From f25fdcee569572add62b4f05ac708dcbf08870b8 Mon Sep 17 00:00:00 2001 From: Paulo Custodio Date: Tue, 17 Sep 2024 16:35:26 +0100 Subject: Add Python solution to challenge 061 --- challenge-061/paulo-custodio/python/ch-1.py | 36 ++++++++++++++++++++++ challenge-061/paulo-custodio/python/ch-2.py | 46 +++++++++++++++++++++++++++++ 2 files changed, 82 insertions(+) create mode 100644 challenge-061/paulo-custodio/python/ch-1.py create mode 100644 challenge-061/paulo-custodio/python/ch-2.py (limited to 'challenge-061/paulo-custodio/python') diff --git a/challenge-061/paulo-custodio/python/ch-1.py b/challenge-061/paulo-custodio/python/ch-1.py new file mode 100644 index 0000000000..47e232e6aa --- /dev/null +++ b/challenge-061/paulo-custodio/python/ch-1.py @@ -0,0 +1,36 @@ +#!/usr/bin/env perl + +# Challenge 061 +# +# TASK #1 > Product SubArray +# Reviewed by: Ryan Thompson +# Given a list of 4 or more numbers, write a script to find the contiguous +# sublist that has the maximum product. The length of the sublist is irrelevant; +# your job is to maximize the product. +# +# Example +# Input: [ 2, 5, -1, 3 ] +# +# Output: [ 2, 5 ] which gives maximum product 10. + +import sys + +def product(nums): + prod = 1 + for x in nums: + prod *= x + return x + + +n = list(map(int, sys.argv[1:])) +max_sublist = [] +max_prod = int(-1e6) + +for i in range(0, len(n)): + for j in range(i, len(n)): + sublist = n[i:j+1] + prod = product(sublist) + if prod > max_prod: + max_sublist, max_prod = sublist, prod + +print(", ".join(list(map(str, max_sublist)))) diff --git a/challenge-061/paulo-custodio/python/ch-2.py b/challenge-061/paulo-custodio/python/ch-2.py new file mode 100644 index 0000000000..0cbd7cd9c9 --- /dev/null +++ b/challenge-061/paulo-custodio/python/ch-2.py @@ -0,0 +1,46 @@ +#!/usr/bin/env python3 + +# Challenge 061 +# +# TASK #2 > IPv4 Partition +# Reviewed by: Ryan Thompson +# You are given a string containing only digits (0..9). The string should have +# between 4 and 12 digits. +# +# Write a script to print every possible valid IPv4 address that can be made by +# partitioning the input string. +# +# For the purpose of this challenge, a valid IPv4 address consists of four +# "octets" i.e. A, B, C and D, separated by dots (.). +# +# Each octet must be between 0 and 255, and must not have any leading zeroes. +# (e.g., 0 is OK, but 01 is not.) +# +# Example +# Input: 25525511135, +# +# Output: +# +# 255.255.11.135 +# 255.255.111.35 + +import re +import sys + +def partition1(prefix, digits): + if re.search(r'^(\d+\.){4}$', prefix): + if digits == "": + prefix = prefix[:-1] + print(prefix) + else: + for l in range(1, 4): + if l <= len(digits): + part = digits[:l] + if int(part) <= 255: + partition1(prefix+part+".", digits[l:]) + +def partition(digits): + partition1("", digits) + +digits = sys.argv[1] +partition(digits) -- cgit