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+                            ━━━━━━━━━━━━━━━
+                             CHALLENGE 078
+                            ━━━━━━━━━━━━━━━
+
+
+Table of Contents
+─────────────────
+
+1 Task 1 - Leader Element
+.. 1.1 Perl
+2 Task 2 - Left Rotation
+.. 2.1 Perl
+
+
+
+
+
+1 Task 1 - Leader Element
+═════════════════════════
+
+  You are given an array @A containing distinct integers.
+
+  Write a script to find all leader elements in the array @A. Print (0)
+  if none found.
+
+  • An element is leader if it is greater than all the elements to its
+    right side.
+
+
+1.1 Perl
+────────
+
+  • Program: [file:perl/ch-1.pl]
+
+  We take input from `@ARGV', loop over it. And then we loop over the
+  elements at right, goto next if `$arg' is less than `$elm'. This will
+  push all the leader elements to `@leader'.
+  ┌────
+  │ my @leader;
+  │ MAIN: while (my $arg = shift @ARGV) {
+  │     foreach my $elm (@ARGV) {
+  │         next MAIN if $arg < $elm;
+  │     }
+  │     push @leader, $arg;
+  │ }
+  └────
+
+
+2 Task 2 - Left Rotation
+════════════════════════
+
+  You are given array @A containing positive numbers and @B containing
+  one or more indices from the array @A.
+
+  Write a script to left rotate @A so that the number at the first index
+  of @B becomes the first element in the array. Similary, left rotate @A
+  again so that the number at the second index of @B becomes the first
+  element in the array.
+
+
+2.1 Perl
+────────
+
+  • Program: [file:perl/ch-2.pl]
+
+  Loop over `@B' & then rotate the elements. Same could've been done
+  with Left Rotation, I find this easier to understand.
+  ┌────
+  │ my @A = qw(10 20 30 40 50);
+  │ my @B = qw(3 4);
+  │
+  │ foreach (@B) {
+  │     my @tmp = @A;
+  │     foreach (1 ... scalar @tmp - $_) {
+  │         unshift @tmp, pop @tmp;
+  │     }
+  │     print join(', ', @tmp), "\n";
+  │ }
+  └────