Add ch-1 in Perl, Go, Elixir, Python, Ruby

5 files changed, 323 insertions, 0 deletions

diff --git a/challenge-259/spadacciniweb/elixir/ch-1.exs b/challenge-259/spadacciniweb/elixir/ch-1.exs
new file mode 100644
index 0000000000..434545bfd9
--- /dev/null
+++ b/challenge-259/spadacciniweb/elixir/ch-1.exs
@@ -0,0 +1,57 @@
+# Task 1: Banking Day Offset
+# Submitted by: Lee Johnson
+# 
+# You are given a start date and offset counter. Optionally you also get bank holiday date list.
+# 
+# Given a number (of days) and a start date, return the number (of days) adjusted to take into account non-banking days. In other words: convert a banking day offset to a calendar day offset.
+# 
+# Non-banking days are:
+# a) Weekends
+# b) Bank holidays
+# 
+# Example 1
+# Input: $start_date = '2018-06-28', $offset = 3, $bank_holidays = ['2018-07-03']
+# Output: '2018-07-04'
+# 
+# Thursday bumped to Wednesday (3 day offset, with Monday a bank holiday)
+# 
+# Example 2
+# Input: $start_date = '2018-06-28', $offset = 3
+# Output: '2018-07-03'
+
+defmodule BankingDay do
+
+  def banking_day_offset(start_date, offset, bank_holidays) do
+    out(start_date, offset, bank_holidays,
+        next_working_day( Date.from_iso8601!(start_date), offset, bank_holidays)
+       )
+  end
+
+  def next_working_day(dt, offset, bank_holidays) when offset > 0 do
+    dt1 = Date.add(dt, 1)
+    if Date.day_of_week(dt1) >= 6 || Enum.member?(bank_holidays, "#{dt1}") do
+        next_working_day(dt1, offset, bank_holidays)
+    else
+        next_working_day(dt1, offset-1, bank_holidays)
+    end
+  end
+
+ def next_working_day(dt, 0, _bank_holidays) do
+    Date.to_iso8601(dt)
+  end
+
+  def out(start_date, offset, bank_holidays, ymd) do
+    IO.write( "(start #{start_date} offset #{offset} Bank holidays [" <> Enum.join(bank_holidays, ", ") <> "]) -> ")
+    IO.puts( ymd )
+  end
+end
+
+start_date = "2018-06-28"
+offset = 3
+bank_holidays = ["2018-07-03"]
+BankingDay.banking_day_offset(start_date, offset, bank_holidays)
+
+start_date = "2018-06-28"
+offset = 3
+bank_holidays = []
+BankingDay.banking_day_offset(start_date, offset, bank_holidays)
diff --git a/challenge-259/spadacciniweb/go/ch-1.go b/challenge-259/spadacciniweb/go/ch-1.go
new file mode 100644
index 0000000000..37446a936b
--- /dev/null
+++ b/challenge-259/spadacciniweb/go/ch-1.go
@@ -0,0 +1,80 @@
+/*
+Task 1: Banking Day Offset
+Submitted by: Lee Johnson
+
+You are given a start date and offset counter. Optionally you also get bank holiday date list.
+
+Given a number (of days) and a start date, return the number (of days) adjusted to take into account non-banking days. In other words: convert a banking day offset to a calendar day offset.
+
+Non-banking days are:
+a) Weekends
+b) Bank holidays
+
+Example 1
+Input: $start_date = '2018-06-28', $offset = 3, $bank_holidays = ['2018-07-03']
+Output: '2018-07-04'
+
+Thursday bumped to Wednesday (3 day offset, with Monday a bank holiday)
+
+Example 2
+Input: $start_date = '2018-06-28', $offset = 3
+Output: '2018-07-03'
+*/
+
+package main
+
+import (
+    "fmt"
+    "log"
+    "strings"
+    "time"
+)
+
+func contains(s []string, str string) bool {
+    for _, v := range s {
+        if v == str {
+            return true
+        }
+    }
+    return false
+}
+
+func no_banking_day(t time.Time, bank_holidays []string) bool {
+    ymd := t.Format("2006-01-02")
+    if (t.Weekday() == 0 || t.Weekday() == 6 || contains(bank_holidays, ymd)) {
+        return true
+    }
+    return false
+}
+
+func banking_day_offset(start_date string, offset int, bank_holidays []string) {
+    t, err := time.Parse("2006-01-02", start_date)
+    if err != nil {
+        log.Fatal(err)
+    }
+
+    for i := 1; i <= offset; i++ {
+        t = t.AddDate(0, 0, 1)
+        for no_banking_day(t, bank_holidays) {
+            t = t.AddDate(0, 0, 1)
+        }
+    }
+
+    fmt.Printf("start %s offset %d Bank holidays [%s]) -> %s\n",
+        start_date, offset,
+        strings.Join(bank_holidays, " "),
+        t.Format("2006-01-02"))
+}
+
+func main() {
+    start_date := "2018-06-28"
+    offset := 3;
+    bank_holidays := []string{"2018-07-03"}
+    banking_day_offset(start_date, offset, bank_holidays)
+
+    start_date = "2018-06-28"
+    offset = 3;
+    bank_holidays = []string{}
+    banking_day_offset(start_date, offset, bank_holidays)
+}
+
diff --git a/challenge-259/spadacciniweb/perl/ch-1.pl b/challenge-259/spadacciniweb/perl/ch-1.pl
new file mode 100644
index 0000000000..8f804849b6
--- /dev/null
+++ b/challenge-259/spadacciniweb/perl/ch-1.pl
@@ -0,0 +1,76 @@
+#!/usr/bin/env perl
+
+# Task 1: Banking Day Offset
+# Submitted by: Lee Johnson
+# 
+# You are given a start date and offset counter. Optionally you also get bank holiday date list.
+# 
+# Given a number (of days) and a start date, return the number (of days) adjusted to take into account non-banking days. In other words: convert a banking day offset to a calendar day offset.
+# 
+# Non-banking days are:
+# a) Weekends
+# b) Bank holidays
+# 
+# Example 1
+# Input: $start_date = '2018-06-28', $offset = 3, $bank_holidays = ['2018-07-03']
+# Output: '2018-07-04'
+# 
+# Thursday bumped to Wednesday (3 day offset, with Monday a bank holiday)
+# 
+# Example 2
+# Input: $start_date = '2018-06-28', $offset = 3
+# Output: '2018-07-03'
+
+use strict;
+use warnings;
+use DateTime;
+
+my $start_date = '2018-06-28';
+my $offset = 3;
+my $bank_holidays = ['2018-07-03'];
+banking_day_offset($start_date, $offset, $bank_holidays);
+
+$start_date = '2018-06-28';
+$offset = 3;
+$bank_holidays = [];
+banking_day_offset($start_date, $offset, $bank_holidays);
+
+exit 0;
+
+sub banking_day_offset {
+    my $start_date = shift;
+    my $offset = shift;
+    my $bank_holidays = shift;
+
+    my ($yyyy, $mm, $dd) = split /-/, $start_date;
+    my $dt = DateTime->new(
+        year  => $yyyy,
+        month => $mm,
+        day  => $dd
+    );
+    
+    foreach (1..$offset) {
+        $dt->add( days => 1 );
+        while (no_banking_day($dt)) {
+            $dt->add( days => 1 );
+        }
+    }
+
+    printf "(start %s offset %s Bank holidays [%s]) -> %s\n",
+        $start_date,
+        $offset,
+        (join ' / ', @$bank_holidays ),
+        $dt->ymd;
+
+    return undef;
+}
+
+sub no_banking_day {
+    my $dt = shift;
+    my $ymd = $dt->ymd;
+    return 1
+        if $dt->day_of_week >= 6
+                ||
+               grep /$ymd/, @$bank_holidays;
+    return 0;
+}
diff --git a/challenge-259/spadacciniweb/python/ch-1.py b/challenge-259/spadacciniweb/python/ch-1.py
new file mode 100644
index 0000000000..23cedaa661
--- /dev/null
+++ b/challenge-259/spadacciniweb/python/ch-1.py
@@ -0,0 +1,54 @@
+# Task 1: Banking Day Offset
+# Submitted by: Lee Johnson
+# 
+# You are given a start date and offset counter. Optionally you also get bank holiday date list.
+# 
+# Given a number (of days) and a start date, return the number (of days) adjusted to take into account non-banking days. In other words: convert a banking day offset to a calendar day offset.
+# 
+# Non-banking days are:
+# a) Weekends
+# b) Bank holidays
+# 
+# Example 1
+# Input: $start_date = '2018-06-28', $offset = 3, $bank_holidays = ['2018-07-03']
+# Output: '2018-07-04'
+# 
+# Thursday bumped to Wednesday (3 day offset, with Monday a bank holiday)
+# 
+# Example 2
+# Input: $start_date = '2018-06-28', $offset = 3
+# Output: '2018-07-03'
+
+from dateutil.parser import *
+import datetime
+
+def banking_day_offset(start_date, offset, bank_holidays):
+    dt = parse(start_date, dayfirst = False, yearfirst = True)
+
+    for i in range(1,5):
+        dt += datetime.timedelta(days=1)
+        while no_banking_day(dt):
+            dt += datetime.timedelta(days=1)
+
+    print("(start %s offset %s Bank holidays [%s]) -> %s" %
+            (start_date, offset,
+             ' / '.join(bank_holidays),
+             dt.strftime('%Y-%m-%d')
+            )
+    )
+
+def no_banking_day(dt):
+    if (dt.weekday() >= 6) or (dt.strftime('%Y-%m-%d') in bank_holidays):
+        return 1
+    return 0
+
+if __name__ == "__main__":
+    start_date = '2018-06-28'
+    offset = 3
+    bank_holidays = ['2018-07-03']
+    banking_day_offset(start_date, offset, bank_holidays)
+
+    start_date = '2018-06-28';
+    offset = 3;
+    bank_holidays = [];
+    banking_day_offset(start_date, offset, bank_holidays)
diff --git a/challenge-259/spadacciniweb/ruby/ch-1.rb b/challenge-259/spadacciniweb/ruby/ch-1.rb
new file mode 100644
index 0000000000..b069356a52
--- /dev/null
+++ b/challenge-259/spadacciniweb/ruby/ch-1.rb
@@ -0,0 +1,56 @@
+# Task 1: Banking Day Offset
+# Submitted by: Lee Johnson
+# 
+# You are given a start date and offset counter. Optionally you also get bank holiday date list.
+# 
+# Given a number (of days) and a start date, return the number (of days) adjusted to take into account non-banking days. In other words: convert a banking day offset to a calendar day offset.
+# 
+# Non-banking days are:
+# a) Weekends
+# b) Bank holidays
+# 
+# Example 1
+# Input: $start_date = '2018-06-28', $offset = 3, $bank_holidays = ['2018-07-03']
+# Output: '2018-07-04'
+# 
+# Thursday bumped to Wednesday (3 day offset, with Monday a bank holiday)
+# 
+# Example 2
+# Input: $start_date = '2018-06-28', $offset = 3
+# Output: '2018-07-03'
+
+require 'date'
+
+def banking_day_offset(start_date, offset, bank_holidays)
+    dt = Date.strptime(start_date, '%Y-%m-%d')
+
+    (1..offset).each do |i|
+        dt += 1
+        while no_banking_day(dt, bank_holidays) == true
+            dt += 1
+        end
+    end
+
+    printf "(start %s offset %s Bank holidays [%s]) -> %s\n",
+        start_date,
+        offset,
+        bank_holidays.join(" / "),
+        dt.strftime('%Y-%m-%d');
+end
+
+def no_banking_day(dt, bank_holidays)
+    if dt.strftime("%u").to_i >= 6 or bank_holidays.include?( dt.strftime('%Y-%m-%d') )
+        return true
+    end
+    false
+end
+
+start_date = '2018-06-28'
+offset = 3
+bank_holidays = ['2018-07-03']
+banking_day_offset(start_date, offset, bank_holidays)
+
+start_date = '2018-06-28';
+offset = 3;
+bank_holidays = [];
+banking_day_offset(start_date, offset, bank_holidays);