Check if a binary string has two consecutive occurrences of one everywhere

Given a string str consisting of only characters ‘a’ and ‘b’, the task is to check whether the string is valid or not. In a valid string, every group of consecutive b must be of length 2 and must appear after 1 or more occurrences of character ‘a’ i.e. “abba” is a valid sub-string but “abbb” and aba are not. Print 1 if the string is valid else print -1.

Examples:

Input: str = “abbaaabbabba”
Output: 1



Input: str = “abbaaababb”
Output: -1

Approach: Find every occurrence of ‘b’ in the string and check whether it is a part of the sub-string “abb”. If the condition fails for any sub-string then print -1 else print 1.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function that returns 1 if str is valid
bool isValidString(string str, int n)
{
    // Index of first appearance of 'b'
    int index = find(str.begin(),       
                     str.end(), 'b') - 
                     str.begin();
  
    // If str starts with 'b'
    if (index == 0)
        return false;
  
    // While 'b' occurs in str
    while (index <= n - 1)
    {
        // If 'b' doesn't appear after an 'a'
        if (str[index - 1] != 'a')
            return false;
  
        // If 'b' is not succeeded by another 'b'
        if (index + 1 < n && str[index + 1] != 'b')
            return false;
  
        // If sub-string is of the type "abbb"
        if (index + 2 < n && str[index + 2] == 'b')
            return false;
  
        // If str ends with a single b
        if (index == n - 1)
            return false;
  
        index = find(str.begin() + index + 2, 
                     str.end(), 'b') - str.begin();
    }
    return true;
}
  
// Driver code
int main()
{
    string str = "abbaaabbabba";
    int n = str.length();
    isValidString(str, n) ? cout
                << "true" : cout << "false";
    return 0;
}
  
// This code is contributed by
// sanjeev2552

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Java

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// Java implementation of the approach
class GFG {
  
    // Function that returns 1 if str is valid
    private static boolean isValidString(String str, int n)
    {
  
        // Index of first appearance of 'b'
        int index = str.indexOf("b");
  
        // If str starts with 'b'
        if (index == 0)
            return false;
  
        // While 'b' occurs in str
        while (index != -1) {
  
            // If 'b' doesn't appear after an 'a'
            if (str.charAt(index - 1) != 'a')
                return false;
  
            // If 'b' is not succeeded by another 'b'
            if (index + 1 < n && str.charAt(index + 1) != 'b')
                return false;
  
            // If sub-string is of the type "abbb"
            if (index + 2 < n && str.charAt(index + 2) == 'b')
                return false;
  
            // If str ends with a single b
            if (index == n - 1)
                return false;
  
            index = str.indexOf("b", index + 2);
        }
        return true;
    }
  
    // Driver code
    public static void main(String[] args)
    {
        String str = "abbaaabbabba";
        int n = str.length();
        System.out.println(isValidString(str, n));
    }
}

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Python 3

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# Python 3 implementation of the approach
  
# Function that returns 1 if str is valid
def isValidString(str, n):
  
    # Index of first appearance of 'b'
    idx = str.find("b")
  
    # If str starts with 'b'
    if (idx == 0):
        return False
  
    # While 'b' occurs in str
    while (idx != -1):
  
        # If 'b' doesn't appear after an 'a'
        if (str[idx - 1] != 'a'):
            return False
  
        # If 'b' is not succeeded by another 'b'
        if (idx + 1 < n and str[idx + 1] != 'b'):
            return False
  
        # If sub-string is of the type "abbb"
        if (idx + 2 < n and str[idx + 2] == 'b'):
            return False
  
        # If str ends with a single b
        if (idx == n - 1):
            return False
  
        idx = str.find("b", idx + 2)
  
    return True
  
# Driver code
if __name__ == "__main__":
  
    str = "abbaaabbabba"
    n = len(str)
    print(isValidString(str, n))
  
# This code is contributed by ita_c

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C#

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// C# implementation of the approach 
using System;
  
class GFG 
  
    // Function that returns 1 if str is valid 
    private static bool isValidString(string str, int n) 
    
  
        // Index of first appearance of 'b' 
        int index = str.IndexOf("b"); 
  
        // If str starts with 'b' 
        if (index == 0) 
            return false
  
        // While 'b' occurs in str 
        while (index != -1) 
        
  
            // If 'b' doesn't appear after an 'a' 
            if (str[index - 1] != 'a'
                return false
  
            // If 'b' is not succeeded by another 'b' 
            if (index + 1 < n && str[index + 1] != 'b'
                return false
  
            // If sub-string is of the type "abbb" 
            if (index + 2 < n && str[index + 2] == 'b'
                return false
  
            // If str ends with a single b 
            if (index == n - 1) 
                return false
  
            index = str.IndexOf("b", index + 2); 
        
        return true
    
  
    // Driver code 
    public static void Main() 
    
        string str = "abbaaabbabba"
        int n = str.Length; 
        Console.WriteLine(isValidString(str, n)); 
    
  
// This code is contributed by Ryuga

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Output:

true


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