Check if a string follows a^nb^n pattern or not

Given a string str, return true string follows pattern anbn, i.e., it has a’s followed by b’s such that the number of a’s and b’s are same.

Input : str = "aabb"
Output : Yes

Input : str = "abab"
Output : No

Input : str = "aabbb"
Output : No

The idea is to first count a’s. If number of a’s is not equal to half of string’s length, then return false. Else check if all remaining characters are b’s or not.

Below is C++ implementation of above algorithm.


// C++ program to check if a string is of
// the form a^nb^n.
using namespace std;

// Returns true str is of the form a^nb^n.
bool isAnBn(string str)
   int n = str.length();

   // After this loop 'i' has count of a's
   int i;
   for (i=0; i<n; i++)
      if (str[i] != 'a')

   // Since counts of a's and b's should
   // be equal, a should apear exactly
   // n/2 times
   if (i*2 != n)
      return false;

   // Rest of the characters must be all 'b'
   int j;
   for (j=i; j<n; j++)
      if (str[j] != 'b')
         return false;

   return true;

// Driver code
int main()
    string str = "abab";
    isAnBn(str) ? cout << "Yes"
                : cout << "No";
    return 0;


// Java program to check if a string is of
// the form a^nb^n.
import java.util.*;
import java.lang.*;

class CheckPattern
    public static boolean isAnBn(String s)
        int l = s.length();
        // Only even length strings will have same number of a's and b's
        if (l%2 == 1)
            return false;
        // Set two pointers, one from the left and another from right
        int i = 0;
        int j = l-1;
        // Compare the characters till the center
        while (i<j)
            if(s.charAt(i) != 'a' || s.charAt(j) != 'b')
                return false;
        return true;    
    public static void main (String[] args) throws java.lang.Exception
        String s = "abab";
        // Function call
        boolean value = isAnBn(s);
        if(value == true){

// Code contributed by Shivani Sanjay Shinde.



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