Check if suffix and prefix of a string are palindromes

Given a string ‘s’, the task is to check whether the string has both prefix and suffix substrings of length greater than 1 which are palindromes.
Print ‘YES’ if the above condition is satisfied or ‘NO’ otherwise.

Examples:

Input : s = abartbb
Output : YES
Explanation : The string has prefix substring 'aba' 
and suffix substring 'bb' which are both palindromes, so the output is 'YES'.

Input : s = abcc
Output : NO
Explanation : The string has no prefix substring which is palindrome, 
it only has a suffix substring 'cc' which is a palindrome. 
So the output is 'NO'.


Approach:

  • First, check all the prefix substrings of length > 1 to find if there’s any which is a palindrome.
  • Check all the suffix substrings as well.
  • If both the conditions are true, then the output is ‘YES’.
  • Otherwise, the output is ‘NO’.

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 to check whether
// the string is a palindrome
bool isPalindrome(string r)
{
    string p = r;
  
    // reverse the string to
    // compare with the 
    // original string
    reverse(p.begin(), p.end());
  
    // check if both are same
    return (r == p);
}
  
// Function to check whether the string
// has prefix and suffix substrings
// of length greater than 1
// which are palindromes.
bool CheckStr(string s)
{
    int l = s.length();
  
    // check all prefix substrings
    int i;
    for (i = 2; i <= l; i++) {
  
        // check if the prefix substring
        // is a palindrome
        if (isPalindrome(s.substr(0, i)))
           break;
    }
      
    // If we did not find any palindrome prefix
    // of length greater than 1.
    if (i == (l+1))
      return false;
  
    // check all suffix substrings,
    // as the string is reversed now
    i = 2;
    for (i = 2; i <= l; i++) {
  
        // check if the suffix substring
        // is a palindrome
        if (isPalindrome(s.substr(l-i, i)))
            return true;
    }
    
    // If we did not find a suffix
    return false;    
}
  
// Driver code
int main()
{
    string s = "abccbarfgdbd";
    if (CheckStr(s))
        cout << "YES\n";
    else
        cout << "NO\n";
    return 0;
}

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Python3





# Python3 implementation of the approach 


# Function to check whether

# the string is a palindrome

def isPalindrome(r): 


	# Reverse the string and assign

	# it to new variable for comparison

	p = r[::-1]


	# check if both are same

	return r == p 


# Function to check whether the string

# has prefix and suffix substrings

# of length greater than 1

# which are palindromes.

def CheckStr(s): 


	l = len(s) 


	# check all prefix substrings

	i = 0

	for i in range(2, l + 1): 


		# check if the prefix substring

		# is a palindrome

		if isPalindrome(s[0:i]) == True:

			break


	# If we did not find any palindrome

	# prefix of length greater than 1.

	if i == (l + 1):

		return False


	# check all suffix substrings,

	# as the string is reversed now

	for i in range(2, l + 1):


		# check if the suffix substring

		# is a palindrome

		if isPalindrome(s[l – i : l]) == True:

			return True


	# If we did not find a suffix

	return False	


# Driver code

if __name__ == “__main__”:


	s = “abccbarfgdbd”


	if CheckStr(s) == True:

		print(“YES”)

	else:

		print(“NO”) 


# This code is contributed by Rituraj Jain



PHP














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<?php 


// PHP implementation of the approach 


  


// Function to check whether 


// the string is a palindrome 


function isPalindrome($r) 


{ 


    $p = $r; 


  


    // reverse the string to 


    // compare with the  


    // original string 


    strrev($p); 


  


    // check if both are same 


    return ($r == $p); 


} 


  


// Function to check whether the  


// string has prefix and suffix  


// substrings of length greater  


// than 1 which are palindromes. 


function CheckStr($s) 


{ 


    $l = strlen($s); 


  


    // check all prefix substrings 


    for ($i = 2; $i <= $l; $i++)  


    { 


  


        // check if the prefix substring 


        // is a palindrome 


        if (isPalindrome(substr($s, 0, $i))) 


        break; 


    } 


      


    // If we did not find any palindrome 


    // prefix of length greater than 1. 


    if ($i == ($l + 1)) 


    return false; 


  


    // check all suffix substrings, 


    // as the string is reversed now 


    $i = 2; 


    for ($i = 2; $i <= $l; $i++) 


    { 


  


        // check if the suffix substring 


        // is a palindrome 


        if (isPalindrome(substr($s, $l -  


                                $i, $i))) 


            return true; 


    } 


  


    // If we did not find a suffix 


    return false;  


} 


  


// Driver code 


$s = "abccbarfgdbd"; 


if (CheckStr($s)) 


    echo ("YES\n"); 


else


    echo ("NO\n"); 


  


// This code is contributed  


// by Shivi_Aggarwal  


?> 



















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Output:


YES





Complexity: O(n^2)



          
          
          
          

            

    

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