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Given a number of brackets pairs. I want to display all correct combinations of these brackets. By correct I mean each bracket should be opened before closing in each combination. For example, if the number of brackets is 2 the output should be:
(())
()()
For 3:
((()))
()()()
(()())
(())()
()(())
The order of the output lines doesn't matter.
How can I do it with Python?
Try this code, please:
def __F(l, r, pref):
if r < l or l < 0 or r < 0:
return
if r == 0 and l == 0:
print(pref)
return
__F(l - 1, r, pref + "(")
__F(l, r - 1, pref + ")")
def F(n):
__F(n, n, "")
F(2)
F(3)
Related
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I'm trying to make combinations of two number I have facto defined before in my code now I want to use it to find combinations between n and k
I have this already
def facto(n):
res = 1
if n == 0:
return 1
for i in range(1, n+1):
res *= i
return res
now I need
def comb(n, k):
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I have the following interview question that may require traversing through the entire string.
Problem I searched about find Problem and most people do it like this def palindrome(s): return s==s[::-1] But this task is diffrent?
palindrome is a word the can read from both directions like 'mam', 'dad'. Your task is to get a list of palindrome in a given string.
def palindrome(s):
stack = []
return ['aaa']
exampls
palindrome('aaa') #['aaa']
palindrome('abcbaaaa') #['abcba','aaaa']
palindrome('xrlgabccbaxrlg') #['abccba']
palindrome('abcde') #['']
Let's try to avoid checking all the possible combinations :). My idea is, start from the extremities and converge:
def palindrome(s):
out = [''] #we need the list to not be empty for the following check
for main_start in range(len(s) - 1):
for main_end in range(len(s) - 1, main_start, -1):
start = main_start
end = main_end
while (end - start) > 0:
if s[start] == s[end]: ##may be palindrome
start += 1
end -= 1
else:
break
else:
if s[main_start:main_end + 1] not in out[-1]: #ignore shorter ("inner") findings
out.append(s[main_start:main_end + 1])
return out[1:] #skip the dummy item
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So I am having trouble concatenating this, I am not allowed to use .append(), and right now Im getting the error 'int' object not iterable.
def halveEvens(l):
num = []
for n in l:
if n % 2 == 0:
num += (n // 2)
return num
print(halveEvens([10,21,32,42,55]))```
sum(x//2 for x in numbers if x%2 == 0)
is probably how I would do it
if you just want to collect them (without summing them)
generator (x//2 for x in numbers if x%2 == 0) would be evaluated as you iterate it
or list comprehension that is evaluated immediatly
[x//2 for x in numbers if x%2 == 0]
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>>> 0 < 10 != 1 < 5
True
Why is it?? 0<10 is true. 1<5 is also true.True != True should be false 🤔. Then why the the output is True ???
Because of operations priorities meaning of your expression is different. You need to put parentheses: (0 < 10) != (1 < 5), to have what you want.
Otherwise your original expression means same as (0 < 10) and (10 != 1) and (1 < 5) which is not what you expected. (thanks to #TomKarzes)
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I play to HackNet game and i have to guess a word to bypass a firewall.
The key makes 6 characters long and contains the letters K,K,K,U,A,N.
What is the simplest way to generate all possible combinations either in bash or in python ? (bonus point for bash)
Here is a backtracking-based solution in Python:
db = {"K" : 3, "U" : 1, "A" : 1, "N" : 1}
N = 6
def possibilities(v):
l = []
for k in db.keys():
if v.count(k) < db[k]:
l.append(k)
return l
def generateImpl(a):
if len(a) < N:
lst = []
for c in possibilities(a):
lst += generateImpl(a+[c])
return lst
else:
return [''.join(a)]
def generate():
return generateImpl([])
Just run generate() to obtain a list of possible words.
You have 6 letters and need to find a combination of 6 letters. If you are not using the same character in ['K','K','K','U','A','N'] again and again, then there is only 1 permutation.
If you can use the same character again and again, you can use the following code to generate all possible combinations.
import itertools
y = itertools.combinations_with_replacement(['K','U','A','N'],6)
for i in list(y):
print(i)