Merge pull request #7650 from spadacciniweb/PWC-206

PWC 206 - Perl, Python, Go

5 files changed, 355 insertions, 0 deletions

diff --git a/challenge-206/spadacciniweb/go/ch-1.go b/challenge-206/spadacciniweb/go/ch-1.go
new file mode 100644
index 0000000000..3a97ddde59
--- /dev/null
+++ b/challenge-206/spadacciniweb/go/ch-1.go
@@ -0,0 +1,86 @@
+/*
+Task 1: Shortest Time
+Submitted by: Mohammad S Anwar
+
+You are given a list of time points, at least 2, in the 24-hour clock format HH:MM.
+Write a script to find out the shortest time in minutes between any two time points.
+
+Example 1
+Input: @time = ("00:00", "23:55", "20:00")
+Output: 5
+
+Since the difference between "00:00" and "23:55" is the shortest (5 minutes).
+
+Example 2
+Input: @array = ("01:01", "00:50", "00:57")
+Output: 4
+
+Example 3
+Input: @array = ("10:10", "09:30", "09:00", "09:55")
+Output: 15
+*/
+
+package main
+
+import (
+    "fmt"
+    "log"
+    "os"
+    "strconv"
+    "gonum.org/v1/gonum/stat/combin"
+)
+
+func abs(x int) int {
+    if x < 0 {
+        return -x
+    }
+    return x
+}
+
+func getDistanceMinutes(m1 int, m2 int) int {
+    var dist = abs(m1 - m2)
+    if dist >= 720 {        // 12 hh * 60 = minutes
+        return 1440 - dist  // 24 hh * 60 = minutes
+    }
+    return dist
+}
+
+func getMinutes(hhmm string) int {
+    hh, mm := hhmm[0:2], hhmm[3:5]
+    hhInt, err := strconv.Atoi(hh)
+    if (err != nil) {
+        log.Fatal(err)
+    }
+    mmInt, err := strconv.Atoi(mm)
+    if (err != nil) {
+        log.Fatal(err)
+    }
+    return (hhInt % 24) * 60 + mmInt
+}
+
+func main() {
+    arrStr := os.Args[1:]
+    if (len(arrStr) < 2) {
+        log.Fatal("input error")
+    }
+
+    minutes := make([]int, len(arrStr))
+    for i := 0; i <= len(arrStr)-1; i++ {
+        minutes[i] = getMinutes(arrStr[i])
+    }
+
+    var minPointer *int
+    var pair [2]string
+    cs := combin.Combinations(len(minutes), 2)
+    for _, c := range cs {
+        var dist = getDistanceMinutes(minutes[c[0]], minutes[c[1]])
+        if minPointer == nil || *minPointer > dist {
+            minPointer = &dist
+            pair[0] = arrStr[c[0]]
+            pair[1] = arrStr[c[1]]
+        }
+    }
+
+    fmt.Printf("Output: %d", *minPointer)
+    fmt.Printf("\nSince the difference between \"%s\" and \"%s\" is the shortest (%d minutes).\n", pair[0], pair[1], *minPointer)
+}
diff --git a/challenge-206/spadacciniweb/perl/ch-1.pl b/challenge-206/spadacciniweb/perl/ch-1.pl
new file mode 100644
index 0000000000..481de5d4a1
--- /dev/null
+++ b/challenge-206/spadacciniweb/perl/ch-1.pl
@@ -0,0 +1,74 @@
+#!/usr/bin/env perl
+
+# Task 1: Shortest Time
+# Submitted by: Mohammad S Anwar
+# 
+# You are given a list of time points, at least 2, in the 24-hour clock format HH:MM.
+# Write a script to find out the shortest time in minutes between any two time points.
+#
+# Example 1
+# Input: @time = ("00:00", "23:55", "20:00")
+# Output: 5
+# 
+# Since the difference between "00:00" and "23:55" is the shortest (5 minutes).
+# 
+# Example 2
+# Input: @array = ("01:01", "00:50", "00:57")
+# Output: 4
+# 
+# Example 3
+# Input: @array = ("10:10", "09:30", "09:00", "09:55")
+# Output: 15
+
+use strict;
+use warnings;
+use Math::Combinatorics;
+
+my @input = @ARGV;
+die "Input error\n"
+    if scalar @input < 2
+       or
+       (scalar map { $_ =~ /^\d{2}:\d{2}$/ ? () : 1 }
+                @input) != 0
+       or
+       (scalar map { $_ le '24:00' ? () : 1 }
+                @input) != 0;
+
+my @minutes = map { get_minutes($_) } @input;
+
+my $combinat = Math::Combinatorics->new(count => 2,
+                                        data => [0..$#minutes],
+                                       );
+my %min;
+while (my @pair = $combinat->next_combination) {
+    my $diff_minutes = diff_minutes($minutes[$pair[0]], $minutes[$pair[1]]);
+    if (!defined($min{val})
+        or
+        $min{val} > $diff_minutes
+    ) {
+        $min{val} = $diff_minutes;
+        $min{pair} = \@pair;
+    }
+}
+
+print $min{val};
+printf "\nSince the difference between \"%s\" and \"%s\" is the shortest (%d minutes).\n", $input[ $min{pair}[0] ], $input[ $min{pair}[1] ], $min{val};
+
+exit 0;
+
+sub get_minutes {
+    my $time = shift;
+
+    return 60 * (substr $time, 0, 2) +
+                (substr $time, 3, 2);
+}
+
+sub diff_minutes {
+    my ($m1, $m2) = @_;
+
+    my $diff_minutes = abs($m1-$m2);
+    
+    return ($diff_minutes >= 720)        # 12h * 60 = minutes
+        ? (1440 - $diff_minutes)         # 24h * 60 = minutes
+        : $diff_minutes;
+}
diff --git a/challenge-206/spadacciniweb/perl/ch-2.pl b/challenge-206/spadacciniweb/perl/ch-2.pl
new file mode 100644
index 0000000000..94c3d005c1
--- /dev/null
+++ b/challenge-206/spadacciniweb/perl/ch-2.pl
@@ -0,0 +1,64 @@
+#!/usr/bin/env perl
+
+# Task 2: Array Pairings
+# Submitted by: Mohammad S Anwar
+# 
+# You are given an array of integers having even number of elements..
+# Write a script to find the maximum sum of the minimum of each pairs.
+# 
+# Example 1
+# Input: @array = (1,2,3,4)
+# Output: 4
+# 
+# Possible Pairings are as below:
+# a) (1,2) and (3,4). So min(1,2) + min(3,4) => 1 + 3 => 4
+# b) (1,3) and (2,4). So min(1,3) + min(2,4) => 1 + 2 => 3
+# c) (1,4) and (2,3). So min(1,4) + min(2,3) => 2 + 1 => 3
+# 
+# So the maxium sum is 4.
+# 
+# Example 2
+# Input: @array = (0,2,1,3)
+# Output: 2
+# 
+# Possible Pairings are as below:
+# a) (0,2) and (1,3). So min(0,2) + min(1,3) => 0 + 1 => 1
+# b) (0,1) and (2,3). So min(0,1) + min(2,3) => 0 + 2 => 2
+# c) (0,3) and (2,1). So min(0,3) + min(2,1) => 0 + 1 => 1
+# 
+# So the maximum sum is 2.
+
+use strict;
+use warnings;
+use Algorithm::Combinatorics qw(partitions);
+use List::Util qw(min max sum);
+
+my @input = @ARGV;
+die "Input error\n"
+    if scalar @input < 1
+       or
+       (scalar @input) % 2 != 0
+       or
+       (scalar map { $_ =~ /\D/ ? 1 : () }
+                @input) != 0;
+
+my @pairing;
+my $iter = partitions([map { int($_) } @input], (scalar @input/2));
+while (my $p = $iter->next) {
+    my $bool = 1;
+    foreach my $set (@$p) {
+        if (scalar @$set != 2) {
+            $bool = 0;
+            last;
+        }
+    }
+    push @pairing, $p
+        if $bool;
+}
+
+my @min;
+foreach my $pairs (@pairing) {
+    push @min, sum map { min @{$_} }
+                     values @$pairs;
+}
+printf "%s\n", max @min;
diff --git a/challenge-206/spadacciniweb/python/ch-1.py b/challenge-206/spadacciniweb/python/ch-1.py
new file mode 100644
index 0000000000..14d7e34598
--- /dev/null
+++ b/challenge-206/spadacciniweb/python/ch-1.py
@@ -0,0 +1,58 @@
+# Task 1: Shortest Time
+# Submitted by: Mohammad S Anwar
+# 
+# You are given a list of time points, at least 2, in the 24-hour clock format HH:MM.
+# Write a script to find out the shortest time in minutes between any two time points.
+#
+# Example 1
+# Input: @time = ("00:00", "23:55", "20:00")
+# Output: 5
+# 
+# Since the difference between "00:00" and "23:55" is the shortest (5 minutes).
+# 
+# Example 2
+# Input: @array = ("01:01", "00:50", "00:57")
+# Output: 4
+# 
+# Example 3
+# Input: @array = ("10:10", "09:30", "09:00", "09:55")
+# Output: 15
+
+import re
+import sys
+from itertools import combinations
+
+def get_minutes(t):
+    return 60 * int(t[0:2]) + int(t[3:5])
+
+def get_diff_minutes(m1, m2):
+    res = abs(m1-m2)
+
+    if (res >= 720):        # 12h * 60 = minutes
+        return 1440 - res   # 24h * 60 = minutes
+    return res
+
+if __name__ == "__main__":
+    input = sys.argv[1:]
+    if (len(input) < 2
+        or
+        len(list(filter(lambda x: re.search(r'\d\d:\d\d', x), input))) != len(input)
+        or
+        len(list(filter(lambda x: x > '24:00', input))) != 0 ):
+        sys.exit("Input error\n")
+
+    minutes = [get_minutes(x) for x in input]
+    comb = combinations(range(len(input)), 2)
+    h_minutes = {
+        'val': None,
+        'pair': None
+    }
+
+    for pair in list(comb):
+        diff_minutes = get_diff_minutes(minutes[pair[0]], minutes[pair[1]])
+        if h_minutes['val'] == None or h_minutes['val'] > diff_minutes:
+            h_minutes['val'] = diff_minutes
+            h_minutes['pair'] = pair
+    
+    print(h_minutes['val'])
+    print("Since the difference between \"{:s}\" and \"{:s}\" is the shortest ({:d} minutes).".format(input[ h_minutes['pair'][0] ], input[ h_minutes['pair'][1] ], h_minutes['val'] ))
diff --git a/challenge-206/spadacciniweb/python/ch-2.py b/challenge-206/spadacciniweb/python/ch-2.py
new file mode 100644
index 0000000000..832d866b18
--- /dev/null
+++ b/challenge-206/spadacciniweb/python/ch-2.py
@@ -0,0 +1,73 @@
+# Task 2: Array Pairings
+# Submitted by: Mohammad S Anwar
+# 
+# You are given an array of integers having even number of elements..
+# Write a script to find the maximum sum of the minimum of each pairs.
+# 
+# Example 1
+# Input: @array = (1,2,3,4)
+# Output: 4
+# 
+# Possible Pairings are as below:
+# a) (1,2) and (3,4). So min(1,2) + min(3,4) => 1 + 3 => 4
+# b) (1,3) and (2,4). So min(1,3) + min(2,4) => 1 + 2 => 3
+# c) (1,4) and (2,3). So min(1,4) + min(2,3) => 2 + 1 => 3
+# 
+# So the maxium sum is 4.
+# 
+# Example 2
+# Input: @array = (0,2,1,3)
+# Output: 2
+# 
+# Possible Pairings are as below:
+# a) (0,2) and (1,3). So min(0,2) + min(1,3) => 0 + 1 => 1
+# b) (0,1) and (2,3). So min(0,1) + min(2,3) => 0 + 2 => 2
+# c) (0,3) and (2,1). So min(0,3) + min(2,1) => 0 + 1 => 1
+# 
+# So the maximum sum is 2.
+import re
+import sys
+from functools import reduce
+
+# https://stackoverflow.com/questions/19368375/set-partitions-in-python
+def partition(collection):
+    if len(collection) == 1:
+        yield [ collection ]
+        return
+
+    first = collection[0]
+    for smaller in partition(collection[1:]):
+        # insert `first` in each of the subpartition's subsets
+        for n, subset in enumerate(smaller):
+            yield smaller[:n] + [[ first ] + subset]  + smaller[n+1:]
+        # put `first` in its own subset 
+        yield [ [ first ] ] + smaller
+
+
+
+if __name__ == "__main__":
+    input = sys.argv[1:]
+    if (len(input) < 2
+        or
+        len(input) % 2 != 0
+        or
+        len(list(filter(lambda x: re.search(r'\D', x), input))) != 0):
+        sys.exit("Input error\n")
+
+    input = list(map(int, input))
+    pairing = []
+    for n, p in enumerate(partition(input), 1):
+        sets = sorted(p)
+        b_pair = True
+        for s in sets:
+            if len(s) != 2:
+                b_pair = False
+                break
+        if b_pair == True:
+            pairing.append(sets)
+
+    l_min = []
+    for pairs in list(pairing):
+        l_min.append([min(s) for s in pairs])
+
+    print(max([sum(s) for s in l_min]))