Open In App
Related Articles

Check If every group of a’s is followed by a group of b’s of same length

Improve Article
Improve
Save Article
Save
Like Article
Like

Given string str, the task is to check whether every group of consecutive a’s is followed by a group of consecutive b’s of the same length. If the condition is true for every group then print 1 else print 0.

Examples: 

Input: str = “ababaabb” 
Output:
ab, ab, aabb. All groups are valid

Input: str = “aabbabb” 
Output:
aabb, abb (A single ‘a’ followed by 2 ‘b’) 

Approach: 

  • For every a in the string increment the count.
  • Starting from the first b, decrement the count for every b.
  • If at the end of the above cycle, count != 0 then return false.
  • Else repeat the first two steps for the rest of the string.
  • Return true if the condition is satisfied for all the cycles else print 0.

Below is the implementation of the above approach: 

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to match whether there are always n consecutive b's
// followed by n consecutive a's throughout the string
int matchPattern(string s)
{
    int count = 0;
    int n = s.length();
 
    // Traverse through the string
    int i = 0;
    while (i < n) {
 
        // Count a's in current segment
        while (i < n && s[i] == 'a') {
            count++;
            i++;
        }
 
        // Count b's in current segment
        while (i < n && s[i] == 'b') {
            count--;
            i++;
        }
 
        // If both counts are not same.
        if (count != 0)
            return false;
    }
 
    return true;
}
 
// Driver code
int main()
{
    string s = "bb";
    if (matchPattern(s) == true)
        cout << "Yes";
    else
        cout << "No";
    return 0;
}


Java




// Java implementation of the above approach
 
public class GFG{
 
// Function to match whether there are always n consecutive b's
// followed by n consecutive a's throughout the string
static boolean matchPattern(String s)
{
    int count = 0;
    int n = s.length();
 
    // Traverse through the string
    int i = 0;
    while (i < n) {
 
        // Count a's in current segment
        while (i < n && s.charAt(i) == 'a') {
            count++;
            i++;
        }
 
        // Count b's in current segment
        while (i < n && s.charAt(i) == 'b') {
            count--;
            i++;
        }
 
        // If both counts are not same.
        if (count != 0)
            return false;
    }
 
    return true;
}
 
// Driver code
public static void main(String []args)
{
    String s = "bb";
    if (matchPattern(s) == true)
        System.out.println("Yes");
    else
        System.out.println("No");
}
 
// This code is contributed by Ryuga
}


Python3




# Python 3 implementation of the approach
 
# Function to match whether there are
# always n consecutive b's followed by
# n consecutive a's throughout the string
def matchPattern(s):
 
    count = 0;
    n = len(s);
 
    # Traverse through the string
    i = 0;
    while (i < n) :
 
        # Count a's in current segment
        while (i < n and s[i] == 'a'):
 
            count += 1 ;
            i =+ 1;
 
        # Count b's in current segment
        while (i < n and s[i] == 'b'):
            count -= 1 ;
            i += 1;
     
        # If both counts are not same.
        if (count != 0):
            return False;
 
    return True;
 
# Driver code
s = "bb";
if (matchPattern(s) == True):
    print("Yes");
else:
    print("No");
 
# This code is contributed
# by Akanksha Rai


C#




// C# implementation of the above approach
  
using System;
public class GFG{
  
// Function to match whether there are always n consecutive b's
// followed by n consecutive a's throughout the string
static bool matchPattern(string s)
{
    int count = 0;
    int n = s.Length;
  
    // Traverse through the string
    int i = 0;
    while (i < n) {
  
        // Count a's in current segment
        while (i < n && s[i] == 'a') {
            count++;
            i++;
        }
  
        // Count b's in current segment
        while (i < n && s[i] == 'b') {
            count--;
            i++;
        }
  
        // If both counts are not same.
        if (count != 0)
            return false;
    }
  
    return true;
}
  
// Driver code
public static void Main()
{
    string s = "bb";
    if (matchPattern(s) == true)
        Console.Write("Yes");
    else
        Console.Write("No");
}
 
}


PHP




<?php
//PHP implementation of the approach
 
// Function to match whether there are always n consecutive b's
// followed by n consecutive a's throughout the string
function  matchPattern($s)
{
    $count = 0;
    $n = strlen($s);
 
    // Traverse through the string
    $i = 0;
    while ($i < $n) {
 
        // Count a's in current segment
        while ($i < $n && $s[$i] == 'a') {
            $count++;
            $i++;
        }
 
        // Count b's in current segment
        while ($i < $n && $s[$i] == 'b') {
            $count--;
            $i++;
        }
 
        // If both counts are not same.
        if ($count != 0)
            return false;
    }
 
    return true;
}
 
// Driver code
  
    $s = "bb";
    if (matchPattern($s) == true)
        echo  "Yes";
    else
        echo "No";
     
 
// This code is contributed by ajit
?>


Javascript




<script>
 
    // Javascript implementation of
    // the above approach
     
    // Function to match whether there are
    // always n consecutive b's
    // followed by n consecutive a's
    // throughout the string
    function matchPattern(s)
    {
        let count = 0;
        let n = s.length;
 
        // Traverse through the string
        let i = 0;
        while (i < n)
        {
 
            // Count a's in current segment
            while (i < n && s[i] == 'a')
            {
                count++;
                i++;
            }
 
            // Count b's in current segment
            while (i < n && s[i] == 'b')
            {
                count--;
                i++;
            }
 
            // If both counts are not same.
            if (count != 0)
                return false;
        }
 
        return true;
    }
     
    let s = "bb";
    if (matchPattern(s) == true)
        document.write("Yes");
    else
        document.write("No");
     
</script>


Output

No

Complexity Analysis:

  • Time Complexity : O( | s | ) ,where | s | is length of given string s.
  • Space Complexity : O(1) ,as we are not using any extra space

Feeling lost in the world of random DSA topics, wasting time without progress? It's time for a change! Join our DSA course, where we'll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 geeks!

Last Updated : 12 Sep, 2022
Like Article
Save Article
Previous
Next
Similar Reads
Complete Tutorials