diff options
| -rw-r--r-- | challenge-057/paulo-custodio/python/ch-1.py | 1 | ||||
| -rw-r--r-- | challenge-058/paulo-custodio/python/ch-1.py | 70 | ||||
| -rw-r--r-- | challenge-058/paulo-custodio/python/ch-2.py | 88 |
3 files changed, 158 insertions, 1 deletions
diff --git a/challenge-057/paulo-custodio/python/ch-1.py b/challenge-057/paulo-custodio/python/ch-1.py index f5ef6b8cc9..f852b736bb 100644 --- a/challenge-057/paulo-custodio/python/ch-1.py +++ b/challenge-057/paulo-custodio/python/ch-1.py @@ -85,4 +85,3 @@ def dump_tree(tree): tree = parse_tree(read_input()) invert_tree(tree) dump_tree(tree) - diff --git a/challenge-058/paulo-custodio/python/ch-1.py b/challenge-058/paulo-custodio/python/ch-1.py new file mode 100644 index 0000000000..915851c677 --- /dev/null +++ b/challenge-058/paulo-custodio/python/ch-1.py @@ -0,0 +1,70 @@ +#!/usr/bin/env python3 + +# Challenge 058 +# +# TASK #1 › Compare Version +# Reviewed by Ryan Thompson +# Compare two given version number strings v1 and v2 such that: +# +# If v1 > v2 return 1 +# If v1 < v2 return -1 +# Otherwise, return 0 +# The version numbers are non-empty strings containing only digits, and the dot +# (“.”) and underscore (“_”) characters. (“_” denotes an alpha/development +# version, and has a lower precedence than a dot, “.”). Here are some examples: +# +# v1 v2 Result +# ------ ------ ------ +# 0.1 < 1.1 -1 +# 2.0 > 1.2 1 +# 1.2 < 1.2_5 -1 +# 1.2.1 > 1.2_1 1 +# 1.2.1 = 1.2.1 0 +# Version numbers may also contain leading zeros. You may handle these how you +# wish, as long as it’s consistent. + +import re +import sys + +def split_version(v): + m = re.search(r'^\d+', v) + if not m: + return False + else: + v_1 = int(m.group(0)) + if len(m.group(0)) == len(v): + v_sep = '' + v_rest = '' + else: + v_sep = v[len(m.group(0))] + v_rest = v[len(m.group(0))+1:] + return True, v_1, v_sep, v_rest + +def compare_version(a, b): + if a == b: + return 0 + else: + a_ok, a_1, a_sep, a_rest = split_version(a) + b_ok, b_1, b_sep, b_rest = split_version(b) + if not a_ok or not b_ok: + return 0 + elif a_1 < b_1: + return -1 + elif a_1 > b_1: + return 1 + elif a_sep == b_sep: + return compare_version(a_rest, b_rest) + elif a_sep == '': + return -1 + elif b_sep == '': + return 1 + elif a_sep == '.': + return -1 + elif b_sep == '.': + return 1 + else: + return compare_version(a_rest, b_rest) + +a = sys.argv[1] +b = sys.argv[2] +print(compare_version(a,b)) diff --git a/challenge-058/paulo-custodio/python/ch-2.py b/challenge-058/paulo-custodio/python/ch-2.py new file mode 100644 index 0000000000..b26bc37eda --- /dev/null +++ b/challenge-058/paulo-custodio/python/ch-2.py @@ -0,0 +1,88 @@ +#!/usr/bin/env python3 + +# Challenge 058 +# +# TASK #2 › Ordered Lineup +# Reviewed by Ryan Thompson +# Write a script to arrange people in a lineup according to how many taller +# people are in front of each person in line. You are given two arrays. @H is a +# list of unique heights, in any order. @T is a list of how many taller people +# are to be put in front of the corresponding person in @H. The output is the +# final ordering of people’s heights, or an error if there is no solution. +# +# Here is a small example: +# +# @H = (2, 6, 4, 5, 1, 3) # Heights +# @T = (1, 0, 2, 0, 1, 2) # Number of taller people in front +# The ordering of both arrays lines up, so H[i] and T[i] refer to the same +# person. For example, there are 2 taller people in front of the person with +# height 4, and there is 1 person in front of the person with height 1. +# +# Here is a diagram of the input arrays @H and @T: +# +# +# +# Finally, here is one possible solution that satisfies @H and @T: +# +# +# +# As per the last diagram, your script would then output the ordering +# (5, 1, 2, 6, 3, 4) in this case. (The leftmost element is the “front” of the +# array.) +# +# Here’s a 64-person example, with answer provided: +# +# # Heights +# @H = (27, 21, 37, 4, 19, 52, 23, 64, 1, 7, 51, 17, 24, 50, 3, 2, +# 34, 40, 47, 20, 8, 56, 14, 16, 42, 38, 62, 53, 31, 41, 55, 59, +# 48, 12, 32, 61, 9, 60, 46, 26, 58, 25, 15, 36, 11, 44, 63, 28, +# 5, 54, 10, 49, 57, 30, 29, 22, 35, 39, 45, 43, 18, 6, 13, 33); +# +# # Number taller people in front +# @T = ( 6, 41, 1, 49, 38, 12, 1, 0, 58, 47, 4, 17, 26, 1, 61, 12, +# 29, 3, 4, 11, 45, 1, 32, 5, 9, 19, 1, 4, 28, 12, 2, 2, +# 13, 18, 19, 3, 4, 1, 10, 16, 4, 3, 29, 5, 49, 1, 1, 24, +# 2, 1, 38, 7, 7, 14, 35, 25, 0, 5, 4, 19, 10, 13, 4, 12); +# +# # Expected answer +# @A = (35, 23, 5, 64, 37, 9, 13, 25, 16, 44, 50, 40, 2, 27, 36, 6, +# 18, 54, 20, 39, 56, 45, 12, 47, 17, 33, 55, 30, 26, 51, 42, 53, +# 49, 41, 32, 15, 22, 60, 14, 46, 24, 59, 10, 28, 62, 38, 58, 63, +# 8, 48, 4, 7, 31, 19, 61, 43, 57, 11, 1, 34, 21, 52, 29, 3); +# You’re free to come up with your own inputs. Here is a 1000-person list, +# if you like! + +H = [27, 21, 37, 4, 19, 52, 23, 64, 1, 7, 51, 17, 24, 50, 3, 2, + 34, 40, 47, 20, 8, 56, 14, 16, 42, 38, 62, 53, 31, 41, 55, 59, + 48, 12, 32, 61, 9, 60, 46, 26, 58, 25, 15, 36, 11, 44, 63, 28, + 5, 54, 10, 49, 57, 30, 29, 22, 35, 39, 45, 43, 18, 6, 13, 33] + +T = [ 6, 41, 1, 49, 38, 12, 1, 0, 58, 47, 4, 17, 26, 1, 61, 12, + 29, 3, 4, 11, 45, 1, 32, 5, 9, 19, 1, 4, 28, 12, 2, 2, + 13, 18, 19, 3, 4, 1, 10, 16, 4, 3, 29, 5, 49, 1, 1, 24, + 2, 1, 38, 7, 7, 14, 35, 25, 0, 5, 4, 19, 10, 13, 4, 12] + +# zip height and taller +HT = [[H[x], T[x]] for x in range(len(H))] + +# reverse sort by height +HT.sort(key=lambda x:x[0]) +HT = HT[::-1] + +# iterate through the indices +for i in range(len(HT)): + + # check if we need more people in front than are taller + if HT[i][1] > i: + print("no solution") + exit(1) + + # find position to move person + j = HT[i][1] + if i != j: + # remove from $i and insert in $j + person = HT[i] + HT = HT[:i] + HT[i+1:] + HT = HT[:j] + [person] + HT[j:] + +print(", ".join([str(x[0]) for x in HT])) |