It is a sample problem for practice (here), and is not getting accepted due to it giving 'wrong answer'. It is compiling fine, but might be for different test inputs failing on submission.
I just request comment for the same, as hope that the issue must be small.
The problem statement is:
The program should accept first line of input as no. of strings, s.t. it is less than 10. And the next lines should contain one string each of length less than 100 characters. Need find occurrence of "gfg" in the strings.
If no occurrence is found, -1 should be returned.
#code
t = int(input())
if t > 10 or t<0:
exit()
arr = [[0] for i in range(t)]
total = [[-1] for i in range(t)]
for i in range(t):
arr[i] = [k for k in input().split()[:1]]
for j in arr[i]:
total[i] = j.count("gfg")
if total[i]==0: total[i]=-1
print (total[i])
t = int(input())
if t not in range(10):
exit()
else:
pass
total = []
for i in range(t):
line = input()[:100]
if line.count("gfg") == 0:
total.append(-1)
else:
total.append(line.count("gfg"))
print('\n'.join(map(str, total)))
SOLUTION FOR YOUR TASK:
t = int(input())
total = []
for i in range(1, t + 1):
line = input()
if len(line)<=100:
count = 0
for i in range(0, len(line) - 3 + 1):
if line[i:i + 3] == "gfg":
count += 1
if count != 0:
total.append(count)
else:
total.append(-1)
for i in total:
print (i)
NOTE: your submitting was failing because of special cases
For example:
in string gfgfg, your substring "gfg" occurres 2 times, and in that case you can't use the string count() method
how you can see ,here line[i:i + 3] I am moving index by index and checking next 3 values (because your substing "gfg" have length 3)
You are erasing the value of total[i] each time you iterate in the last loop :
>>> for j in arr[i]:
>>> total[i] = j.count("gfg")
>>> if total[i]==0:
>>> total[i]=-1
>>> print (total[i])
Do you want to count the occurences in each word or in the whole sentence ? Because then you just have to write :
>>> for i in range(t):
>>> n_occ = input().count("gfg")
>>> if n_occ != 0:
>>> total[i] = n_occ
If there's no occurence you don't have to do anything because the value in total is -1 already.
Also, write:
>>> total = [-1]*t
not:
>>> total = [[-1] for i in range(t)]
Related
I wrote the following code. It should return to me the length of the longest subscript in a string without a repeat in letters.
def lengthOfLongestSubstring(s):
lst = []
y = 0
final = 0
count = len(s)
while len(s) > 0:
s = s[y:]
for i in range(len(s)):
if s[i] in lst:
y += 1
count = len(lst)
lst =[]
break
else:
lst.append(s[i])
if count > final:
final=count
return(final)
when entering the string "tmmzuxt" i expect to get an output of 5 (length of "mzuxt") but instead get 4. I have debugged to figure out the problem seems to be that my function skips over the second 'm' when indexing but I can't figure out why. Any suggestions?
Realized I somehow missed a line. Hope this makes more sense.
Your issue here is that you are modifying s while you are running your code.
Consider that in the first iteration, you are getting s = s[0:], so s will now be 'tmmzuxt'. In your next iteration, you are getting s = s[1:], from the modified s. This is still not a problem, because you just get 'mmzuxt'. However, in your third iteration, you are getting s = s[2:], which is now 'zuxt'.
So you need a different variable than s to hold the substring of s that you are actually testing.
here, in your code(line 7) you are updating your string value inside function, everytime your for loop iterates.
for e.g., after every break inside for loop. you string(which is "tmmzuxt") is becoming short and short.
i created a new variable which contains your original string.
def lengthOfLongestSubstring(s):
lst = []
y = 0
final = 0
count = len(s)
main_string = s;#change done here
while len(s) > 0:
s = main_string[y:] #change done here
for i in range(len(s)):
if s[i] in lst:
y += 1
count = len(lst)
lst =[]
break
else:
lst.append(s[i])
if count > final:
final =count
print(final)
return(final)
lengthOfLongestSubstring("tmmzuxt")
The main problem with your code is that you incremented y, even though it should only ever remove the first character. There is no need for a variable y. Try this:
def lengthOfLongestSubstring(s):
final = 0
while len(s) > 0:
count = len(s)
lst = []
for i in range(len(s)):
if s[i] in lst:
count = i - 1
break
lst.append(s[i])
if count > final:
final = count
s = s[1:]
return final
print(lengthOfLongestSubstring("tmmzuxt"))
Here is an edited code. removing #lst =[] and #break lines.
[Code]
def lengthOfLongestSubstring(s):
lst = []
y = 0
final = 0
count = len(s)
while len(s) > 0:
s = s[y:]
for i in range(len(s)):
if s[i] in lst:
y += 1
count = len(lst)
#lst =[]
#break
else:
lst.append(s[i])
if count > final:
final=count
return(final)
s="tmmzuxt"
print(lengthOfLongestSubstring(s))
[Output]
5
I'm not sure if I understand your code, or if the while loop is needed here, actually. Try this instead:
def lengthOfLongestSubstring(s):
max_length = 0
length = 0
previous = ''
for thisCharacter in s:
if thisCharacter != previous:
length += 1
else:
max_length = max(length, max_length)
length = 1
return max_length
I'm here to ask help about my program.
I realise a program that raison d'être is to find the most occured four letters string on a x letters bigger string which have been generated randomly.
As example, if you would know the most occured sequence of four letters in 'abcdeabcdef' it's pretty easy to understand that is 'abcd' so the program will return this.
Unfortunately, my program works very slow, I mean, It take 119.7 seconds, for analyze all possibilities and display the results for only a 1000 letters string.
This is my program, right now :
import random
chars = ['a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z']
string = ''
for _ in range(1000):
string += str(chars[random.randint(0, 25)])
print(string)
number = []
for ____ in range(0,26):
print(____)
for ___ in range(0,26):
for __ in range(0, 26):
for _ in range(0, 26):
test = chars[____] + chars[___] + chars[__] + chars[_]
print('trying :',test, end = ' ')
number.append(0)
for i in range(len(string) -3):
if string[i: i+4] == test:
number[len(number) -1] += 1
print('>> finished')
_max = max(number)
for i in range(len(number)-1):
if number[i] == _max :
j, k, l, m = i, 0, 0, 0
while j > 25:
j -= 26
k += 1
while k > 25:
k -= 26
l += 1
while l > 25:
l -= 26
m += 1
Result = chars[m] + chars[l] + chars[k] + chars[j]
print(str(Result),'occured',_max, 'times' )
I think there is ways to optimize it but at my level, I really don't know. Maybe the structure itself is not the best. Hope you'll gonna help me :D
You only need to loop through your list once to count the 4-letter sequences. You are currently looping n*n*n*n. You can use zip to make a four letter sequence that collects the 997 substrings, then use Counter to count them:
from collections import Counter
import random
chars = ['a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z']
s = "".join([chars[random.randint(0, 25)] for _ in range(1000)])
it = zip(s, s[1:], s[2:], s[3:])
counts = Counter(it)
counts.most_common(1)
Edit:
.most_common(x) returns a list of the x most common strings. counts.most_common(1) returns a single item list with the tuple of letters and number of times it occurred like; [(('a', 'b', 'c', 'd'), 2)]. So to get a string, just index into it and join():
''.join(counts.most_common(1)[0][0])
Even with your current approach of iterating through every possible 4-letter combination, you can speed up a lot by keeping a dictionary instead of a list, and testing whether the sequence occurs at all first before trying to count the occurrences:
counts = {}
for a in chars:
for b in chars:
for c in chars:
for d in chars:
test = a + b + c + d
print('trying :',test, end = ' ')
if test in s: # if it occurs at all
# then record how often it occurs
counts[test] = sum(1 for i in range(len(s)-4)
if test == s[i:i+4])
The multiple loops can be replaced with itertools.permutations, though this improves readability rather than performance:
length = 4
for sequence in itertools.permutations(chars, length):
test = "".join(sequence)
if test in s:
counts[test] = sum(1 for i in range(len(s)-length) if test == s[i:i+length])
You can then display the results like this:
_max = max(counts.values())
for k, v in counts.items():
if v == _max:
print(k, "occurred", _max, "times")
Provided that the string is shorter or around the same length as 26**4 characters, then it is much faster still to iterate through the string rather than through every combination:
length = 4
counts = {}
for i in range(len(s) - length):
sequence = s[i:i+length]
if sequence in counts:
counts[sequence] += 1
else:
counts[sequence] = 1
This is equivalent to the Counter approach already suggested.
I am trying to make a string alternate between upper and lower case letters. My current code is this:
def skyline (str1):
result = ''
index = 0
for i in str1:
result += str1[index].upper() + str1[index + 1].lower()
index += 2
return result
When I run the above code I get an error saying String index out of range. How can I fix this?
One way using below with join + enumerate:
s = 'asdfghjkl'
''.join(v.upper() if i%2==0 else v.lower() for i, v in enumerate(s))
#'AsDfGhJkL'
This is the way I would rewrite your logic:
from itertools import islice, zip_longest
def skyline(str1):
result = ''
index = 0
for i, j in zip_longest(str1[::2], islice(str1, 1, None, 2), fillvalue=''):
result += i.upper() + j.lower()
return result
res = skyline('hello')
'HeLlO'
Explanation
Use itertools.zip_longest to iterate chunks of your string.
Use itertools.islice to extract every second character without building a separate string.
Now just iterate through your zipped iterable and append as before.
Try for i in range(len(str1)): and substitute index for i in the code. After, you could do
if i % 2 == 0: result += str1[i].upper()
else: result += str1[i].lower()
For every character in your input string, you are incrementing the index by 2. That's why you are going out of bounds.
Try using length of string for that purpose.
you do not check if your index is still in the size of your string.
It would be necessary to add a condition which verifies if the value of i is always smaller than the string and that i% 2 == 0 and that i == 0 to put the 1st character in Upper
with i% 2 == 0 we will apply the upper one letter on two
for i, __ in enumerate(str1):
if i+1 < len(str1) and i % 2 == 0 or i == 0:
result += str1[i].upper() + str1[i + 1].lower()
I tried to modify as minimal as possible in your code, so that you could understand properly. I just added a for loop with step 2 so that you wouldn't end up with index out of range. And for the final character in case of odd length string, I handled separately.
def skyline (str1):
result = ''
length = len(str1)
for index in range(0, length - 1, 2):
result += str1[index].upper() + str1[index + 1].lower()
if length % 2 == 1:
result += str1[length - 1].upper()
return result
You can use the following code:
def myfunc(str1):
result=''
for i in range(0,len(str1)):
if i % 2 == 0:
result += str1[i].upper()
else:
result += str1[i].lower()
return result
in your code you are get 2 word by one time so you should divide your loop by 2 because your loop work by depending your input string so make an variable like peak and equal it to len(your input input) then peak = int(peak/2) it will solve your pr
def func(name):
counter1 = 0
counter2 = 1
string = ''
peak = len(name)
peak = int(peak/2)
for letter in range(1,peak+1):
string += name[counter1].lower() + name[counter2].upper()
counter1 +=2
counter2 +=2
return string
Not sure if it's the best title. The explanation of what the program is suposed to do is below, my version only works with the first example but it doesn't work in the second when you get for example 1 1 3 1 1 2 because i can't figure out a good way to handle so much variations especially if K is bigger than 3 and the limit is 50. My version:
N, K, M = map(int, input().split())
niz = list(map(int, input().split()))
nizk = list(range(1, K+1))
izlazi = []
for r in range(0, M):
operacija = list(map(int, input().split()))
index = 0
if operacija[0] == 2:
nizkk = []
for z in range(0, len(nizk)):
if nizk[z] in niz:
continue
else:
izlazi.append(-1)
break
for p in range(0, N):
if niz[p] not in nizkk:
nizkk.append(niz[p])
nizkk.sort()
if nizkk == nizk:
index = p
izlazi.append(index+1)
break
else:
continue
else:
index, repl = map(int, operacija[1:])
niz[index - 1] = repl
print(izlazi)
In the first line of the input there should be N, K, M (1 <= N, M <= 100k, 1 <= K <= 50, you don't need to actually check this the numbers that are tested will always be in those ranges). In the second line of input you put a list of numbers which are the lenght of N you entered earlier. M is the number of operations you will do in the following lines of input. There can be 2 operations. If you enter 1 p v(p = index of number you want to replace, v the number you replace it with) or if you enter 2 it needs to find the shortest array of numbers defined by range(1, K+1) in the list of numbers you entered in line 2 and possibly changed with operation 1. If it doesn't exist it should output -1 if it does it should output lenght of numbers in the array you look in(numbers can be like 2, 1, 3 if you're looking for 1, 2, 3, also if you're looking for 1, 2, 3 etc and you have 2, 1, 1, 3 as the shortest one that is the solution and it's lenght is 4). Also the replacement operation doesnt count from 0 but from 1. So watch out when managing lists.
These are the examples you can input in the program ulaz = input, izlaz = ouput:
I have the following idea:
Min length sequence either starts from first element or does not contain first element and hence equals to min length of the same sequence without first element.
So we have recursion here.
For sequence [1,1,3,2,1,1] and [1,2,3] we will have:
Min length from start element [1,1,3,2,1,1] is 4
Min length from start element __[1,3,2,1,1] is 3
Min length from start element ____[3,2,1,1] is 3
Min length from start element ______[2,1,1] is -1
Can stop here.
Result is minimum for [4,3,3] = 3
You have already implemented the part for min length, if it starts from the first element. Need now extract it as a function and create a recursive function.
Some metacode:
function GetMinLength(seq)
{
minLengthFromFirstElement = GetMinLenthFromFirstElement(seq)
minLengthFromRest = GetMinLength(seq[1:]) //recusive call
return Min(minLengthFromFirstElement, minLengthFromRest )//-1 results should not count, add extra code to handle it
}
Unfortunately I don't know python, but I can provide code on F# in case you need it.
EDIT:
Try this:
N, K, M = map(int, input().split())
niz = list(map(int, input().split()))
nizk = list(range(1, K+1))
izlazi = []
def GetMinLengthStartingFromFirstElement(seq):
nizkk = []
for z in range(0, len(seq)):
if seq[z] in nizk:
continue
else:
return -1
for p in range(0, len(seq)):
if seq[p] not in nizkk:
nizkk.append(seq[p])
nizkk.sort()
if nizkk == nizk:
index = p
return index+1
else:
continue
return -1
def GetMinLength(seq):
if len(seq) == 0:
return -1
else:
curMinLength = GetMinLengthStartingFromFirstElement(seq)
if curMinLength == -1:
return -1
minLengthFromRest = GetMinLength(seq[1:])
if minLengthFromRest > -1:
return min(curMinLength,minLengthFromRest)
else:
return curMinLength;
for r in range(0, M):
operacija = list(map(int, input().split()))
index = 0
if operacija[0] == 2:
minLength = GetMinLength(niz)
izlazi.append(minLength)
else:
index, repl = map(int, operacija[1:])
niz[index - 1] = repl
print(izlazi)
def is_prime(x):
count = 1
my_list = []
while count > 0 and count < x:
if x % count == 0:
my_list.append(x/count)
count += 1
return my_list
my_list = is_prime(18)
def prime(x):
my_list2 = []
for number in my_list:
if number <= 2:
my_list2.append(number)
else:
count = 2
while count < number:
if number % count == 0:
break
else:
my_list2.append(number)
count += 1
return my_list2
print prime(18)
Just started out with Python. I have a very simple question.
This prints: [9, 3, 2].
Can someone please tell me why the loop inside my else stops at count = 2? In other words, the loop inside my loop doesn't seem to loop. If I can get my loop to work, hopefully this should print [2, 3]. Any insight is appreciated!
Assuming that my_list2 (not a very nice name for a list) is supposed to contain only the primes from my_list, you need to change your logic a little bit. At the moment, 9 is being added to the list because 9 % 2 != 0. Then 9 % 3 is tested and the loop breaks but 9 has already been added to the list.
You need to ensure that each number has no factors before adding it to the list.
There are much neater ways to do this but they involve things that you may potentially find confusing if you're new to python. This way is pretty close to your original attempt. Note that I've changed your variable names! I have also made use of the x that you are passing to get_prime_factors (in your question you were passing it to the function but not using it). Instead of using the global my_list I have called the function get_factors from within get_prime_factors. Alternatively you could pass in a list - I have shown the changes this would require in comments.
def get_factors(x):
count = 1
my_list = []
while count > 0 and count < x:
if x % count == 0:
my_list.append(x/count)
count += 1
return my_list
# Passing in the number # Passing in a list instead
def get_prime_factors(x): # get_prime_factors(factors):
prime_factors = []
for number in get_factors(x): # for number in factors:
if number <= 2:
prime_factors.append(number)
else:
count = 2
prime = True
while count < number:
if number % count == 0:
prime = False
count += 1
if prime:
prime_factors.append(number)
return prime_factors
print get_prime_factors(18)
output:
[3, 2]
Just to give you a taste of some of the more advanced ways you could go about doing this, get_prime_factors could be reduced to something like this:
def get_prime_factors(x):
prime_factors = []
for n in get_factors(x):
if n <= 2 or all(n % count != 0 for count in xrange(2, n)):
prime_factors.append(n)
return prime_factors
all is a built-in function which would be very useful here. It returns true if everything it iterates through is true. xrange (range on python 3) allows you to iterate through a list of values without manually specifying a counter. You could go further than this too:
def get_prime_factors(x):
return [n for n in get_factors(x) if n <= 2 or all(n % c != 0 for c in xrange(2, n))]