Check if suffix and prefix of a string are palindromes

Last Updated : 05 Dec, 2022

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 are 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++

// 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;
}

Java

// Java implementation of the approach
import java.util.*;

class GFG 
{

    static String reverse(String input) 
    {
        char[] a = input.toCharArray();
        int l, r = 0;
        r = a.length - 1;
        for (l = 0; l < r; l++, r--) 
        {
            // Swap values of l and r 
            char temp = a[l];
            a[l] = a[r];
            a[r] = temp;
        }
        return String.valueOf(a);
    }

    // Function to check whether
    // the string is a palindrome
    static boolean isPalindrome(String r) 
    {
        String p = r;

        // reverse the string to
        // compare with the 
        // original string
        p = reverse(p);
        
        // check if both are same
        return (r.equals(p));
    }

    // Function to check whether the string
    // has prefix and suffix substrings
    // of length greater than 1
    // which are palindromes.
    static boolean 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.substring(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.substring(l-i,l))) 
            {
                return true;
            }
        }

        // If we did not find a suffix
        return false;
    }

    // Driver code
    public static void main(String args[]) 
    {
        String s = "abccbarfgdbd";
        if (CheckStr(s)) 
        {
            System.out.println("Yes");
        } 
        else
        {
            System.out.println("No");
        }
    }
}

// This code is contributed by Rajput-Ji

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

// C# implementation of the approach 
using System;

class GFG 
{ 

    static String reverse(String input) 
    { 
        char[] a = input.ToCharArray(); 
        int l, r = 0; 
        r = a.Length - 1; 
        for (l = 0; l < r; l++, r--) 
        { 
            // Swap values of l and r 
            char temp = a[l]; 
            a[l] = a[r]; 
            a[r] = temp; 
        } 
        return String.Join("",a); 
    } 

    // Function to check whether 
    // the string is a palindrome 
    static Boolean isPalindrome(String r) 
    { 
        String p = r; 

        // reverse the string to 
        // compare with the 
        // original string 
        p = reverse(p); 
        
        // check if both are same 
        return (r.Equals(p)); 
    } 

    // Function to check whether the string 
    // has prefix and suffix substrings 
    // of length greater than 1 
    // which are palindromes. 
    static Boolean 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.Substring(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.Substring(l-i,i))) 
            { 
                return true; 
            } 
        } 

        // If we did not find a suffix 
        return false; 
    } 

    // Driver code 
    public static void Main(String []args) 
    { 
        String s = "abccbarfgdbd"; 
        if (CheckStr(s)) 
        { 
            Console.WriteLine("Yes"); 
        } 
        else
        { 
            Console.WriteLine("No"); 
        } 
    } 
} 

// This code is contributed by 29AjayKumar

PHP

<?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 
?>

JavaScript

<script>

// JavaScript implementation of the approach

// Function to check whether
// the string is a palindrome
function isPalindrome(r)
{
    var p = r;

    // reverse the string to
    // compare with the 
    // original string
    p.split('').reverse().join('');

    // 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)
{
    var l = s.length;

    // check all prefix substrings
    var i;
    for (i = 2; i <= l; i++) {

        // check if the prefix substring
        // is a palindrome
        if (isPalindrome(s.substring(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.substring(l-i, l)))
            return true;
    }
  
    // If we did not find a suffix
    return false;    
}

// Driver code
var s = "abccbarfgdbd";
if (CheckStr(s))
    document.write( "YES");
else
    document.write( "NO");

</script>

Output

YES

Time Complexity: O(n^2), where n is the length of the given string.
Auxiliary Space: O(n)

Analysis of Algorithms

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