Program to check if binary string has at most one segment of ones or not using Python

Binary strings often need validation to check specific patterns. A common requirement is to verify if a binary string has at most one contiguous segment of ones. This means all the '1's should be grouped together without any '0's in between.

For example, "111000" has one segment of ones, while "11010" has two separate segments.

Problem Understanding

Given a binary string without leading zeros, we need to determine if it contains at most one contiguous segment of ones. The string should have all '1's grouped together, followed by all '0's (if any).

Examples

• "111000" ? True (one segment: "111")

• "1" ? True (single character)

• "0" ? True (no ones)

• "11010" ? False (two segments: "11" and "1")

Solution Approach

We can solve this by tracking when we encounter zeros. Once we find a '0', any subsequent '1' indicates a second segment ?

def solve(s):
    count = -1
    if len(s) == 1:
        return True
    
    for i in s:
        if i == "1" and count > -1:
            return False
        elif i == "0":
            count += 1
    
    return True

# Test with different examples
test_cases = ["11100", "1", "0", "11010", "111", "101"]

for s in test_cases:
    result = solve(s)
    print(f"'{s}' ? {result}")

'11100' ? True
'1' ? True
'0' ? True
'11010' ? False
'111' ? True
'101' ? False

How It Works

1. Initialize counter: Set count = -1 to track zero encounters

2. Handle edge case: Single character strings always return True

3. Iterate through string: For each character:

   • If we find '1' after encountering zeros (count > -1), return False

   • If we find '0', increment the counter

4. Return True: If no violations found, the string is valid

Alternative Solution

A simpler approach using Python string methods ?

def check_one_segment(s):
    # Find the first occurrence of '0'
    try:
        zero_index = s.index('0')
        # Check if there are any '1's after the first '0'
        return '1' not in s[zero_index:]
    except ValueError:
        # No '0' found, so all are '1's
        return True

# Test the alternative solution
test_cases = ["11100", "1", "0", "11010", "111", "101"]

for s in test_cases:
    result = check_one_segment(s)
    print(f"'{s}' ? {result}")

'11100' ? True
'1' ? True
'0' ? True
'11010' ? False
'111' ? True
'101' ? False

Comparison

Conclusion

Both approaches efficiently check for at most one contiguous segment of ones. The counter method tracks zero encounters, while the string method finds the first zero and checks for subsequent ones. Choose based on readability preference and coding style.