# 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
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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/C++

```// C++ program to check if a string is of
// the form a^nb^n.
#include<iostream>
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')
break;

// 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

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

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;
}
i++;
j--;
}
return true;
}

public static void main (String[] args) throws java.lang.Exception
{
String s = "abab";

// Function call
boolean value = isAnBn(s);
if(value == true){
System.out.println("Yes");
}
else{
System.out.println("No");
}
}
}

// Code contributed by Shivani Sanjay Shinde.
```

Output:

```No
```

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