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Python | Check if there are K consecutive 1’s in a binary number

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Given K and a binary number, check if there exists k consecutive 1’s in the binary number. Examples:

Input: binary number = 101010101111
            k = 4 
Output: yes
Explanation: at the last 4 index there exists 4 consecutive 1’s

Input: binary number = 11100000 k=5 
Output: no
Explanation: There is a maximum of 3 consecutive 1’s in the given binary.

Approach: Create a new string with k 1’s. Using if condition check if there is new in s. In python if new in s: checks if there is any existence if new in s, hence returns true if there is else it returns a false. Below is the Python implementation of the above approach: 

Python

# Python program to check if there
# is k consecutive 1's in a binary number
 
# function to check if there are k
# consecutive 1's
def check(s,k):
     
    # form a new string of k 1's
    new = "1"*k
     
    # if there is k 1's at any position
    if new in s:
        print "yes"
    else:
        print "no"
 
# driver code
s = "10101001111"
k = 4
check(s, k)

                    

Output
yes

Time Complexity : O(N)
Auxiliary Space : O(1)

Approach : Using find() method

Python3

# Python program to check if there
# is k consecutive 1's in a binary number
 
s = "10101001111"
k = 4
 
res = "NO"
 
if(s.find("1"*k) != -1):
    res = "YES"
 
print(res)

                    

Output
YES

Time complexity: O(n), where n is the length of s. 
Auxiliary space: O(1), as we are only using a constant amount of memory for the string s, the integer k, and the string res.

Another approach to solve this problem is to use a sliding window approach. In this approach, we can check for the presence of k consecutive 1’s by considering a window of size k and sliding it through the entire binary number.

Here is an example of how this approach can be implemented in Python:

Python3

def check_consecutive_ones(s, k):
    for i in range(len(s) - k + 1):
        if all(s[i + j] == "1" for j in range(k)):
            return True
    return False
 
s = "10101001111"
k = 4
 
if check_consecutive_ones(s, k):
    print("Yes")
else:
    print("No")

                    

Output
Yes

This approach uses a for loop to iterate through each possible starting index of a sequence of k 1’s in the binary string and then checks if all the subsequent characters in the string at those indices are 1’s using the all() function. If any sequence of k 1’s is found, the function returns True, otherwise, it returns False.

The time complexity of the check_consecutive_ones() function is O(n*k), where n is the length of the input string. This is because the function involves iterating through all the characters in the string and checking if there is a consecutive subsequence of k ones.
The Auxiliary Space of the check_consecutive_ones() function is O(k), The all() function returns True if all the elements in an iterable are True, and False otherwise. 

Approach : Using replace() and len() methods

Python3

# Python program to check if there
# is k consecutive 1's in a binary number
 
s = "10101001111"
k = 4
a="1"*k
res = "NO"
b=s.replace(a,"")
if(len(b)!=len(s)):
    res = "YES"
print(res)

                    

Output
YES

Time Complexity : O(N)
Auxiliary Space : O(N)

Approach : Using operator.contains() method

Check whether k consecutive 1’s are present in given binary string using operator.contains() method

Python3

# Python program to check if there
# is k consecutive 1's in a binary number
 
s = "10101001111"
k = 4
a="1"*k
res = "NO"
import operator
if operator.contains(s,a):
    res="YES"
print(res)

                    

Output
YES

Time Complexity : O(N) where N is the length of s. 

Auxiliary Space : O(1) as we are only using a constant amount of memory for the string s, the integer k, and the string res

Approach: Using  a regular expression

  • Imports the re module
  • Sets the input binary string and the number of consecutive 1’s to check using tuple unpacking
  • Searches for a pattern of k consecutive 1’s in the binary string using a regular expression with the re.search() function
  • Sets the result to “YES” if the pattern is found, and “NO” if it is not, using a ternary operator
  • Prints the result to the console.

Python3

import re
 
# Input binary string and number of consecutive 1's to check
s = "10101001111"
k = 4
 
# Create regular expression pattern to match k consecutive 1's
pattern = '1{' + str(k) + '}'
 
# Initialize result to "NO"
res = "NO"
 
# Search for pattern in binary string
if re.search(pattern, s):
   # If pattern is found, set result to "YES"
   res = "YES"
 
# Output the result
print(res)

                    

Output
YES

Time Complexity: O(N) where N is the length of s.

Auxiliary Space: O(N) because the regular expression pattern is created based on the input value of k,.



Last Updated : 31 Mar, 2023
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