### Robbie Hatley's Solutions To The Weekly Challenge #271

For those not familiar with "The Weekly Challenge", it is a weekly programming puzzle with two parts, cycling every Sunday. You can find it here:

The Weekly Challenge for the week of 2024-05-26 through 2024-06-01 is #271. Its tasks are as follows:

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Task 271-1: Maximum Ones Submitted by: Mohammad Sajid Anwar You are given a m x n binary matrix. Write a script to return the row number containing maximum ones. In case of more than one row, then return smallest row number. # Example 1 input: [ [0, 1], [1, 0], ], # Expected output: 1 (Row 1 and Row 2 have the same number of ones, so return 1.) # Example 2 input: [ [0, 0, 0], [1, 0, 1], ], # Expected output: 2 (Row 2 has the maximum ones, so return 2.) # Example 3 input: [ [0, 0], [1, 1], [0, 0], ], # Expected output: 2 (Row 2 have the maximum ones, so return 2.)

I attack this problem by sorting the row indices in reverse order of row sums, then returning 1 + 0th element of sorted indices:

# Return the 1-based index of the first row of a binary matrix with maximum number of "1" elements: sub max_ones ($mref) {(sort {sum0(@{$$mref[$b]}) <=> sum0(@{$$mref[$a]}) || $a <=> $b} 0..$#$mref)[0]+1}

Robbie Hatley's Perl Solution to The Weekly Challenge 271-1

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Task 271-2: Sort by 1 bits Submitted by: Mohammad Sajid Anwar Given an array of non-negative integers, write a script to sort the integers in ascending order by the number of 1 bits in their binary representation. In case more than one integers have the same number of 1 bits then sort them in ascending order. # Example 1 input: [0, 1, 2, 3, 4, 5, 6, 7, 8], # Expected output: (0, 1, 2, 4, 8, 3, 5, 6, 7) 0 = 0 one bits 1 = 1 one bits 2 = 1 one bits 4 = 1 one bits 8 = 1 one bits 3 = 2 one bits 5 = 2 one bits 6 = 2 one bits 7 = 3 one bits # Example 2 input: [1024, 512, 256, 128, 64], # Expected output: (64, 128, 256, 512, 1024) All integers in the given array have one 1-bits, so just sort them in ascending order.

I first made a sub "ones" that counts the number of "1" bits of a non-negative integer. Then I make a sub "sort_by_ones" that sorts an array of non-negative integers primarily by ascending number-of-ones and secondarily by ascending value:

# Return number of "1" digits in the binary # representation of a non-negative integer: sub ones ($x) { my $ones = 0; while ($x) { 1 == $x%2 and ++$ones; $x >>= 1;} return $ones} # Sort an array of non-negative integers, # primarily by number of ones, # and secondarily by value: sub sort_by_ones :prototype(@) (@a) { sort {ones($a)<=>ones($b) || $a<=>$b} @a}

Robbie Hatley's Perl Solution to The Weekly Challenge 271-2

That's it for challenge 271; see you on challenge 272!

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