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The following function should delete a element in an list if it has an overlap with a first list.
However it only works with the first example(a and b1). With others it does not even send an error message and I have no idea where there problem lies. Can someone point me in the right direction?
def funct(firstone, secondone):
counter = 0
while secondone != [] and counter < len(firstone):
if firstone[counter] in secondone:
del(secondone[secondone.index(firstone[counter])])
counter += 1
return secondone
a = [0, 1, 2]
b1 = [1, 2, 0]
b2 = [-1, 1, 1]
b3 = [0, 0, 2]
print(funct(a, b1))
print(funct(a, b2))
print(funct(b3, a))
i think last line in your code "return" must be in the same level with "while" loop like that
def funct(firstone, secondone ):
counter = 0
while secondone != [] and counter < len(firstone):
if firstone [counter] in secondone :
del(secondone[ secondone .index(firstone[counter ])])
counter += 1
return secondone
You need to continue the for loop when the condition is False otherwise you will always return on the first iteration
while secondone != [] and counter < len(firstone):
if firstone[counter] in secondone :
del(secondone[secondone.index(firstone[counter])])
counter += 1
else:
continue
return secondone
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This is for the Third Problem on Google's Codejam Qualification Round 2020
The Judging System says this solution gives a wrong answer, but I could not figure out why. Any insights would be much appreciated.
num_test_cases = int(input())
def notOverlap(activity, arr):
# returns true if we have no overlapping activity in arr
for act in arr:
if not (act[0] >= activity[1] or act[1] <= activity[0]):
return False
return True
def decide(act, num_act):
C, J = [], []
result = [None]*num_act
for i in range(num_act):
if notOverlap(act[i], C):
C.append(act[i])
result[i] = "C"
elif notOverlap(act[i], J):
J.append(act[i])
result[i] = "J"
else:
return "IMPOSSIBLE"
return "".join(result)
for i in range(num_test_cases):
num_act = int(input())
act = []
for _ in range(num_act):
act.append(list(map(int, input().split())))
print("Case #" + str(i+1) + ": " + decide(act, num_act))
You implemented a brute force way to solve it. Your code runs slow, Time complexity O(N^2), but you can do it in O(N*log(N))
Instead of check with notOverlap(activity, arr), sort the array and check with the last ending time activity of C or J. ( Greedy Way to solve it )
You have to store the index of activity before sorting the array.
Here is my solution, but before reading the solution try to implement it yourself
for testcasei in range(1, 1 + int(input())):
n = int(input())
acts = []
for index in range(n):
start, end = map(int, input().split())
acts.append((start, end, index)) # store also the index
acts.sort(reverse=True) # sort by starting time reversed
# so the first activity go to the last
d = ['']*n # construct the array for the answer
cnow = jnow = 0 # next time C or J are available
impossible = False # not impossible for now
while acts: # while there is an activity
start_time, end_time, index = acts.pop()
if cnow <= start_time: # C is available to do the activity
cnow = end_time
d[index] = 'C'
elif jnow <= start_time:
jnow = end_time
d[index] = 'J'
else: # both are'nt available
impossible = True
break
if impossible:
d = 'IMPOSSIBLE'
else:
d = ''.join(d) # convert the array to string
print("Case #%d: %s" % (testcasei, d))
I hope you find this informative and helped you to understand, and keep the hard work.
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I am given a user entered string 'aaabbbccaa'.
I want to find the duplicates and print the string back as 'a3b3c2a2'
Maybe with this way:
from itertools import groupby
s = "aaabbbccaa"
# group by characters
groups = groupby(s)
# process result
result = "".join([label + str(len(list(group))) for label, group in groups])
print(result)
Output:
a3b3c2a2
def process_string(source):
new = ''
while source:
counter = 0
first_char = source[0]
while source and source[0] == first_char:
counter += 1
source = source[1:]
new += f'{first_char}{counter}'
return new
print(process_string('aaabbbccaa'))
'a3b3c2a2'
this kind of solution could mabye solve it, it does what you specify, however if you are able to put it into your context, no idea :)
hope it helps!
c = 0
foo = "aaabbbccaa"
bar = ""
prev = None
for counter, index in enumerate(foo):
print(c)
if prev == None:
print("first")
elif prev == index:
print("second")
elif prev != index:
c = 0
c += 1
prev = index
try:
if index in foo[counter+1]:
print("third")
else:
print("fourth")
bar += index + str(c)
except:
print("fifth")
bar += index + str(c)
print("foo is {}".format(foo)) # will output aaabbbccaa
print("bar is {}".format(bar)) # will output a3b3c2a2
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def HydrogenCount(Compound):
HydrogenNo = 0
for i in range(0, len(Compound)):
Compound[i] == "H":
print(Compound[i+1])
Temp = Compound[i+1]
Temp = int(Temp)
HydrogenNo = HydrogenNo + Temp
return HydrogenNo
HydrogenNo = HydrogenCount(Compound)
print ("HydrogenCount = ", HydrogenNo)
for an input like CH3CH2CH3 it should output hydrogen count = 8
but instead it outputs hydrogen count = 3 as it stops at the first h
Unindent the return statement. It's currently inside of the for loop and needs to be executed after. Otherwise it will only count the first.
def HydrogenCount(Compound):
HydrogenNo = 0
for i in range(0, len(Compound)):
Compound[i] == "H":
print(Compound[i+1])
Temp = Compound[i+1]
Temp = int(Temp)
HydrogenNo += Temp
return HydrogenNo
What if the H in the molecule has more than 9 atoms, say sugar compound C12H22O11 or glucose C6H12O6?
May I suggest you revamp the code this way:
import re
regex = re.compile('H([0-9]*)')
def HydrogenCount(Compound):
try:
return sum([int(i) for i in regex.findall(Compound)])
except:
return(0)
You may run this as:
print(HydrogenCount("CH3CH2CH3"))
print(HydrogenCount("C6H12O6"))
I still see one more flaw in the question and therefore all answers, which is how about molecules like CH3COOH, where H followed by no number implies 1 atom. So, this is the revised code to handle that too:
import re
regex = re.compile('H([0-9]*)')
def HydrogenCount_v2(Compound):
try:
res = [i if i != '' else '1' for i in regex.findall(Compound)]
return sum([int(i) for i in res])
except:
return(0)
print(HydrogenCount_v2("CH3CH2CH3"))
print(HydrogenCount_v2("C6H12O6"))
print(HydrogenCount_v2("CH3COOH"))
You can refactor your code like this:
def calculate_hydrogen_count(compound):
hydrogen_count = 0
for i in range(0, len(compound) - 1):
if compound[i] == "H":
hydrogen_count += int(compound[i + 1])
return hydrogen_count
compound = "CH3CH2CH3"
hydrogen_count = calculate_hydrogen_count(compound)
print ("HydrogenCount = ", hydrogen_count)
Outputting
8
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I just started taking classes of python and I am trying to make a code in where the value of the first month = 1 and multiply its value by 2 next month, and then by 3 next month, and 2 next month and so on. Until it reaches 6 months. I am using this code but it only gives me Month 1 = 1 which is the initial value.
P = 1
count = 12
print ("month 1: ",P)
for month in range(count-1):
if month %2 == 0:
P = P*2
else:
P = P*3
print:("month", month+2 ,":",P)
Change
print:("month", month+2 ,":",P)
to
print("month", month+2 ,":",P)
I'm not sure why python didn't complain about the colon. You can actually put anything there
weird: ("month", month+2 ,":",P)
And it won't complain. Awesome mistake, thanks!
Remember that range(count-1) will return numbers between 0 and 11 (the last number is non-inclusive)
Your logic could be like this:
(a) A counter that starts at 1 (cnt)
(b) A loop that watches for the counter to reach 6, then exits
(c) A running total (month_val or some such)
cnt = 1
month_val = 1
while cnt < 7:
month_val = month_val * cnt
print(month_val)
cnt += 1
The above assumes that you retain the new value of month_num -- but re-reading your question, you might just want to print the values 1 to 6, in which case the month_num value should just remain 1 all the time:
cnt = 1
month_val = 1
while cnt < 7:
print(month_val * cnt)
cnt += 1
You can define a dictionary for your problem like
month_values={}
for i in range(1,7):
if i == 1:
month_values['Month'+str(i)]=1
elif i%2 == 0:
month_values['Month'+str(i)]=i*2
elif i%2 == 1:
month_values['Month'+str(i)]=i*3
print(month_values)
prints
{'Month1': 1, 'Month2': 4, 'Month3': 9, 'Month4': 8, 'Month5': 15, 'Month6': 12}
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I am trying to extract what's between parenthesis (including parenthesis) recursively.
This is my solution:
def paren(txt):
if txt[1] == ")":
return ''
if txt[0] == "(":
if len(txt) > 2:
return txt[1] + paren(txt[:1] + txt[2:])
return txt[1]
if len(txt) > 2:
return paren(txt[1:])
return ""
But it doesn't include any parenthesis. How can I fix it?
Example:
print paren("h(ello)o")
Output: (ello)
print paren("(hello)")
Output: (hello)
If you have a single pair of parenthesis, I would recommend to go with Halcyon Abraham Ramirez's answer. Otherwise, try this method:
def paren(text):
pstack = 0
start = 0
end = len(text)
for i, c in enumerate(text):
if c == '(':
if pstack == 0:
start = i
pstack += 1
elif c == ')':
pstack -= 1
if pstack == 0:
end = i
break
return text[start:end]
And here is an example:
>>> paren("h(ello)")
'(ello)'
If you do not need the root parenthesis, you can modify the return statement like this:
return text[start+1:end-1]
And again:
>>> paren("h(ello)")
'ello'
use index
word = "hehlllllllll(ooooo)jejejeje"
def extract(word):
return word[word.index("("):word.index(")") + 1]
output:
(ooooo)
taking it further.
if there are multiple parenthesis:
a = "h((el(l))o"
def extract_multiple_parenthesis(word):
closing_parenthesis = word[::-1].index(")")
last_parenthesis_index = (len(word) - closing_parenthesis)
return word[word.index("("):last_parenthesis_index]
output:
((el(l))