Python - Check if there are K consecutive 1's in a binary number

Last Updated : 06 Jan, 2025

The task is to check if a binary number (a string of 0s and 1s) has k consecutive 1s in it. The goal is to figure out if this pattern exists in the easiest and fastest way possible.

Using for loop

This method works by going through the string once and keeping track of how many '1's appear in a row. If the count reaches k more, it means we found what we were looking for, and we can stop early. If we see a character that’s not '1', we simply reset the count to zero and continue.

Python

s = "10101001111" k = 4   # Initialize counter count = 0    for char in s:      # If '1' found     if char == '1':           count += 1                  # If k consecutive 1's found         if count >= k:               print(True)             break     else:               # Reset counter if '0' found         count = 0  else:     print(False)

Output

True

Explanation:

This loop starts, and as it encounters '1' at positions 1, 3, and 6, the count is incremented.
When the loop reaches the substring '1111' at positions 7, 8, 9, and 10, the count becomes 4.
At this point, count >= 4 is true, so True is printed and the loop exits.

Table of Content

Using sliding window

This method involves moving a "window" of size k across the string s. For each position of the window, we check if the substring within the window contains k consecutive '1's. If a match is found, we terminate early; otherwise, we continue sliding the window until the end of the string.

Python

s = "11100000" k = 5    # Loop through the string, considering every substring of length k for i in range(len(s) - k + 1):        # Check if the current substring of length k is made of only '1's     if s[i:i+k] == '1' * k:         print(True)         break else:     print(False)

Output

False

Explanation:

for i in range(len(s) - k + 1): This iterates through the string, checking all substrings of length k (5). This loop checks 4 substrings: "11100", "11000", "10000", and "00000".
Substring Comparison: This loop compares each substring s[i:i+k] with "11111". If a match is found, it prints True and exits the loop.
Else Block: If no match is found, the else block executes and prints False.

Using find ()

find () quickly locate a substring of k consecutive '1's in the string. The find method searches the string and returns the starting index of the substring if it exists or -1 if it doesn’t. Based on this, we determine whether the condition is satisfied.

Python

s = "10101001111" k = 4 if s.find("1" * k) != -1: print(True) else: print(False)

Output

True

Explanation:

s.find("1" * k): This checks if a substring of k consecutive '1's (e.g., "1111" when k=4) exists in the string s.
if s.find("1" * k) != -1: If the find() method returns an index (i.e., the substring is found), it prints True otherwise False .

Python - Check if there are K consecutive 1's in a binary number

Striver

Improve

Article Tags :

Practice Tags :

Python - Check if there are K consecutive 1's in a binary number

Using for loop

Using sliding window

Using find ()

Similar Reads