I came across this exercise of checking whether or not the simple brackets "(", ")" in a given string are matched evenly.
I have seen examples here using the stack command which I haven't encountered yet. So I attempted a different approach. Can anyone tell me where I am going wrong?
def matched(str):
ope = []
clo = []
for i in range(0,len(str)):
l = str[i]
if l == "(":
ope = ope + ["("]
else:
if l == ")":
clo = clo + [")"]
else:
return(ope, clo)
if len(ope)==len(clo):
return True
else:
return False
The idea is to pile up "(" and ")" into two separate lists and then compare the length of the lists. I also had another version where I had appended the lists ope and clo with the relevant I which held either ( or ) respectively.
A very slightly more elegant way to do this is below. It cleans up the for loop and replaces the lists with a simple counter variable. It also returns false if the counter drops below zero so that matched(")(") will return False.
def matched(str):
count = 0
for i in str:
if i == "(":
count += 1
elif i == ")":
count -= 1
if count < 0:
return False
return count == 0
This checks whether parentheses are properly matched, not just whether there is an equal number of opening and closing parentheses. We use a list as a stack and push onto it when we encounter opening parentheses and pop from it when we encounter closing parentheses.
The main problem with your solution is that it only counts the number of parentheses but does not match them. One way of keeping track of the current depth of nesting is by pushing opening parentheses onto a stack and popping them from the stack when we encounter a closing parenthesis.
def do_parentheses_match(input_string):
s = []
balanced = True
index = 0
while index < len(input_string) and balanced:
token = input_string[index]
if token == "(":
s.append(token)
elif token == ")":
if len(s) == 0:
balanced = False
else:
s.pop()
index += 1
return balanced and len(s) == 0
My solution here works for brackets, parentheses & braces
openList = ["[", "{", "("]
closeList = ["]", "}", ")"]
def balance(myStr):
stack = []
for i in myStr:
if i in openList:
stack.append(i)
elif i in closeList:
pos = closeList.index(i)
if stack and (openList[pos] == stack[-1]):
stack.pop()
else:
return "Unbalanced"
if len(stack) == 0:
return "Balanced"
print(balance("{[()](){}}"))
Most blatant error done by you is:
if l == ")":
clo = clo + [")"]
else:
return(ope, clo) # here
By using return, you exit from function when first char not equal to "(" or ")" is encountered. Also some indentation is off.
Minimal change which allows your code to run (although it won't give correct answers for all possible input strings) is:
def matched(str):
ope = []
clo = []
for i in range(0,len(str)):
l = str[i]
if l == "(":
ope = ope + ["("]
elif l == ")":
clo = clo + [")"]
if len(ope)==len(clo):
return True
else:
return False
The problem with your approach is that you don't consider the order. Following line would pass: ))) (((.
I'd suggest to keep the count of open and closed parenthesis:
counter starts from 0
every ( symbol increments counter
every ) symbol decrements counter
if at any moment counter is negative it is an error
if at the end of the line counter is 0 - string has matching parenthesis
a = "((a+b)*c)+(b*a))"
li = list(a)
result = []
for i in range(0, len(a)):
if a[i] == "(":
result.append(i)
elif a[i] == ")":
if len(result) > 0:
result.pop()
else:
li.pop(i)
for i in range(0, len(result)):
li.pop(result[i])
print("".join(li))
this code works fine
def matched(s):
p_list=[]
for i in range(0,len(s)):
if s[i] =='(':
p_list.append('(')
elif s[i] ==')' :
if not p_list:
return False
else:
p_list.pop()
if not p_list:
return True
else:
return False
You can do this in a couple of lines using accumulate (from itertools). The idea is to compute a cumulative parenthesis level going through the string with opening parentheses counting as level+1 and closing parentheses counting as level-1. If, at any point, the accumulated level falls below zero then there is an extra closing parenthesis. If the final level is not zero, then there is a missing closing parenthesis:
from itertools import accumulate
def matched(s):
levels = list(accumulate((c=="(")-(c==")") for c in s))
return all( level >= 0 for level in levels) and levels[-1] == 0
An alternative to check for balanced nested parentheses:
def is_balanced(query: str) -> bool:
# Alternative: re.sub(r"[^()]", "", query)
query = "".join(i for i in query if i in {"(", ")"})
while "()" in query:
query = query.replace("()", "")
return not query
for stmt in [
"(()()()())", # True
"(((())))", # True
"(()((())()))", # True
"((((((())", # False
"()))", # False
"(()()))(()", # False
"foo", # True
"a or (b and (c or d)", # False
"a or (b and (c or d))" # True
"a or (b and (c or (d and e)))", # True
]:
print(stmt)
print("Balanced:", is_balanced(stmt))
print()
It works by:
Removing everything but parentheses
Recursively remove innermost parentheses pairs
If you're left with anything besides the empty string, the statement is not balanced. Otherwise, it is.
if the parenthesis sequence is not an issue (strings like )( ) this code is faster :
def matched_parenthesis(s):
return s.count('(') == s.count(')')
Tested with 15KB string, it is ~20μs v.s. 1ms iterating over the whole string.
And for me the order is not an issue as the underlying protocol guaranties that the string is well-formed.
In case u also need to find the position of the first mismatching bracket from left u can use the below code which also cover certain edge cases:
def isBalanced(expr):
opening=set('([{')
new=set(')]}{[(')
match=set([ ('(',')'), ('[',']'), ('{','}') ])
stack=[]
stackcount=[]
for i,char in enumerate(expr,1):
if char not in new:
continue
elif char in opening:
stack.append(char)
stackcount.append(i)
else:
if len(stack)==0:
print(i)
return False
lastOpen=stack.pop()
lastindex=stackcount.pop()
if (lastOpen, char) not in match:
print (i)
return False
length=len(stack)
if length!=0:
elem=stackcount[0]
print (elem)
return length==0
string =input()
ans=isBalanced(string)
if ans==True:
print("Success")
if "(" ,")" these two characters are not present then we don't want to return true or false just return no matching found. if matching found i just checking the count of both characters are same then return true, else return false
def matched(str):
count1=0
count2=1
for i in str:
if i =="(":
count1+=1:
elif i==")":
count2+=1:
else:
print "no matching found for (,)"
if count1==count2:
return True
else:
return False
Simplest of all , though all of you guys have done good:
def wellbracketed(s):
left=[]
right=[]
for i in range(0,len(s)):``
if s[i]=='(':
left=left+['(']
elif s[i]==')':
if len(left)!=0:
right=right+[')']
else:
return False
return(len(left)==len(right))
here's another way to solve it by having a counter that tracks how many open parentheses that are difference at this very moment.
this should take care all of the cases.
def matched(str):
diffCounter = 0
length = len(str)
for i in range(length):
if str[i] == '(':
diffCounter += 1
elif str[i] == ')':
diffCounter -= 1
if diffCounter == 0:
return True
else:
return False
input_str = "{[()](){}}"
strblance=""
for i in input_str:
if not strblance:
strblance = strblance+i
elif (i is '}' and strblance[len(strblance)-1] is '{') \
or ( i is']'and strblance[len(strblance)-1] is '[') \
or ( i is ')'and strblance[len(strblance)-1] is '('):
strblance = strblance[:len(strblance)-1]
else:
strblance = strblance+i
if not strblance:
print ("balanced")
else:
print ("Not balanced")
More advanced example in which you additionally need to check a matching of square brackets '[]' and braces '{}' pars.
string = '([]{})'
def group_match(string):
d = {
')':'(',
']':'[',
'}':'{'
}
list_ = []
for index, item in enumerate(string):
if item in d.values():
list_.append(item)
elif (item in d.keys()) and (d.get(item) in list_):
list_.pop()
return len(list_) == 0
The simplest code ever!!
def checkpar(x):
while len(''.join([e for e in x if e in "()"]).split('()'))>1: x=''.join(x.split('()'))
return not x
you can check this code.
This code don't use stack operations.
def matched(s):
count = 0
for i in s:
if i is "(":
count += 1
elif i is ")":
if count != 0:
count -= 1
else:
return (False)
if count == 0:
return (True)
else:
return (False)
#function to check if number of closing brackets is equal to the number of opening brackets
#this function also checks if the closing bracket appears after the opening bracket
def matched(str1):
if str1.count(")")== str1.count("("):
p1=str1.find("(")
p2=str1.find(")")
if p2 >= p1:
str1=str1[p1+1:p2]+ str1[p2+1:]
if str1.count(")")>0 and str1.count("(")>0:
matched(str1)
return True
else:
return False
else:
return False
matched(str1)
parenthesis_String = input("Enter your parenthesis string")
parenthesis_List = []
for p in parenthesis_String:
parenthesis_List.append(p)
print(parenthesis_List)
if len(parenthesis_List)%2 != 0:
print("Not Balanced Wrong number of input")
for p1 in parenthesis_List:
last_parenthesis = parenthesis_List.pop()
print(last_parenthesis)
if (p1 == '{' and last_parenthesis == '}' or p1 == '[' and last_parenthesis == ']' or p1 == '(' and last_parenthesis == ')'):
print("Balanced")
else:
print("Not balanced")
A little different one.
expression = '{(){({)}}'
brackets = '[](){}'
stack = []
balanced = False
for e in expression:
if e in brackets and stack: # Popping from the stack if it is closing bracket
if stack [-1] == brackets[brackets.index(e)-1]:
stack.pop()
balanced = True
continue # it will go to the new iteration skipping the next if below
if e in brackets: # Push to stack if new bracket in the expression
stack .append(e)
balanced = False
balanced = 'Balanced' if balanced and not stack else 'Unbalanced'
print(balanced, stack)
just modified Henry Prickett-Morgan's code a little bit to handle it more sensibly, namely taking into account that the number of "(" matches that of ")" but string starts with ")" or ends with "(" which are apparently not right.
def ValidParenthesis(s):
count = 0
if s[0] == ')' or s[-1] == '(':
return False
else:
for c in s:
if c == '(':
count += 1
elif c == ')':
count -= 1
else:
continue
return count == 0
The best way to understand this snippet is to follow along with all kind of scenarios.
in_data = ['{','[','(']
out_data = ['}',']',')']
def check_match(statements):
stack = []
for ch in statements:
if ch in in_data:
stack.append(ch)
if ch in out_data:
last = None
if stack:
last = stack.pop()
if last is '{' and ch is '}':
continue
elif last is '[' and ch is ']':
continue
elif last is '(' and ch is ')':
continue
else:
return False
if len(stack) > 0:
return False
else:
return True
print(check_match("{www[eee}ee)eee"))
print(check_match("(ee)(eee[eeew]www)"))
print(check_match("(ss(ss[{ss}]zs)zss)"))
print(check_match("([{[[]]}])"))
def matched(str):
braces = {"{": "}", "(": ")", "[": "]"}
stack = []
for c in str:
if c in braces.keys():
stack.append(c)
elif c in braces.values():
if not stack:
return False
last_brace = stack.pop()
if braces[last_brace] != c:
return False
if stack:
return False
return True
print(matched("()"))
>> True
print(matched("(}"))
>> False
print(matched("}{"))
>> False
print(matched("}"))
>> False
print(matched("{"))
>> False
print(matched("(ff{fgg} [gg]h)"))
>> True
Given a string s containing just the characters '(', ')', '{', '}', '[' and ']',
determine if the input string is valid.
def isValid(s):
stack = []
for i in s:
if i in open_list:
stack.append(i)
elif i in close_list:
pos = close_list.index(i)
if open_list[pos] == stack[len(stack)-1]:
stack.pop()
else:
return False
if len(stack) == 0:
return True
else:
return False
print(isValid("{[(){}]}"))
s='{[]{()}}}{'
t=list(s)
cntc=0
cnts=0
cntp=0
cntc=min(t.count("{"),t.count("}"))
cnts=min(t.count("["),t.count("]"))
cntp=min(t.count("("),t.count(")"))
print(cntc+cnts+cntp)
for a balanced string, we can find an opening brace followed by it closing brace. if you do this basic check you could remove the checked substring and check the remaining string. At the end, if the string is not empty then it is not balanced.
def is_balanced(s: str) -> bool:
while any([x in s for x in ["", "", ""]]):
s=s.replace("{}", "").replace("[]","").replace("()","")
return s==""
def parenthesis_check(parenthesis):
chars = []
matches = {')':'(',']':'[','}':'{'}
for i in parenthesis:
if i in matches:
if chars.pop() != matches[i]:
return False
else:
chars.append(i)
return chars == []
foo1="()()())("
def bracket(foo1):
count = 0
for i in foo1:
if i == "(":
count += 1
else:
if count==0 and i ==")":
return False
count -= 1
if count == 0:
return True
else:
return False
bracket(foo1)
Although I'm not proposing a fix to your implementation, I suggest a cleaner and more pythonic version of the #kreld solution:
def check_parentheses(expr):
s = []
for c in expr:
if c in '(':
s.append(c)
elif c in ')':
if not len(s):
break
else:
s.pop()
else:
return not len(s)
return False
# test -----------------------------------------------------------------
test_expr = [')(', '(()', '())', '(', ')', '((', '))', '(()())', '(())',
'()', '()(())']
for i, t in enumerate(test_expr, 1):
print '%i\t%s\t%s' % (i, t, check_parentheses(t))
# output ---------------------------------------------------------------
1 )( False
2 (() False
3 ()) False
4 ( False
5 ) False
6 (( False
7 )) False
8 (()()) True
9 (()) True
10 () True
11 ()(()) True
I am writing a program to reverse the substrings enclosed in parenthesis in python. The resultant string should not contain any parenthesis. I am printing b1 and b2 and ch for testing purposes. It seems that in the second iteration of the for loop inside the while loop, the b1 variable is not updated with the correct index.
I tried to write a solution like below:
def reverseParentheses(s):
r = s
sstring = ''
astring = ''
b1 = b2 = 0
count = 0
for ch in s:
if ch == '(':
count+=1
elif ch ==')':
count+=1
else:
pass
while True:
b1 = b2 = 0
for ch in r:
if ch == '(':
b1 = r.index(ch)
print("b1= ",b1, ch)
if ch == ')':
b2 = r.index(ch)
print("b2= ",b2, ch)
sstring = r[b2-1:b1:-1]
print(r)
print(sstring)
astring = r[0:b1]+sstring+r[b2+1:]
print(astring)
r = astring
break
if len(astring)+count == len(s):
break
return r
s = "a(bcdefghijkl(mno)p)q"
print(reverseParentheses(s))
This is the output that I get:
aonmpbcdefghijklq
This is the output I expect:
apmnolkjihgfedcbq
A nice way to deal with nested delimiters is to use a stack. When you encounter an opening delimiter push a new collection to the stack. pop() when you find a closing. This will keep the order of nesting correct.
Here's one way to do this (it doesn't check for balanced parenthesis, but it's not hard to add):
s = "a(bcdefghijkl(mno)p)q"
stack = [[]] # accumulate letters in stack[0]
for l in s:
if l == '(':
stack.append([]) # start a new level
elif l == ')':
sub = stack.pop()[::-1] # pop the last level and reverse
stack[-1].extend(sub) # add to current
else:
stack[-1].append(l) # add to current
''.join(stack[0]) #'apmnolkjihgfedcbq'
A method by finding the position of the parenthesis and reversing from inside out (so the ones that are contained in-between an even number of parenthesis stay the same) and finally gets rid of the parenthesis:
s = "a(bcdefghijkl(mno)p)q"
leftp = reversed([pos for pos, char in enumerate(s) if char == "("])
rightp = [pos for pos, char in enumerate(s) if char == ")"]
for i in zip(leftp,rightp):
subs = s[i[0]+1:i[1]][::-1]
s = s[:i[0]+1]+subs+s[i[1]:]
for c in ["(", ")"]:
s = s.replace(c, "")
print(s) # Outputs "apmnolkjihgfedcbq"
EDIT
For parenthesis that are not nested, as pointed out by .#Mark Meyer, you can find them as described here and same rule applies
def find_parens(s):
toret = {}
pstack = []
for i, c in enumerate(s):
if c == '(':
pstack.append(i)
elif c == ')':
if len(pstack) == 0:
raise IndexError("No matching closing parens at: " + str(i))
toret[pstack.pop()] = i
if len(pstack) > 0:
raise IndexError("No matching opening parens at: " + str(pstack.pop()))
return toret
s = "a(bcd)efghijkl(mno)pq"
parens = find_parens(s)
for leftp, rightp in parens.items():
subs = s[leftp+1:rightp][::-1]
s = s[:leftp+1]+subs+s[rightp:]
for c in ["(", ")"]:
s = s.replace(c, "")
print(s) # Outputs "adcbefghijklonmpq"
I have list:
words = ["ALI", "SIN", "ASI", "LIR", "IRI", "INI", "KAR"]
I want to check if they form matrix such as this:
and return my solution as a list like:
solution = ["ALI", "SIN", "IRI"]
I have come up with this code:
words=["ALI", "SIN", "ASI", "LIR", "IRI", "INI", "KAR"]
solution =[]
failedsolutions = []
def Get_next_word():
while True:
for word in words:
if (word in solution) == False:
solution.append(word)
if (solution in failedsolutions) == False:
return False
else:
solution.pop(len(solution) - 1 )
return True
def Check_word_order():
checker = True
for i in range(len(solution)):
for j in range(len(words[0])):
for word in words:
if solution[i][j] == word[j]:
checker = False
else:
checker = True
if checker == False:
return checker
def main():
while True:
Get_next_word()
check = Check_word_order()
if check is False:
#Backtrack
failedsolutions.append(solution)
solution.pop(len(solution) - 1 )
# else:
# solution.append()
print(solution)
main()
I am tired and no longer be able to debug my code. Help will be appreciated.
Thank you
ps: I advice not fixing my code if there is a better way to it all.
You can use this simple recursive function which will analyze all possible groups:
import itertools
import copy
words = ["ALI", "SIN", "ASI", "LIR", "IRI", "INI", "KAR"]
def get_pairs(word_group, current_words, found):
if not current_words:
return found
new_group = list(word_group)+[current_words[0]]
if all(''.join(i) in words and ''.join(i) not in new_group for i in zip(*new_group)):
return get_pairs(word_group, current_words[1:], new_group)
return get_pairs(word_group, current_words[1:], found)
starting_pairs = [i for i in itertools.combinations(words, 2)]
final_listing = filter(lambda x:x, [get_pairs(i, copy.deepcopy(words), []) for i in starting_pairs])
Output:
[['ALI', 'SIN', 'IRI'], ['ASI', 'LIR', 'INI']]
Which yields all combinations of valid matrices.
Or, without using itertools:
def get_combos(word, new_words):
if new_words[1:]:
new_list = [(yield (word, i)) for i in new_words if i != word]
for b in get_combos(new_words[0], new_words[1:]):
yield b
starting_pairs = get_combos(words[0], words[1:])
final_listing = filter(lambda x:x, [get_pairs(i, copy.deepcopy(words), []) for i in starting_pairs])
I need a python function that can do the following:
Given an input of 't' and 'tattle', it should return a list like so:
['Tattle','taTtle','tatTle']
Or with 'z' and 'zzzzz':
['Zzzzz','zZzzz','zzZzz','zzzZz','zzzzZ']
I coded the following, but it does not work with the second example because the current function checks to see if the basestr matches what is already in the resulting list, R, and can pick up false positives due to words with multiple basestr's already in the word. Anyone have any advice?
def all_variants(wrapletter,word):
L,R,WLU,basestr=list(word),[],wrapletter.upper(),''
if L.count(wrapletter)==1:
for char in L:
if wrapletter==char:
basestr=basestr+WLU
else:
basestr=basestr+char
R.append(basestr)
return(R)
else:
for i in range(L.count(wrapletter)):
basestr=''
if i==0 and L[0]==wrapletter:
basestr=WLU
for char in range(1,len(L)):
basestr=basestr+L[char]
R.append(basestr)
else:
for char in L:
if wrapletter==char:
if WLU in basestr:
basestr=basestr+char
elif basestr in str(R):
basestr=basestr+char
else:
basestr=basestr+WLU
else:
basestr=basestr+char
R.append(basestr)
R.remove(R[0])
return(R)
It's not elegant, but maybe it's what you need?
target = "daaddaad"
def capitalize(target_letter, word):
return [word[:i] + word[i].upper() + word[i + 1:]
for i in xrange(len(word)) if word[i] == target_letter]
print capitalize("d", target)
Outputs:
['Daaddaad', 'daaDdaad', 'daadDaad', 'daaddaaD']
inp = 't'
word = 'tattle'
inds = (i for i,ele in enumerate(word) if ele == inp)
print([word[:i]+word[i].upper()+word[i+1:] for i in inds])
['Tattle', 'taTtle', 'tatTle']
Try this. I iterate through each letter, shift it to uppercase, and sandwich it with the other parts of the original string.
def all_variants(wrapletter, word):
variants = []
for i, letter in enumerate(word):
if letter == wrapletter:
variants.append(word[:i] + letter.upper() + word[i+1:])
return variants
print all_variants('z', 'zzzzz')
print all_variants('t', 'tattle')
def all_variants(wrapletter, word):
list = []
for i in range(len(word)):
if(word[i]==wrapletter):
start = word[0:i].lower()
str = word[i].upper()
end = word[i+1::].lower()
list.append(start+str+end)
return list
These returned when I ran this function:
>>>all_variants("t", "tattle")
['Tattle', 'taTtle', 'tatTle']
>>>all_variants("z", "zzzzz")
['Zzzzz', 'zZzzz', 'zzZzz', 'zzzZz', 'zzzzZ']