1440 ) +1 ) * 1440 -1;
        push @out, [ int2dhm( $s ), int2dhm( $newe ), $in->[2], $s, $n
+ewe ];
        $s = $newe +1;
    }
    ( @out, [ int2dhm( $s ), $in->[1], $in->[2], $s, $e ] );
} @listone, @listtwo;

pp \@res;

__END__
c:\test>junk33
[
  ["010000", "010004", 2],
  ["010005", "010015", 1],
  ["010200", "010206", 5],
  ["010207", "010211", 4],
  ["012359", "012359", 3],
  ["020000", "020001", 3],
]

c:\test>junk33
[
  ["010000", "010004", 6],
  ["010005", "010159", 1],
  ["010200", "010206", 5],
  ["010207", "012349", 4],
  ["012350", "012354", 3],
  ["012355", "012359", 1],
  ["020004", "020012", 3],
  ["020000", "020003", 1],
]
[download]

Note: The second set of results apply to the second (uncommented) set of data

One consequence is that the output is not longer in strictly sorted order. If this is a requirement a second sort will be required.

Where does this '010159' come from?

0                  110        200   210        2350   12
|-------------------| (6)      |-----| (5)      |------| (3)
   5   15                         207  211      2355  3
   |----| (1)                      |----| (4)    |----| (1)
[download]

Good question!

I was concentrating on the new requirement so much I didn't notice that I'd screwed the old one up!

---

With the rise and rise of 'Social' network sites: 'Computers are making people easier to use everyday'

Examine what is said, not who speaks -- Silence betokens consent -- Love the truth but pardon error.

"Science is about questioning the status quo. Questioning authority".

In the absence of evidence, opinion is indistinguishable from prejudice.

it is possible for the smaller duration one to have a lower value.

So for

111111111111111111111
    6666666
[download]

The smaller interval would vanish completely because it has a higher value than the overlapping interval?

Correct, because the smaller duration window has a higher value, the longer lower value would take priority and effectively completely remove the shorter window.

Hm. That's a curve ball for my adjacent pairs method.

I'd need to retain knowledge from the previous cycle in order to conditionally discard the shorter interval.

Need to thunk some more now. (Thunk! Thunk! Twang!).

---

With the rise and rise of 'Social' network sites: 'Computers are making people easier to use everyday'

Examine what is said, not who speaks -- Silence betokens consent -- Love the truth but pardon error.

"Science is about questioning the status quo. Questioning authority".

In the absence of evidence, opinion is indistinguishable from prejudice.