Add Python solution to challenge 127

4 files changed, 104 insertions, 6 deletions

diff --git a/challenge-127/paulo-custodio/perl/ch-1.pl b/challenge-127/paulo-custodio/perl/ch-1.pl
index dd8de28b85..60134894d8 100644
--- a/challenge-127/paulo-custodio/perl/ch-1.pl
+++ b/challenge-127/paulo-custodio/perl/ch-1.pl
@@ -8,7 +8,7 @@
 #
 # Write a script to figure out if they are disjoint.
 #
-# The two sets are disjoint if they don’t have any common members.
+# The two sets are disjoint if they don't have any common members.
 #
 # Example
 # Input: @S1 = (1, 2, 5, 3, 4)
diff --git a/challenge-127/paulo-custodio/perl/ch-2.pl b/challenge-127/paulo-custodio/perl/ch-2.pl
index 3fefaca3e5..39f76f858d 100644
--- a/challenge-127/paulo-custodio/perl/ch-2.pl
+++ b/challenge-127/paulo-custodio/perl/ch-2.pl
@@ -37,7 +37,7 @@ use Modern::Perl;
 
 my @intervals = parse_intervals(@ARGV);
 my @conflicts = find_conflicts(@intervals);
-print_conflicts(@conflicts);
+print_intervals(@conflicts);
 
 
 sub parse_intervals {
@@ -71,12 +71,11 @@ sub find_conflicts {
     return @conflicts;
 }
 
-sub print_conflicts {
-    my(@conflicts) = @_;
-    use Data::Dump 'dump';
+sub print_intervals {
+    my(@intervals) = @_;
     print "[ ";
     my $sep = "";
-    for (@conflicts) {
+    for (@intervals) {
         my($s, $e) = @$_;
         print $sep,"($s,$e)";
         $sep = ", ";
diff --git a/challenge-127/paulo-custodio/python/ch-1.py b/challenge-127/paulo-custodio/python/ch-1.py
new file mode 100644
index 0000000000..7313b831b1
--- /dev/null
+++ b/challenge-127/paulo-custodio/python/ch-1.py
@@ -0,0 +1,28 @@
+#!/usr/bin/env python3
+
+# Challenge 127
+#
+# TASK #1 > Disjoint Sets
+# Submitted by: Mohammad S Anwar
+# You are given two sets with unique integers.
+#
+# Write a script to figure out if they are disjoint.
+#
+# The two sets are disjoint if they don't have any common members.
+#
+# Example
+# Input: @S1 = (1, 2, 5, 3, 4)
+#        @S2 = (4, 6, 7, 8, 9)
+# Output: 0 as the given two sets have common member 4.
+#
+# Input: @S1 = (1, 3, 5, 7, 9)
+#        @S2 = (0, 2, 4, 6, 8)
+# Output: 1 as the given two sets do not have common member.
+
+s1 = set([int(x) for x in input().split()])
+s2 = set([int(x) for x in input().split()])
+z = s1.intersection(s2)
+if len(z)==0:
+    print(1)
+else:
+    print(0)
diff --git a/challenge-127/paulo-custodio/python/ch-2.py b/challenge-127/paulo-custodio/python/ch-2.py
new file mode 100644
index 0000000000..e72a3f39b9
--- /dev/null
+++ b/challenge-127/paulo-custodio/python/ch-2.py
@@ -0,0 +1,71 @@
+#!/usr/bin/env python3
+
+# TASK #2 > Conflict Intervals
+# Submitted by: Mohammad S Anwar
+# You are given a list of intervals.
+#
+# Write a script to find out if the current interval conflicts with any of the
+# previous intervals.
+#
+# Example
+# Input: @Intervals = [ (1,4), (3,5), (6,8), (12, 13), (3,20) ]
+# Output: [ (3,5), (3,20) ]
+#
+#     - The 1st interval (1,4) do not have any previous intervals to compare
+#       with, so skip it.
+#     - The 2nd interval (3,5) does conflict with previous interval (1,4).
+#     - The 3rd interval (6,8) do not conflicts with any of the previous intervals
+#       (1,4) and (3,5), so skip it.
+#     - The 4th interval (12,13) again do not conflicts with any of the previous
+#       intervals (1,4), (3,5) and (6,8), so skip it.
+#     - The 5th interval (3,20) conflicts with the first interval (1,4).
+#
+# Input: @Intervals = [ (3,4), (5,7), (6,9), (10, 12), (13,15) ]
+# Output: [ (6,9) ]
+#
+#     - The 1st interval (3,4) do not have any previous intervals to compare
+#       with, so skip it.
+#     - The 2nd interval (5,7) do not conflicts with the previous interval (3,4),
+#       so skip it.
+#     - The 3rd interval (6,9) does conflict with one of the previous intervals (5,7).
+#     - The 4th interval (10,12) do not conflicts with any of the previous
+#       intervals (3,4), (5,7) and (6,9), so skip it.
+#     - The 5th interval (13,15) do not conflicts with any of the previous
+#       intervals (3,4), (5,7), (6,9) and (10,12), so skip it.
+
+import sys
+
+def parse_intervals(args):
+    args = [int(x) for x in args]
+    out = []
+    while args:
+        out.append([args.pop(0), args.pop(0)])
+    return out
+
+def find_conflicts(intervals):
+    conflicts = []
+    for i in range(1, len(intervals)):
+        for j in range(0, i):
+            i1s = intervals[i][0]
+            i1e = intervals[i][1]
+            i2s = intervals[j][0]
+            i2e = intervals[j][1]
+            if (i1s >= i2s and i1s <  i2e) or \
+               (i1e >= i2s and i1e <  i2e) or \
+               (i1s <  i2s and i1e >= i2e):
+                conflicts.append(intervals[i])
+                break;
+    return conflicts
+
+def intervals_as_string(intervals):
+    out = "[ "
+    sep = ""
+    for interval in intervals:
+        out += f"{sep}({interval[0]},{interval[1]})"
+        sep = ", "
+    out += " ]"
+    return out
+
+intervals = parse_intervals(sys.argv[1:])
+conflicts = find_conflicts(intervals)
+print(intervals_as_string(conflicts))