Sorting

Merge Sort (Top-Down)

Split the array recursively, sort each half, then merge two sorted runs into one sorted result.

Algorithm

The checked-in replay follows the same small input and final output across all 21 DSA books, so this Perl DSA implementation can be compared directly with the other languages.

divide and conquer Each recursive call solves a smaller sorted subproblem.
merge step Two sorted halves are combined by repeatedly taking the smaller front item.

Visual walkthrough

The pinned input is [5, 1, 4, 2, 8]. The diagrams show the split into recursive halves, the sorted subarrays, and the final merge choices.

Step 1 - Split the input

The first midpoint splits [5, 1, 4, 2, 8] into left [5, 1] and right [4, 2, 8].

Top-down split used by merge_sort.[5,1,4,2,8]mid = 2[5,1]left[4,2,8]right

Step 2 - Sorted halves return

Recursive calls return [1, 5] and [2, 4, 8] before the final merge begins.

Returned subarrays before the final merge.sidebefore sortafter sortleft[5, 1][1, 5]right[4, 2, 8][2, 4, 8]

Step 3 - Merge by taking smaller fronts

Take 1 from left, then 2 and 4 from right, then the remaining 5 and 8.

Final merge produces [1, 2, 4, 5, 8].choiceleft frontright frontmergedtake 112[1]take 252[1, 2]take 454[1, 2, 4]extend58[1, 2, 4, 5, 8]

Basic Implementation

basic.pl 
use strict; use warnings;
sub merge_sort {
	my @values = @_;
	return @values if @values <= 1;
	my $mid = int(@values / 2);
	my @left = merge_sort(@values[0 .. $mid - 1]);
	my @right = merge_sort(@values[$mid .. $#values]);
	my @merged;
	while (@left && @right) {
		push @merged, ($left[0] <= $right[0]) ? shift @left : shift @right;
	}
	return (@merged, @left, @right);
}

my @arr = merge_sort(5, 1, 4, 2, 8);
print "[" . join(", ", @arr) . "]
";

Complexity

  • Time: O(n log n)
  • Space: O(n)
  • Stable: yes

Implementation notes

  • sub merge_sort receives its input as the argument list @_, then copies it into my @values = @_.
  • Perl arrays use the @ sigil here: @values, @left, @right, @merged, and the final @arr.
  • The base case is return @values if @values <= 1; in that numeric comparison, @values is in scalar context, so it means the list length.
  • my $mid = int(@values / 2) uses integer truncation to split the current list.
  • The recursive calls use Perl slice syntax: @values[0 .. $mid - 1] for the left half and @values[$mid .. $#values] for the right half.
  • The checked input is passed as a list in my @arr = merge_sort(5, 1, 4, 2, 8).
  • The trace shows the first split as left [5, 1] and right [4, 2, 8].
  • Recursive calls return sorted lists, so the replay then has left [1, 5] and right [2, 4, 8].
  • my @merged holds the output during the merge.
  • The merge loop runs while both halves still have elements: while (@left && @right).
  • The comparison is numeric and stable on ties: ($left[0] <= $right[0]) ? shift @left : shift @right.
  • shift removes the chosen front value from @left or @right, and push @merged, ... appends it to the merged result.
  • return (@merged, @left, @right) appends whichever half still has leftover values after the loop.
  • The final replayed merge returns [1, 2, 4, 5, 8].
  • The final print concatenates "[", join(", ", @arr), and a closing bracket string that includes the newline, producing [1, 2, 4, 5, 8].