17 May 2024  Challenge 269 
Bits and Bins
Task 1: Bitwise OR
Submitted by: Mohammad Sajid Anwar
You are given an array of positive integers, @ints
.
Write a script to find out if it is possible to select two or more elements of the given array such that the bitwise OR of the selected elements has atlest one trailing zero in its binary representation.
Example 1
Input: @ints = (1, 2, 3, 4, 5)
Output: true
Say, we pick 2 and 4, thier bitwise OR is 6. The binary representation of 6 is 110.
Return true since we have one trailing zero.
Example 2
Input: @ints = (2, 3, 8, 16)
Output: true
Say, we pick 2 and 8, thier bitwise OR is 10. The binary representation of 10 is 1010.
Return true since we have one trailing zero.
Example 3
Input: @ints = (1, 2, 5, 7, 9)
Output: false
Solution
There is no need to perform any OR operation in this task. The question may be formulated as:
Are there two or more elements having a trailing zero in their binary representation?
If so, the bitOR of these elements will have a trailing zero, too.
Thus all we need to do is pick the least significant bit from each number and count the zeroes thereof.
Using grep
in scalar context for the selection and count.
use strict;
use warnings;
sub mtz {
1 < grep !($_ & 1), @_;
}
See the full solution to task 1.
Task 2: Distribute Elements
Submitted by: Mohammad Sajid Anwar
You are given an array of distinct integers, @ints
.
Write a script to distribute the elements as described below:
1) Put the 1st element of the given array to a new array @arr1
.
2) Put the 2nd element of the given array to a new array @arr2
.
Once you have one element in each arrays, @arr1
and @arr2
, then follow the rule below:
If the last element of the array @arr1
is greater than the last
element of the array @arr2
then add the first element of the
given array to @arr1
otherwise to the array @arr2
.
When done distribution, return the concatenated arrays. @arr1
and @arr2
.
Example 1
Input: @ints = (2, 1, 3, 4, 5)
Output: (2, 3, 4, 5, 1)
1st operation:
Add 1 to @arr1 = (2)
2nd operation:
Add 2 to @arr2 = (1)
3rd operation:
Now the last element of @arr1 is greater than the last element
of @arr2, add 3 to @arr1 = (2, 3).
4th operation:
Again the last element of @arr1 is greate than the last element
of @arr2, add 4 to @arr1 = (2, 3, 4)
5th operation:
Finally, the last element of @arr1 is again greater than the last
element of @arr2, add 5 to @arr1 = (2, 3, 4, 5)
Mow we have two arrays:
@arr1 = (2, 3, 4, 5)
@arr2 = (1)
Concatenate the two arrays and return the final array: (2, 3, 4, 5, 1).
Example 2
Input: @ints = (3, 2, 4)
Output: (3, 4, 2)
1st operation:
Add 1 to @arr1 = (3)
2nd operation:
Add 2 to @arr2 = (2)
3rd operation:
Now the last element of @arr1 is greater than the last element
of @arr2, add 4 to @arr1 = (3, 4).
Mow we have two arrays:
@arr1 = (3, 4)
@arr2 = (2)
Concatenate the two arrays and return the final array: (3, 4, 2).
Example 3
Input: @ints = (5, 4, 3 ,8)
Output: (5, 3, 4, 8)
1st operation:
Add 1 to @arr1 = (5)
2nd operation:
Add 2 to @arr2 = (4)
3rd operation:
Now the last element of @arr1 is greater than the last element
of @arr2, add 3 to @arr1 = (5, 3).
4th operation:
Again the last element of @arr2 is greate than the last element
of @arr1, add 8 to @arr2 = (4, 8)
Mow we have two arrays:
@arr1 = (5, 3)
@arr2 = (4, 8)
Concatenate the two arrays and return the final array: (5, 3, 4, 8).
Solution
The numbers will be distributed onto the arrays @{$arr[0]}
and @{$arr[1]}
.
Developing a solution backwards from the middle:

When both arrays are occupied. The comparison of the arrays’ last elements can be used as an index for
@arr
. For the correct assignment of the target array, the result has to be inverted, leading to a starting expression as:push @{$arr[$arr[0][1] <= $arr[1][1]]}, $_ for @ints;
Next trying to find default values for $arr[$i][1]
in case of an empty array, such that the first two elements’ handling fits into the schema.

When distributing the second element from the list, the first given integer is the only element in the first array and the second array is empty. With the default for the second array, the inequality shall be true for all first values. We may choose
Inf
, as$i <= Inf
is true for all integers. 
When distributing the first element from the list, both arrays are empty and there is
Inf
as the default for the second array. With the default for the first array, the inequality shall be false. Though it looks strange to have a value that is not<= Inf
, the expressionNaN <= Inf
is false.
See perldata.
 Finally the array refs are dereferenced and concatenated.
use strict;
use warnings;
use POSIX qw(NaN Inf);
sub de {
my @arr;
push @{$arr[($arr[0][1] // NaN) <= ($arr[1][1] // Inf)]}, $_ for @_;
map @$_, @arr;
}
See the full solution to task 2.
If you have a question about this post or if you like to comment on it, feel free to open an issue in my github repository.