Given string str. The task is to check whether it is a binary string or not.
Examples:
Input: str = "01010101010" Output: Yes Input: str = "Lazyroar101" Output: No
Approach 1: Using Set
- Insert the given string in a set
- Check if the set characters consist of 1 and/or 0 only.
Example:
Python3
# Python program to check# if a string is binary or not# function for checking the# string is accepted or notdef check(string): # set function convert string # into set of characters . p = set(string) # declare set of '0', '1' . s = {'0', '1'} # check set p is same as set s # or set p contains only '0' # or set p contains only '1' # or not, if any one condition # is true then string is accepted # otherwise not . if s == p or p == {'0'} or p == {'1'}: print("Yes") else: print("No")# driver codeif __name__ == "__main__": string = "101010000111" # function calling check(string) |
Yes
Time Complexity: O(n), Here n is the length of the string.
Auxiliary Space: O(1), As constant extra space is used.
Approach 2: Simple Iteration
- Iterate for every character and check if the character is 0 or 1.
- If it is not then set a counter and break.
- After iteration check whether the counter is set or not.
Python3
# Python program to check# if a string is binary or not# function for checking the# string is accepted or notdef check2(string): # initialize the variable t # with '01' string t = '01' # initialize the variable count # with 0 value count = 0 # looping through each character # of the string . for char in string: # check the character is present in # string t or not. # if this condition is true # assign 1 to the count variable # and break out of the for loop # otherwise pass if char not in t: count = 1 break else: pass # after coming out of the loop # check value of count is non-zero or not # if the value is non-zero the en condition is true # and string is not accepted # otherwise string is accepted if count: print("No") else: print("Yes")# driver codeif __name__ == "__main__": string = "001021010001010" # function calling check2(string) |
No
Time Complexity: O(n), Here n is the length of the string.
Auxiliary Space: O(1), As constant extra space is used.
Approach 3: Regular Expressions
- Compile a regular expression using compile() for “character is not 0 or 1”.
- Use re.findall() to fetch the strings satisfying the above regular expression.
- Print output based on the result.
Python3
#import libraryimport resampleInput = "1001010"# regular expression to find the strings# which have characters other than 0 and 1c = re.compile('[^01]')# use findall() to get the list of strings# that have characters other than 0 and 1.if(len(c.findall(sampleInput))): print("No") # if length of list > 0 then it is not binaryelse: print("Yes") # if length of list = 0 then it is binary |
Yes
Time complexity: O(n)
Auxiliary space: O(1)
Approach 4: Using exception handling and int
Python has a built in method to convert a string of a specific base to a decimal integer, using int(string, base). If the string passed as an argument is not of the specified base, then a ValueError is raised.
Python3
# Python program to check# if a string is binary or not# function for checking the# string is accepted or notdef check(string): try: # this will raise value error if # string is not of base 2 int(string, 2) except ValueError: return "No" return "Yes"# driver codeif __name__ == "__main__": string1 = "101011000111" string2 = "201000001" # function calling print(check(string1)) print(check(string2))# this code is contributed by phasing17 |
Yes No
Time Complexity: O(1),
Auxiliary Space: O(1)
Approach 5 : Using count() function
Python3
string = "01010101010"if(string.count('0')+string.count('1')==len(string)): print("Yes")else: print("No") |
Yes
Time Complexity: O(n), where n is number of characters in string.
Auxiliary Space: O(1)
Approach 6 : Using replace() and len() methods
Python3
#Python program to check string is binary or notstring = "01010121010"binary="01"for i in binary: string=string.replace(i,"")if(len(string)==0): print("Yes")else: print("No") |
No
Approach 7: Using all()
The all() method in Python3 can be used to evaluate if all the letters in the string are 0 or 1.
Python3
# Python program to check# if a string is binary or not# function for checking the# string is accepted or notdef check(string): if all((letter in "01") for letter in string): return "Yes" return "No"# driver codeif __name__ == "__main__": string1 = "101011000111" string2 = "201000001" # function calling print(check(string1)) print(check(string2))# this code is contributed by phasing17 |
Yes No
Time Complexity: O(n), Here n is the length of the string.
Auxiliary Space: O(1), As constant extra space is used.
Approach 8: Using issubset() method
We can use a issubset method in comprehension to check if all characters in the string are either 0 or 1.
Python3
def is_binary_string(s): # use set comprehension to extract all unique characters from the string unique_chars = {c for c in s} # check if the unique characters are only 0 and 1 return unique_chars.issubset({'0', '1'})# driver codeif __name__ == "__main__": string1 = "101011000111" string2 = "201000001" # function calling print(is_binary_string(string1)) print(is_binary_string(string2)) |
True False
Time complexity: O(n), where n is the length of the input string, and
Auxiliary space: O(1).
Method: Using re.search()
- Import the re module.
- Define a regular expression that matches any character other than ‘0’ and ‘1’.
- Use the re.search() method to search for the regular expression in the given string.
- If a match is found, then the given string is not a binary string. Otherwise, it is.
Python3
import redef is_binary_string(str): # Define regular expression regex = r"[^01]" # Search for regular expression in string if re.search(regex, str): return False else: return True#Examplesprint(is_binary_string("01010101010")) # Output: Yesprint(is_binary_string("Lazyroar101")) # Output: No |
True False
Time Complexity: O(n), where n is the length of the given string (as we are iterating over the string once).
Auxiliary Space: O(1)
