I've been having problems with this simple hackerrank question. My code works in the compiler but hackerrank test is failing 6 test cases. One of which my output is correct for (I didn't pay premium). Is there something wrong here?
Prompt:
Steve has a string of lowercase characters in range ascii[‘a’..’z’]. He wants to reduce the string to its shortest length by doing a series of operations in which he selects a pair of adjacent lowercase letters that match, and then he deletes them. For instance, the string aab could be shortened to b in one operation.
Steve’s task is to delete as many characters as possible using this method and print the resulting string. If the final string is empty, print Empty String
Ex.
aaabccddd → abccddd → abddd → abd
baab → bb → Empty String
Here is my code:
def super_reduced_string(s):
count_dict = {}
for i in s:
if (i in count_dict.keys()):
count_dict[i] += 1
else:
count_dict[i] = 1
new_string = ''
for char in count_dict.keys():
if (count_dict[char] % 2 == 1):
new_string += char
if (new_string is ''):
return 'Empty String'
else:
return new_string
Here is an example of output for which it does not work.
print(super_reduced_string('abab'))
It outputs 'Empty String' but should output 'abab'.
By using a counter, your program loses track of the order in which it saw characters. By example with input 'abab', you program sees two a's and two b's and deletes them even though they are not adjacent. It then outputs 'Empty String' but should output 'abab'.
O(n) stack-based solution
This problem is equivalent to finding unmatched parentheses, but where an opening character is its own closing character.
What this means is that it can be solved in a single traversal using a stack.
Since Python can return an actual empty string, we are going to output that instead of 'Empty String' which could be ambiguous if given an input such as 'EEEmpty String'.
Code
def super_reduced_string(s):
stack = []
for c in s:
if stack and c == stack[-1]:
stack.pop()
else:
stack.append(c)
return ''.join(stack)
Tests
print(super_reduced_string('aaabab')) # 'abab'
print(super_reduced_string('aabab')) # 'bab'
print(super_reduced_string('abab')) # 'abab'
print(super_reduced_string('aaabccddd ')) # 'abd'
print(super_reduced_string('baab ')) # ''
I solved it with recursion:
def superReducedString(s):
if not s:
return "Empty String"
for i in range(0,len(s)):
if i < len(s)-1:
if s[i] == s[i+1]:
return superReducedString(s[:i]+s[i+2:])
return s
This code loops over the string and checks if the current and next letter/position in the string are the same. If so, these two letters/positions I get sliced from the string and the newly created reduced string gets passed to the function.
This occurs until there are no pairs in the string.
TESTS:
print(super_reduced_string('aaabccddd')) # 'abd'
print(super_reduced_string('aa')) # 'Empty String'
print(super_reduced_string('baab')) # 'Empty String'
I solved it by creating a list and then add only unique letters and remove the last letter that found on the main string. Finally all the tests passed!
def superReducedString(self, s):
stack = []
for i in range(len(s)):
if len(stack) == 0 or s[i] != stack[-1]:
stack.append(s[i])
else:
stack.pop()
return 'Empty String' if len(stack) == 0 else ''.join(stack)
I used a while loop to keep cutting down the string until there's no change:
def superReducedString(s):
repeat = set()
dups = set()
for char in s:
if char in repeat:
dups.add(char + char)
else:
repeat.add(char)
s_old = ''
while s_old != s:
s_old = s
for char in dups:
if char in s:
s = s.replace(char, '')
if len(s) == 0:
return 'Empty String'
else:
return s
Related
I am trying to emulate a circumstance where i send information and only get a true or false as a return. So i can check each character and if it is true, that means that character is in the string. I would know there would be a position 0 to some number x. I would receive a true result and eventually only receive false result and then I would know the string has been solved. In my circumstance i would not know the target string.
I am trying to iterate through all characters and see if it matches the string character. if it does, I add the character to a list until the list contains all the characters of the string. but for some reason, this isn't working.
import string
hi = list()
swoll = "dkjfksjdfksjdkfjksdjfsjkdfjsjreuvnslei"
characters = string.ascii_lowercase + string.ascii_uppercase + string.digits
for ch in characters:
print(''.join(hi) + ch)
for i in swoll:
if i == ch:
hi.append(ch)
print(''.join(hi))
break
else:
continue
results:
a
b
c
d
d
de
de
def
def
defg
defh
defi
defi
defij
defij
defijk
defijk
defijkl
defijkl
defijklm
defijkln
defijkln
defijklno
defijklnp
defijklnq
defijklnr
defijklnr
defijklnrs
defijklnrs
defijklnrst
defijklnrsu
defijklnrsu
defijklnrsuv
defijklnrsuv
defijklnrsuvw
defijklnrsuvx
defijklnrsuvy
defijklnrsuvz
defijklnrsuvA
defijklnrsuvB
defijklnrsuvC
defijklnrsuvD
defijklnrsuvE
defijklnrsuvF
defijklnrsuvG
defijklnrsuvH
defijklnrsuvI
defijklnrsuvJ
defijklnrsuvK
defijklnrsuvL
defijklnrsuvM`
As you can see, it does not match the string
When I tried the code above, I was expecting the string to come out the same as the other string.
Based on my understanding of the question, I've implemented a function which I believe emulates the interface you are talking to:
spos = 0
def in_swoll(ch):
global spos
if spos == len(swoll) or ch != swoll[spos]:
return False
spos += 1
return True
This will return True and increment the counter into swoll when a character matches, otherwise it will return False.
You can then use this function in a loop which iterates until False is returned for all characters in characters. Inside the loop characters is iterated until a match is found, at which point it is added to hi:
hi = []
while True:
for ch in characters:
if in_swoll(ch):
hi.append(ch)
print(''.join(hi))
break
else:
# no matches, we're done
break
Output for your sample data:
d
dk
dkj
dkjf
dkjfk
dkjfks
dkjfksj
dkjfksjd
dkjfksjdf
dkjfksjdfk
dkjfksjdfks
dkjfksjdfksj
dkjfksjdfksjd
dkjfksjdfksjdk
dkjfksjdfksjdkf
dkjfksjdfksjdkfj
dkjfksjdfksjdkfjk
dkjfksjdfksjdkfjks
dkjfksjdfksjdkfjksd
dkjfksjdfksjdkfjksdj
dkjfksjdfksjdkfjksdjf
dkjfksjdfksjdkfjksdjfs
dkjfksjdfksjdkfjksdjfsj
dkjfksjdfksjdkfjksdjfsjk
dkjfksjdfksjdkfjksdjfsjkd
dkjfksjdfksjdkfjksdjfsjkdf
dkjfksjdfksjdkfjksdjfsjkdfj
dkjfksjdfksjdkfjksdjfsjkdfjs
dkjfksjdfksjdkfjksdjfsjkdfjsj
dkjfksjdfksjdkfjksdjfsjkdfjsjr
dkjfksjdfksjdkfjksdjfsjkdfjsjre
dkjfksjdfksjdkfjksdjfsjkdfjsjreu
dkjfksjdfksjdkfjksdjfsjkdfjsjreuv
dkjfksjdfksjdkfjksdjfsjkdfjsjreuvn
dkjfksjdfksjdkfjksdjfsjkdfjsjreuvns
dkjfksjdfksjdkfjksdjfsjkdfjsjreuvnsl
dkjfksjdfksjdkfjksdjfsjkdfjsjreuvnsle
dkjfksjdfksjdkfjksdjfsjkdfjsjreuvnslei
def checkPattern(x, string):
e = len(string)
if len(x) < e:
return False
for i in range(e - 1):
x = string[i]
y = string[i + 1]
last = x.rindex(x)
first = x.index(y)
if last == -1 or first == -1 or last > first:
return False
return True
if __name__ == "__main__":
x = str(input())
string = "hello"
if checkPattern(x, string) is True:
print('YES')
if checkPattern(x, string) is False:
print('NO')
So basically the code is supposed to identify a substring when the number of characters between the substring's letters don't matter. string = "hello" is supposed to be the substring. While the characters in between don't matter the order still matters so If I type "h.e.l.l.o" for example it's a YES, but if it's something like "hlelo" it's a NO. I sorta copied the base of the code and I'm still a little new to python so sorry if the question and code aren't clear.
Assuming I understand, and the input hlelo is No and the input h.e..l.l.!o is Yes, then the following code should work:
def checkPattern(x, string):
assert x and string, "Error. Both inputs should be non-empty. "
count_idx = 0 # index which counts where you are.
for letter in x:
if letter == string[count_idx]:
count_idx += 1 # increment to check the next string
if count_idx == len(string):
return True
# pattern was found if counter found matches equal to string length
return False
if __name__ == "__main__":
inp = input()
string = "hello"
if checkPattern(inp, string) is True:
print('YES')
if checkPattern(inp, string) is False:
print('NO')
Explaination: Regardless of the input string, x, you want to loop through each character of the search-string hello, to check if you find each character in the correct order. What my solution does is that it counts how many of the characters h, e, l, l, o it has found, starting from 0. If it finds a match for h, it moves on to check for a match for e, and so on. Ultimately, if you search through the entire string x, and the counter does not equal to the length of the search string (i.e. you could not find all the hello characters), it returns false.
EDIT: Small debug in the way the return worked. Instead returns if ever the counter goes over the length. Also added more examples given in comments
Here is my solution to this problem:
pattern = "hello"
def patternCheck(word, pattern) -> bool:
plist = list(pattern)
wlist = list(word)
for p in plist:
if p in wlist:
for _ in range(wlist.index(p) , -1, -1):
wlist.pop(_)
else:
return False
return True
print(patternCheck("h.e.l.l.o", pattern))
print(patternCheck("aalohel", pattern))
print(patternCheck("hhhhheeelllooo", pattern))
Explanation
First we convert our strings to a list
plist = list(pattern)
wlist = list(word)
Now we check using a for loop if every element in our pattern list is in the word list.
for p in plist:
if p in wlist:
If yes then we remove all the elements from index 0 to the index of that element.
for _ in range(wlist.index(p) , -1, -1):
wlist.pop(_)
We are removing elements in decreasing order of there indices to protect ourself from the IndexError: pop index out of range.
If the for loop ends normally then there was a match and we return True. Else if the element was not found in the word list in the first place then we return false as there is no match.
Suppose you have a given string and an integer, n. Every time a character appears in the string more than n times in a row, you want to remove some of the characters so that it only appears n times in a row. For example, for the case n = 2, we would want the string 'aaabccdddd' to become 'aabccdd'. I have written this crude function that compiles without errors but doesn't quite get me what I want:
def strcut(string, n):
for i in range(len(string)):
for j in range(n):
if i + j < len(string)-(n-1):
if string[i] == string[i+j]:
beg = string[:i]
ends = string[i+1:]
string = beg + ends
print(string)
These are the outputs for strcut('aaabccdddd', n):
n
output
expected
1
'abcdd'
'abcd'
2
'acdd'
'aabccdd'
3
'acddd'
'aaabccddd'
I am new to python but I am pretty sure that my error is in line 3, 4 or 5 of my function. Does anyone have any suggestions or know of any methods that would make this easier?
This may not answer why your code does not work, but here's an alternate solution using regex:
import re
def strcut(string, n):
return re.sub(fr"(.)\1{{{n-1},}}", r"\1"*n, string)
How it works: First, the pattern formatted is "(.)\1{n-1,}". If n=3 then the pattern becomes "(.)\1{2,}"
(.) is a capture group that matches any single character
\1 matches the first capture group
{2,} matches the previous token 2 or more times
The replacement string is the first capture group repeated n times
For example: str = "aaaab" and n = 3. The first "a" is the capture group (.). The next 3 "aaa" matches \1{2,} - in this example a{2,}. So the whole thing matches "a" + "aaa" = "aaaa". That is replaced with "aaa".
regex101 can explain it better than me.
you can implement a stack data structure.
Idea is you add new character in stack, check if it is same as previous one or not in stack and yes then increase counter and check if counter is in limit or not if yes then add it into stack else not. if new character is not same as previous one then add that character in stack and set counter to 1
# your code goes here
def func(string, n):
stack = []
counter = None
for i in string:
if not stack:
counter = 1
stack.append(i)
elif stack[-1]==i:
if counter+1<=n:
stack.append(i)
counter+=1
elif stack[-1]!=i:
stack.append(i)
counter = 1
return ''.join(stack)
print(func('aaabbcdaaacccdsdsccddssse', 2)=='aabbcdaaccdsdsccddsse')
print(func('aaabccdddd',1 )=='abcd')
print(func('aaabccdddd',2 )=='aabccdd')
print(func('aaabccdddd',3 )=='aaabccddd')
output
True
True
True
True
The method I would use is creating a new empty string at the start of the function and then everytime you exceed the number of characters in the input string you just not insert them in the output string, this is computationally efficient because it is O(n) :
def strcut(string,n) :
new_string = ""
first_c, s = string[0], 0
for c in string :
if c != first_c :
first_c, s= c, 0
s += 1
if s > n : continue
else : new_string += c
return new_string
print(strcut("aabcaaabbba",2)) # output : #aabcaabba
Simply, to anwer the question
appears in the string more than n times in a row
the following code is small and simple, and will work fine :-)
def strcut(string: str, n: int) -> str:
tmp = "*" * (n+1)
for char in string:
if tmp[len(tmp) - n:] != char * n:
tmp += char
print(tmp[n+1:])
strcut("aaabccdddd", 1)
strcut("aaabccdddd", 2)
strcut("aaabccdddd", 3)
Output:
abcd
aabccdd
aaabccddd
Notes:
The character "*" in the line tmp = "*"*n+string[0:1] can be any character that is not in the string, it's just a placeholder to handle the start case when there are no characters.
The print(tmp[n:]) line simply removes the "*" characters added in the beginning.
You don't need nested loops. Keep track of the current character and its count. include characters when the count is less or equal to n, reset the current character and count when it changes.
def strcut(s,n):
result = '' # resulting string
char,count = '',0 # initial character and count
for c in s: # only loop once on the characters
if c == char: count += 1 # increase count
else: char,count = c,1 # reset character/count
if count<=n: result += c # include character if count is ok
return result
Just to give some ideas, this is a different approach. I didn't like how n was iterating each time even if I was on i=3 and n=2, I still jump to i=4 even though I already checked that character while going through n. And since you are checking the next n characters in the string, you method doesn't fit with keeping the strings in order. Here is a rough method that I find easier to read.
def strcut(string, n):
for i in range(len(string)-1,0,-1): # I go backwards assuming you want to keep the front characters
if string.count(string[i]) > n:
string = remove(string,i)
print(string)
def remove(string, i):
if i > len(string):
return string[:i]
return string[:i] + string[i+1:]
strcut('aaabccdddd',2)
More specifically:
Given a string and a non-empty word string, return a version of the original String where all chars have been replaced by pluses ("+"), except for appearances of the word string which are preserved unchanged.
def(base,word):
plusOut("12xy34", "xy") → "++xy++"
plusOut("12xy34", "1") → "1+++++"
plusOut("12xy34xyabcxy", "xy") → "++xy++xy+++xy"
My original thought was this:
def main():
x = base.split(word)
y = ''.join(x)
print(y.replace(y,'+')*len(y))
From here I have trouble reinserting the word back into the str in the correct places. Any help is appreciated.
You can use any string to join (instead of the empty string '' like you have).
def plusOut(s, word):
x = s.split(word)
y = ['+' * len(z) for z in x]
final = word.join(y)
return final
Edit: I've removed the regex, but I'm keeping the function across multiple lines to more closely match your original code.
A regex is not required. We can solve this without any libraries, iterating through exactly once.
We want to iterate through the indices i of the string, yielding the word and jumping ahead by len(word) if the slice of len(word) starting at i matches the word, and by yielding '+' and incrementing by one otherwise.
def replace_chars_except_word(string, word):
def generate_chars():
i = 0
while i < len(string):
if string[i:(i+len(word))] == word:
i += len(word)
yield word
else:
yield '+'
i+= 1
return ''.join(generate_chars())
if __name__ == '__main__':
test_string = 'stringabcdefg string11010string1'
result = replace_chars_except_word(test_string, word = 'string')
assert result == 'string++++++++string+++++string+'
I use an internal generator function to yield the strings, but you could use a buffer to replace the internal function. (This is slightly less memory efficient).
buffer = []
if (condition)
buffer.append(word)
else:
buffer.append'+'
return ''.join(buffer)
The aim of the program is to print the longest substring within variable s that is in alphabetical order.
s ='abchae'
currentlen = 0
longestlen = 0
current = ''
longest = ''
alphabet = 'abcdefghijklmnopqrstuvwxyz'
for char in s:
for number in range(0,len(s)):
if s[number] == char:
n = number
nxtchar = 1
alphstring = s[n]
while alphstring in alphabet == True and n+nxtchar <= 5:
alphstring += s[n+nxtchar]
nxtchar += 1
currentlen = len(alphstring)
current = alphstring
if currentlen > longestlen:
longest = current
print longest
When run, the program doesn't print anything. I don't seem to see what's wrong with the code. Any help would be appreciated.
I'd use regex for this
import re
string = 'abchae'
alphstring = re.compile(r'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*', re.I)
longest = ''
for match in alphstring.finditer(string):
if len(match.group()) > len(longest):
longest = match.group()
print(longest)
Output:
abch
Note: The flag re.I in the regex expression causes the regex to ignore case. If this is not the desired behavior you can delete the flag and it will only match lowercase characters.
Like Kasramvd said, I do not understand the logic behind your code. You sure your code can run without raise IndentationError? As I concerned, the following part (the second row, have wrong indentation).
for number in range(0,len(s)):
if s[number] == char:
n = number
If you fixed that indentation error, you can run you code without error, and the last row (print longest) does work, it just does not work as you expect, it only prints a blank line.
I think I understood what you meant.
First you need to fix the indentation problem in your code, that would make it run:
for number in range(0,len(s)):
if s[number] == char:
n = number
Second, that condition will return two numbers 0 and 4 since a appears two times in s. I believe you only want the first so you should probably add a break statement after you find a match.
for number in range(0,len(s)):
if s[number] == char:
n = number
break
Finally, alphstring in alphabet == True will always return False. Because alphabet will never be True, you need parentheses to make this work or remove the == True.
ex: while (alphstring in alphabet) == True and n+nxtchar <= 5:
I believe that you were looking for the string abch which is what I managed to obtain with these changes
This is my solution:
result = ""
s = 'abchae'
alphabet = 'abcdefghijklmnopqrstuvwxyz'
max_length=0
for i in range(len(s)):
for j in range(len(s)):
if s[i:j] in alphabet and len(s[i:j])>max_length:
max_length = len(s[i:j])
result = s[i:j]
print result