Develop Reference

identifying the substring when the number of characters in between don't matter

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.

Remove string character after run of n characters in string

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)

Super Reduced String python

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

Check special symbols in string endings

How to check special symbols such as !?,(). in the words ending? For example Hello??? or Hello,, or Hello! returns True but H!??llo or Hel,lo returns False.
I know how to check the only last symbol of string but how to check if two or more last characters are symbols?

You may have to use regex for this.
import re
def checkword(word):
m = re.match("\w+[!?,().]+$", word)
if m is not None:
return True
return False
That regex is:
\w+ # one or more word characters (a-zA-z)
[!?,().]+ # one or more of the characters inside the brackets
# (this is called a character class)
$ # assert end of string
Using re.match forces the match to begin at the beginning of the string, or else we'd have to use ^ before the regular expression.

You can try something like this:
word = "Hello!"
def checkSym(word):
return word[-1] in "!?,()."
print(checkSym(word))
The result is:
True
Try giving different strings as input and check the results.
In case you want to find every symbol from the end of the string, you can use:
def symbolCount(word):
i = len(word)-1
c = 0
while word[i] in "!?,().":
c = c + 1
i = i - 1
return c
Testing it with word = "Hello!?.":
print(symbolCount(word))
The result is:
3

If you want to get a count of the 'special' characters at the end of a given string.
special = '!?,().'
s = 'Hello???'
count = 0
for c in s[::-1]:
if c in special:
count += 1
else:
break
print("Found {} special characters at the end of the string.".format(count))

You can use re.findall:
import re
s = "Hello???"
if re.findall('\W+$', s):
pass

You could try this.
string="gffrwr."
print(string[-1] in "!?,().")

Count vowels from raw input

I have a homework question which asks to read a string through raw input and count how many vowels are in the string. This is what I have so far but I have encountered a problem:
def vowels():
vowels = ["a","e","i","o","u"]
count = 0
string = raw_input ("Enter a string: ")
for i in range(0, len(string)):
if string[i] == vowels[i]:
count = count+1
print count
vowels()
It counts the vowels fine, but due to if string[i] == vowels[i]:, it will only count one vowel once as i keeps increasing in the range. How can I change this code to check the inputted string for vowels without encountering this problem?

in operator
You probably want to use the in operator instead of the == operator - the in operator lets you check to see if a particular item is in a sequence/set.
1 in [1,2,3] # True
1 in [2,3,4] # False
'a' in ['a','e','i','o','u'] # True
'a' in 'aeiou' # Also True
Some other comments:
Sets
The in operator is most efficient when used with a set, which is a data type specifically designed to be quick for "is item X part of this set of items" kind of operations.*
vowels = set(['a','e','i','o','u'])
*dicts are also efficient with in, which checks to see if a key exists in the dict.
Iterating on strings
A string is a sequence type in Python, which means that you don't need to go to all of the effort of getting the length and then using indices - you can just iterate over the string and you'll get each character in turn:
E.g.:
for character in my_string:
if character in vowels:
# ...
Initializing a set with a string
Above, you may have noticed that creating a set with pre-set values (at least in Python 2.x) involves using a list. This is because the set() type constructor takes a sequence of items. You may also notice that in the previous section, I mentioned that strings are sequences in Python - sequences of characters.
What this means is that if you want a set of characters, you can actually just pass a string of those characters to the set() constructor - you don't need to have a list one single-character strings. In other words, the following two lines are equivalent:
set_from_string = set('aeiou')
set_from_list = set(['a','e','i','o','u'])
Neat, huh? :) Do note, however, that this can also bite you if you're trying to make a set of strings, rather than a set of characters. For instance, the following two lines are not the same:
set_with_one_string = set(['cat'])
set_with_three_characters = set('cat')
The former is a set with one element:
'cat' in set_with_one_string # True
'c' in set_with_one_string # False
Whereas the latter is a set with three elements (each one a character):
'c' in set_with_three_characters` # True
'cat' in set_with_three_characters # False
Case sensitivity
Comparing characters is case sensitive. 'a' == 'A' is False, as is 'A' in 'aeiou'. To get around this, you can transform your input to match the case of what you're comparing against:
lowercase_string = input_string.lower()

You can simplify this code:
def vowels():
vowels = 'aeiou'
count = 0
string = raw_input ("Enter a string: ")
for i in string:
if i in vowels:
count += 1
print count
Strings are iterable in Python.

for i in range(0, len(string)):
if string[i] == vowels[i]:
This actually has a subtler problem than only counting each vowel once - it actually only tests if the first letter of the string is exactly a, if the second is exactly e and so on.. until you get past the fifth. It will try to test string[5] == vowels[5] - which gives an error.
You don't want to use i to look into vowels, you want a nested loop with a second index that will make sense for vowels - eg,
for i in range(len(string)):
for j in range(len(vowels)):
if string[i] == vowels[j]:
count += 1
This can be simplified further by realising that, in Python, you very rarely want to iterate over the indexes into a sequence - the for loop knows how to iterate over everything that you can do string[0], string[1] and so on, giving:
for s in string:
for v in vowels:
if s == v:
count += 1
The inner loop can be simplified using the in operation on lists - it does exactly the same thing as this code, but it keeps your code's logic at a higher level (what you want to do vs. how to do it):
for s in string:
if s in vowels:
count += 1
Now, it turns out that Python lets do math with booleans (which is what s in vowels gives you) and ints - True behaves as 1, False as 0, so True + True + False is 2. This leads to a one liner using a generator expression and sum:
sum(s in vowels for s in string)
Which reads as 'for every character in string, count how many are in vowels'.

you can use filter for a one liner
print len(filter(lambda ch:ch.lower() in "aeiou","This is a String"))

Here's a more condensed version using sum with a generator:
def vowels():
string = raw_input("Enter a string: ")
print sum(1 for x in string if x.lower() in 'aeiou')
vowels()

Option on a theme
Mystring = "The lazy DOG jumped Over"
Usestring = ""
count=0
for i in Mystring:
if i.lower() in 'aeiou':
count +=1
Usestring +='^'
else:
Usestring +=' '
print (Mystring+'\n'+Usestring)
print ('Vowels =',count)
The lazy DOG jumped Over
^ ^ ^ ^ ^ ^ ^
Vowels = 7