Check if a string has m consecutive 1s or 0s in Python

Suppose we have a binary string s and another value m, we have to check whether the string has m consecutive 1's or m consecutive 0's.

So, if the input is like s = "1110111000111", m = 3, then the output will be True as there are three consecutive 0s and 1s.

Algorithm

To solve this, we will follow these steps ?

• str_size := size of s
• count_0 := 0, count_1 := 0
• for i in range 0 to str_size − 1, do
  • if s[i] is same as '0', then
    • count_1 := 0
    • count_0 := count_0 + 1
  • otherwise,
    • count_0 := 0
    • count_1 := count_1 + 1
  • if count_0 is same as m or count_1 is same as m, then
    • return True
• return False

Example

Let us see the following implementation to get better understanding ?

def solve(s, m):
    str_size = len(s)
    count_0 = 0
    count_1 = 0
    for i in range(0, str_size):
        if (s[i] == '0'):
            count_1 = 0
            count_0 += 1
        else:
            count_0 = 0
            count_1 += 1
        if (count_0 == m or count_1 == m):
            return True
    return False

s = "1110111000111"
m = 3
print(solve(s, m))

The output of the above code is ?

True

Using Regular Expressions

We can also solve this problem using regular expressions to find patterns of consecutive characters ?

import re

def solve_regex(s, m):
    pattern_0 = '0' * m
    pattern_1 = '1' * m
    return pattern_0 in s or pattern_1 in s

s = "1110111000111"
m = 3
print(solve_regex(s, m))

The output of the above code is ?

True

Comparison

Conclusion

Both approaches effectively check for m consecutive 0s or 1s in a binary string. The counter approach is more memory-efficient, while the string matching approach using the in operator is more readable and concise.