Add Python solution to challenge 065

5 files changed, 94 insertions, 4 deletions

diff --git a/challenge-064/paulo-custodio/python/ch-2.py b/challenge-064/paulo-custodio/python/ch-2.py
index 12034bea57..deff42f193 100644
--- a/challenge-064/paulo-custodio/python/ch-2.py
+++ b/challenge-064/paulo-custodio/python/ch-2.py
@@ -1,4 +1,4 @@
-#!/usr/bin/env perl
+#!/usr/bin/env python3
 
 # Challenge 064
 #
diff --git a/challenge-065/paulo-custodio/perl/ch-1.pl b/challenge-065/paulo-custodio/perl/ch-1.pl
index 67d2eb12e6..a4d26d30c4 100644
--- a/challenge-065/paulo-custodio/perl/ch-1.pl
+++ b/challenge-065/paulo-custodio/perl/ch-1.pl
@@ -2,7 +2,7 @@
 
 # Challenge 065
 #
-# TASK #1 › Digits Sum
+# TASK #1 > Digits Sum
 # Submitted by: Mohammad S Anwar
 # Reviewed by: Ryan Thompson
 #
diff --git a/challenge-065/paulo-custodio/perl/ch-2.pl b/challenge-065/paulo-custodio/perl/ch-2.pl
index 1adaf4eb34..085f793168 100644
--- a/challenge-065/paulo-custodio/perl/ch-2.pl
+++ b/challenge-065/paulo-custodio/perl/ch-2.pl
@@ -2,7 +2,7 @@
 
 # Challenge 065
 #
-# TASK #2 › Palindrome Partition
+# TASK #2 > Palindrome Partition
 # Submitted by: Mohammad S Anwar
 # Reviewed by: Ryan Thompson
 #
@@ -10,7 +10,7 @@
 # gives Palindrome. Return -1 if none found.
 #
 # Please make sure, partition should not overlap. For example, for given string
-# “abaab”, the partition “aba” and “baab” would not be valid, since they overlap.
+# "abaab", the partition "aba" and "baab" would not be valid, since they overlap.
 #
 # Example 1
 # Input: $S = 'aabaab'
diff --git a/challenge-065/paulo-custodio/python/ch-1.py b/challenge-065/paulo-custodio/python/ch-1.py
new file mode 100644
index 0000000000..344143f905
--- /dev/null
+++ b/challenge-065/paulo-custodio/python/ch-1.py
@@ -0,0 +1,34 @@
+#!/usr/bin/env python3
+
+# Challenge 065
+#
+# TASK #1 > Digits Sum
+# Submitted by: Mohammad S Anwar
+# Reviewed by: Ryan Thompson
+#
+# You are given two positive numbers $N and $S.
+#
+# Write a script to list all positive numbers having exactly $N digits where sum
+# of all digits equals to $S.
+#
+# Example
+# Input:
+#     $N = 2
+#     $S = 4
+#
+# Output:
+#     13, 22, 31, 40
+
+import sys
+from itertools import product
+
+def digits_sum(N, S):
+    out = []
+    for n in range(10**(N-1), 10**N):
+        if sum(int(d) for d in str(n)) == S:
+            out.append(n)
+    return out
+
+N, S = map(int, sys.argv[1:])
+out = digits_sum(N, S)
+print(", ".join(map(str, out)))
diff --git a/challenge-065/paulo-custodio/python/ch-2.py b/challenge-065/paulo-custodio/python/ch-2.py
new file mode 100644
index 0000000000..702ed4d413
--- /dev/null
+++ b/challenge-065/paulo-custodio/python/ch-2.py
@@ -0,0 +1,56 @@
+#!/usr/bin/env perl
+
+# Challenge 065
+#
+# TASK #2 > Palindrome Partition
+# Submitted by: Mohammad S Anwar
+# Reviewed by: Ryan Thompson
+#
+# You are given a string $S. Write a script print all possible partitions that
+# gives Palindrome. Return -1 if none found.
+#
+# Please make sure, partition should not overlap. For example, for given string
+# "abaab", the partition "aba" and "baab" would not be valid, since they overlap.
+#
+# Example 1
+# Input: $S = 'aabaab'
+# Ouput:
+#  There are 3 possible solutions.
+#  a) 'aabaa'
+#  b) 'aa', 'baab'
+#  c) 'aba'
+# Example 2
+# Input: $S = 'abbaba'
+# Output:
+#  There are 3 possible solutions.
+#  a) 'abba'
+#  b) 'bb', 'aba'
+#  c) 'bab'
+
+import sys
+
+def is_palindrome(s):
+    return len(s) >= 2 and s == s[::-1]
+
+def partitions(s):
+    seen = set()
+    length = len(s)
+    for lead in range(length + 1):
+        for p1 in range(2, length + 1):
+            for p2 in [0] + list(range(2, length + 1)):
+                for tail in range(length + 1):
+                    if lead + p1 + p2 + tail != length:
+                        continue
+                    s1 = s[lead:lead+p1]
+                    s2 = s[lead+p1:lead+p1+p2]
+                    if is_palindrome(s1) and is_palindrome(s2):
+                        key = f"{s1}, {s2}"
+                        if key not in seen:
+                            print(key)
+                            seen.add(key)
+                    elif is_palindrome(s1):
+                        if s1 not in seen:
+                            print(s1)
+                            seen.add(s1)
+
+partitions(sys.argv[1] if len(sys.argv) > 1 else "")