From 369e3fb38abf3ee3e2dbde3ad919591d6004a390 Mon Sep 17 00:00:00 2001
From: Pavel Kuptsov <pavel@kuptsov.info>
Date: Mon, 17 Aug 2020 16:50:35 +0300
Subject: Add solution for Task-1 at Ch-074

---
 challenge-074/pavel_kuptsov/perl/ch-1.pl | 39 ++++++++++++++++++++++++++++++++
 1 file changed, 39 insertions(+)
 create mode 100644 challenge-074/pavel_kuptsov/perl/ch-1.pl

diff --git a/challenge-074/pavel_kuptsov/perl/ch-1.pl b/challenge-074/pavel_kuptsov/perl/ch-1.pl
new file mode 100644
index 0000000000..c9a66cb4a5
--- /dev/null
+++ b/challenge-074/pavel_kuptsov/perl/ch-1.pl
@@ -0,0 +1,39 @@
+#!/usr/bin/perl -w
+use strict;
+use warnings;
+use v5.26;
+use Test::More;
+use POSIX qw(floor);
+# TASK #1 › Majority Element
+# Submitted by: Mohammad S Anwar
+# You are given an array of integers of size $N.
+
+# Write a script to find the majority element. If none found then print -1.
+
+# Majority element in the list is the one that appears more than floor(size_of_list/2).
+
+# Example 1
+# Input: @A = (1, 2, 2, 3, 2, 4, 2)
+# Output: 2, as 2 appears 4 times in the list which is more than floor(7/2).
+
+# Example 2
+# Input: @A = (1, 3, 1, 2, 4, 5)
+# Output: -1 as none of the elements appears more than floor(6/2).
+
+my @A = (1, 2, 2, 3, 2, 4, 2);
+my @B = (1, 3, 1, 2, 4, 5);
+
+sub majority_el
+{
+  my $arr_ref = shift;
+  my $o = {};
+  map { $o->{$_}++ } @$arr_ref;
+  my ($max) = sort { $o->{$b} <=> $o->{$a} } keys %$o;
+
+  return $o->{$max} >= floor(@$arr_ref/2) ? $max : -1;
+}
+
+ok(majority_el(\@A) == 2 );
+ok(majority_el(\@B) == -1);
+
+done_testing(2);
\ No newline at end of file
-- 
cgit