Challenge 083

Table of Contents

Task 1 - Words Length

You are given a string $S with 3 or more words.

Write a script to find the length of the string except the first and last words ignoring whitespace.


We split the input by space character & remove the first & last word with shift & pop respectively.

my @words = split / /, $ARGV[0];
shift @words;
pop @words;

Then we loop over @words & add the length of each word to $len & print it.

my $len;
$len += length($_) foreach @words;

print $len, "\n";

Task 2 - Flip Array

You are given an array @A of positive numbers.

Write a script to flip the sign of some members of the given array so that the sum of the all members is minimum non-negative.

Given an array of positive elements, you have to flip the sign of some of its elements such that the resultant sum of the elements of array should be minimum non-negative(as close to zero as possible). Return the minimum no. of elements whose sign needs to be flipped such that the resultant sum is minimum non-negative.


Note: The solution is incomplete.

We start by eliminating possible values, here by value I mean the minimum non-negative sum of those numbers.

Input is sorted & stored in @nums.

my @nums = @ARGV;
@nums = sort { $a <=> $b} @nums;

If the input contains a single number then the answer is 0 because flipping nothing will give us the minimum non-negative sum. If the input contains 2 numbers then the answer is 1 because flipping the smaller number will give us the minimum non-negative sum.

print "0\n" and exit 0 if scalar @nums == 1;
print "1\n" and exit 0 if scalar @nums == 2;

The sum of all inputs will be the ceiling for the value. We can eliminate more numbers by subtracting rest of the numbers from the largest one, for example we subtract [1, 2, 3] from 5 if input is [1, 2, 3, 5]. We'll get -1, just make it positive & we have our new ceiling.

In that example the value has to be 1 or less than 1. The value will be one in that case but it's not the answer, for example it would fail in [1, 2, 3, 4]. We'll get -2 which we mod to get 2 when the value is 0.

So, if we get a positive number when subtracting the rest from the largest one then the number is the value. Because if you make the largest one negative then adding the rest will give a negative number so the largest one must be positive. And you cannot make any other number positive because that'll just increase the sum.

# Multiplied by 2 because we'll subtract it once.
my $tmp = 2 * $nums[$#nums];
$tmp -= $_ foreach @nums;

If the sum is 0 then we just need to flip the largest number to get minimum non-negative sum, if it's greater than 0 then we need to flip all the numbers except the largest one.

print "1\n" and exit 0 if $tmp == 0;
print scalar @nums - 1, "\n" and exit 0 if $tmp > 0;

I haven't yet completed it.

die "ch-2.pl: cannot solve\n";

Further thinking

The value will always be zero if we have 4n consecutive numbers, where n is a natural number. This is true because when we add pop @input & shift @input we get same number & because there were 4n consecutive numbers we get even number of same numbers, we can just subtract half of same numbers from rest to get 0.

For example, [2, 3, 4, 5]. 5 + 2 equals 3 + 4, just subtract one set to get 0.

I had missed the "minimum number of elements" part. We need minimum number of elements with flipped sign to get minimum non-negative sum. But that can be solved once we get minimum non-negative sum. What this means is that if we have to flip a thousand signs to get the value then the solution is thousand but if we get the same value in 10 flips then the solution is 10.

Back to minimum non-negative sum. We could just write down all permutations & get the value but that's bad solution.

Also, till now I just assumed that the numbers are not repeated because it's easy to work around that. We will just keep count of how many numbers are repeated & add that to the final count because those numbers will cancel themselves. This does affect the set of numbers so this might lead us to wrong solution.

Actually that will lead us to the wrong solution, instead of cancelling each other some those numbers could be used to bring down the sum close to zero. For example, if in [1, 1, 1, 1, 5] we make the 1's cancel each other then the value we get is 5 & the answer 2 but that's wrong. We could've made the 1's bring down 5 to 1 (5 -1 -1 -1 -1) & the value would've been 1, the answer 4.

I think I'll leave it there for now.

Andinus / 2020-10-23 / Modified: 2022-10-04 Tue 21:34 Emacs 27.2 (Org mode 9.4.4)