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#!/usr/bin/python3
# https://theweeklychallenge.org/blog/perl-weekly-challenge-245/#TASK2
#
# Task 2: Largest of Three
# ========================
#
# You are given an array of integers >= 0.
#
# Write a script to return the largest number formed by concatenating some of
# the given integers in any order which is also multiple of 3. Return -1 if
# none found.
#
## Example 1
##
## Input: @ints = (8, 1, 9)
## Output: 981
##
## 981 % 3 == 0
#
## Example 2
##
## Input: @ints = (8, 6, 7, 1, 0)
## Output: 8760
#
## Example 3
##
## Input: @ints = (1)
## Output: -1
#
############################################################
##
## discussion
##
############################################################
#
# While all examples in the description use single-digit numbers,
# there is nothing that would require this. So in order to catch all
# solutions, we need all possible permutations of all subsets of the
# array and of the numbers created out of those we need the biggest
# one that is divisible by 3.
from itertools import chain, combinations, permutations
def powerset(iterable):
"powerset([1,2,3]) → () (1,) (2,) (3,) (1,2) (1,3) (2,3) (1,2,3)"
s = list(iterable)
return chain.from_iterable(combinations(s, r) for r in range(len(s)+1))
def largest_of_three(ints: list):
print("Input: (", ", ".join(str(x) for x in ints), ")")
result = int()
for myset in powerset(ints):
for c in permutations(myset, len(myset)):
v = str("".join(str(x) for x in c))
# print("-> ", v)
if len(v) > 0:
# print("Considering", v, "as a value")
value = int(v)
if value % 3 == 0:
if len(str(result)) == 0:
result = value
if value > result:
result = value
if len(str(result)) == 0:
print("Output: -1")
else:
print("Output:", str(result))
largest_of_three([8, 1, 9]);
largest_of_three([8, 6, 7, 1, 0]);
largest_of_three([1]);
largest_of_three([8, 60, 7]);
largest_of_three([80, 6, 7]);
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