Build Lowest Number by Removing n digits from a given number - arrays

Question

It takes as input a number(string) and then the task is to remove n numbers to give a resulting lowest possible number, but you have to take care of the order, that is the constraint. You cannot change the order of the original numbers.
I wanted it to work in O(n), so I did this:
#!/usr/bin/perl
#lowestNS.pl
#Date: 2016-06-28
use warnings;
use strict;
use utf8;
(#ARGV == 2) or die "2 args needed";
my $num = $ARGV[0];
my $d = $ARGV[1];
my #arr;
int($num) > 0 or die "Enter positive number";
print "Number in: $num\nDel: $d\n";
if(int($num) == 0) {
print "Result: 0\n";
exit;
}
else {
my $str = $num;
#arr = split(//, $str); #/Split on each number
#Split and multiply by reverse index, to give precedence to the order of numbers
for(my $i = 0; $i < #arr; $i++) {
$arr[$i] *= (#arr - $i);
}
}
print "arr: " . join(',' , #arr) . "\n";
for (my $j = 0; $j < $d; $j++) {
my $max = $arr[0];
my $m_index = -1;
#replace nth maximum with -1
for (my $i = 0; $i < #arr; $i++) {
if($max <= $arr[$i]) {
$max = $arr[$i];
$m_index = $i;
}
}
$arr[$m_index] = -1;
}
#return all numbers with value other than -1
my $result = "";
for (my $i = 0; $i < #arr; $i++) {
if($arr[$i] != -1){
$result = $result . "" . $arr[$i]/(#arr - $i);
}
}
print "Result: $result\n";
It works in all cases, except, cases like:
Number = 765028321
Delete = 5
The problem is the removal of 765028321 when it should have removed the 765028321.
because 2*5 > 3*3.

Answer 1

I think, the algorithm is straightforward:
Suppose, N is the number of digits to delete;
1. Find the first smallest digit in the first N digits, delete digits to the left of it, decrease N by the number of digits deleted.
2. if N>0 take the digits to the right and repeat the steps above.
Of course, we need to check for marginal cases, and to ensure the final number does not start with 0.
Here is the code draft:
#!/usr/bin/perl
use strict;
use warnings;
my ($number, $del)=#ARGV;
my #num = $number=~m{.}g;
my $curpos=0;
my #finalnum;
for(;;) {
last if $del <=0 || $curpos+$del>=#num;
my $minpos=$curpos;
for (my $i=$curpos;$i<$curpos+$del+1 && $i < #num;$i++) {
if ($num[$i]<$num[$minpos] && !($curpos==0 && $num[$i]==0)) {
$minpos=$i;
}
}
push #finalnum, $num[$minpos];
$del-=($minpos-$curpos);
$curpos=$minpos+1;
}
push #finalnum, #num[$curpos+$del..$#num] if $curpos+$del < #num;
print join '', #finalnum;

Answer 2

A possible solution can be:
Create an array having count of all digits, eg. in your case, this array will look like: 1,1,2,1,0,1,1,1,1,0
Then, in a greedy manner, remove all big digits, as many as can be done. So, you remove 9s, but since its count is 0, you still need to remove 5 digits. Remove 8s, so you still need to remove 4 digits and so on.
This case is kind of trivial, so you will be left with 2 2's, 1 1's and 1 0's. In there were, say 3 4's also in the number, you will then need to remove 1 4. So, you will remove the leftmost of partial removals required.
Its a greedy approach, and works in O(n).