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Is there a better way in which this can be written? I'm new to python coding and I want to improve below lines of code. I have included a few lines of my code here but the original code is much longer than this(multiple if statements). Any guidance/feedback would be helpful.
upperchar = #generate a char
pwdlist.append(upperchar)
if len(pwdlist) < length:
upperchar2 = #generate a char
pwdlist.append(upperchar2)
if len(pwdlist) < length:
lowerchar = #generate a char
pwdlist.append(lowerchar)
if len(pwdlist) < length:
lowerchar2 = #generate a char
pwdlist.append(lowerchar2)
if len(pwdlist) < length:
char3 = #generate a char
pwdlist.append(char3)
Posting my first question here so if you need anything more please ask me.
Assuming you are always generating a character the same way, then this would work
pwdlist = []
while len(pwdlist) < length:
pwdlist.append(generate_char())
Or
pwdlist = [generate_char() for _ in range(length)]
Then you can add logic into this generate function to return something based on random values, for example
def generate_char():
from random import randrange
x = randrange(4)
if x == 0:
return 'value a-z'
elif x == 1:
return 'value A-Z'
elif x == 2:
return 'value 0-9'
elif x == 3:
return 'punctuation'
Related
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I want to know what's wrong in the below code. Will it consume more time to in some certain scenarios?
Expected time complexity: O(n)
def subArraySum(self,arr, n, s):
if sum(arr[0:n]) == s:
return [1, n]
if sum(arr[0:n]) < s:
return [-1]
start = 0
i =1
sum_elements = 0
while i < n:
sum_elements = sum(arr[start:i+1])
if sum_elements == s:
return [start+1, i+1]
if sum_elements < s:
i += 1
continue
if sum_elements > s:
start += 1
continue
if sum_elements < s:
return [-1]
Instead of running sum(arr[start:i+1]) in each iteration of the while loop, you should use a variable and add or subtract the respective value that is included or excluded from the subarray in each iteration. That way you can avoid the O(n^2) complexity and stay within O(n).
Currently there is a lot of overhead for calculating the sum of a (potentially large) subarray that has only changed by one single value at the beginning or the end during each iteration.
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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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I'm new to programming in general... I have to check all the rows and columns of an matrix and in case any of them are complete with zeros, return a True value. I made this code in a silly attempt but it doesn't work for the purpose of the question itself. The matrix will always be a square and a list of lists.
def determinanteEhNulo(matriz):
contador = 0
for i in matriz:
for j in range(len(matriz)):
if i[j] == 0:
contador += 1
if contador >= int(len(matriz)):
return True
return False
This seems to do what you need.
def determinanteEhNulo(matriz):
for i in matriz:
if all(x==0 for x in i):
return True
for i in zip(*matriz):
if all(x==0 for x in i):
return True
return False
count = 0
def determinanteEhNulo(matrix):
global count
for i in matrix:
if all([j ==0 for j in i]):
count += 1
if count == len(matrix):
return True
else:
return False
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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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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)