Print the longest palindromic prefix of a given string

Given a string str, the task is to find the longest palindromic prefix of the given string.

Examples:

Input: str = “abaac”
Output: aba
Explanation:
The longest prefix of the given string which is palindromic is “aba”.

Input: str = “abacabaxyz”
Output: abacaba
Explanation:
The prefixes of the given string which is palindromic are “aba” and “abacabaxyz”.
But the longest of among two is “abacabaxyz”.

Naive Approach: The idea is to generate all the substring of the given string from the starting index and check whether the substrings are palindromic or not. The palindromic string with a maximum length is the resultant string.



Below is the implementation of the above approach:

C++

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// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to find the longest prefix
// which is palindromic
void LongestPalindromicPrefix(string s)
{
  
    // Find the length of the given string
    int n = s.length();
  
    // For storing the length of longest
    // prefix palindrome
    int max_len = 0;
  
    // Loop to check the substring of all
    // length from 1 to N which is palindrome
    for (int len = 1; len <= n; len++) {
  
        // String of length i
        string temp = s.substr(0, len);
  
        // To store the reversed of temp
        string temp2 = temp;
  
        // Reversing string temp2
        reverse(temp2.begin(), temp2.end());
  
        // If string temp is palindromic
        // then update the length
        if (temp == temp2) {
            max_len = len;
        }
    }
  
    // Print the palindromic string of
    // max_len
    cout << s.substr(0, max_len);
}
  
// Driver Code
int main()
{
  
    // Given string
    string str = "abaab";
  
    // Function Call
    LongestPalindromicPrefix(str);
}

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Java

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// Java program for the above approach
import java.util.*;
class GFG{
  
// Function to find the longest prefix
// which is palindromic
static void LongestPalindromicPrefix(String s)
{
  
    // Find the length of the given String
    int n = s.length();
  
    // For storing the length of longest
    // prefix palindrome
    int max_len = 0;
  
    // Loop to check the subString of all
    // length from 1 to N which is palindrome
    for (int len = 1; len <= n; len++)
    {
  
        // String of length i
        String temp = s.substring(0, len);
  
        // To store the reversed of temp
        String temp2 = temp;
  
        // Reversing String temp2
        temp2 = reverse(temp2);
  
        // If String temp is palindromic
        // then update the length
        if (temp.equals(temp2))
        {
            max_len = len;
        }
    }
  
    // Print the palindromic String of
    // max_len
    System.out.print(s.substring(0, max_len));
}
  
static String reverse(String input) 
{
    char[] a = input.toCharArray();
    int l, r = a.length - 1;
    for (l = 0; l < r; l++, r--) 
    {
        char temp = a[l];
        a[l] = a[r];
        a[r] = temp;
    }
    return String.valueOf(a);
  
// Driver Code
public static void main(String[] args)
{
  
    // Given String
    String str = "abaab";
  
    // Function Call
    LongestPalindromicPrefix(str);
}
}
  
// This code is contributed by Rajput-Ji

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Python3

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# Python3 program for the above approach 
  
# Function to find the longest prefix
# which is palindrome
def LongestPalindromicPrefix(string):
      
    # Find the length of the given string
    n = len(string)
      
    # For storing the length of longest 
    # Prefix Palindrome
    max_len = 0
      
    # Loop to check the substring of all 
    # length from 1 to n which is palindrome
    for length in range(0, n + 1):
          
        # String of length i
        temp = string[0:length]
          
        # To store the value of temp
        temp2 = temp
          
        # Reversing the value of temp 
        temp3 = temp2[::-1]
          
        # If string temp is palindromic 
        # then update the length
        if temp == temp3:
            max_len = length
      
    # Print the palindromic string 
    # of max_len
    print(string[0:max_len])
  
# Driver code
if __name__ == '__main__' :
      
    string = "abaac";
      
    # Function call
    LongestPalindromicPrefix(string)
      
# This code is contributed by virusbuddah_

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C#

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// C# program for the above approach
using System;
  
class GFG{
  
// Function to find the longest prefix
// which is palindromic
static void longestPalindromicPrefix(String s)
{
  
    // Find the length of the given String
    int n = s.Length;
  
    // For storing the length of longest
    // prefix palindrome
    int max_len = 0;
  
    // Loop to check the subString of all
    // length from 1 to N which is palindrome
    for (int len = 1; len <= n; len++)
    {
  
        // String of length i
        String temp = s.Substring(0, len);
  
        // To store the reversed of temp
        String temp2 = temp;
  
        // Reversing String temp2
        temp2 = reverse(temp2);
  
        // If String temp is palindromic
        // then update the length
        if (temp.Equals(temp2))
        {
            max_len = len;
        }
    }
  
    // Print the palindromic String of
    // max_len
    Console.Write(s.Substring(0, max_len));
}
  
static String reverse(String input) 
{
    char[] a = input.ToCharArray();
    int l, r = a.Length - 1;
    for (l = 0; l < r; l++, r--) 
    {
        char temp = a[l];
        a[l] = a[r];
        a[r] = temp;
    }
    return String.Join("",a);
  
// Driver Code
public static void Main(String[] args)
{
  
    // Given String
    String str = "abaab";
  
    // Function Call
    longestPalindromicPrefix(str);
}
}
  
// This code is contributed by amal kumar choubey

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Output:

aba

Time Complexity: O(N2), where N is the length of the given string.

Efficient Approach: The idea is to use preprocessing algorithm KMP Algorithm. Below are the steps:

  1. Create a temporary string(say str2) which is:
    str2 = str + '?' reverse(str);
    
  2. Create an array(say lps[]) of size of length of the string str2 which will store the longest palindromic prefix which is also a suffix of string str2.
  3. Update the lps[] by using preprocessing algorithm of KMP Search Algorithm.
  4. lps[length(str2) – 1] will give the length of the longest palindromic prefix string of the given string str.

Below is the implementation of the above approach:

C++

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// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to find the longest prefix
// which is palindromic
void LongestPalindromicPrefix(string str)
{
  
    // Create temporary string
    string temp = str + '?';
  
    // Reverse the string str
    reverse(str.begin(), str.end());
  
    // Append string str to temp
    temp += str;
  
    // Find the length of string temp
    int n = temp.length();
  
    // lps[] array for string temp
    int lps[n];
  
    // Intialise every value with zero
    fill(lps, lps + n, 0);
  
    // Iterate the string temp
    for (int i = 1; i <= n; i++) {
  
        // Length of longest prefix
        // till less than i
        int len = lps[i - 1];
  
        // Calculate length for i+1
        while (len > 0
               && temp[len] != temp[i]) {
            len = lps[len - 1];
        }
  
        // If character at current index
        // len are same then increament
        // length by 1
        if (temp[i] == temp[len]) {
            len++;
        }
  
        // Update the length at current
        // index to len
        lps[i] = len;
    }
  
    // Print the palindromic string of
    // max_len
    cout << temp.substr(0, lps[n - 1]);
}
  
// Driver's Code
int main()
{
  
    // Given string
    string str = "abaab";
  
    // Function Call
    LongestPalindromicPrefix(str);
}

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Java

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// Java program for the above approach
import java.util.*;
  
class GFG{
  
// Function to find the longest 
// prefix which is palindromic
static void LongestPalindromicPrefix(String str)
{
  
    // Create temporary String
    String temp = str + '?';
  
    // Reverse the String str
    str = reverse(str);
  
    // Append String str to temp
    temp += str;
  
    // Find the length of String temp
    int n = temp.length();
  
    // lps[] array for String temp
    int []lps = new int[n];
  
    // Intialise every value with zero
    Arrays.fill(lps, 0);
  
    // Iterate the String temp
    for(int i = 1; i < n; i++)
    {
          
       // Length of longest prefix
       // till less than i
       int len = lps[i - 1];
         
       // Calculate length for i+1
       while (len > 0 && temp.charAt(len) != 
                         temp.charAt(i)) 
       {
           len = lps[len - 1];
       }
         
       // If character at current index
       // len are same then increament
       // length by 1
       if (temp.charAt(i) == temp.charAt(len))
       {
           len++;
       }
         
       // Update the length at current
       // index to len
       lps[i] = len;
    }
  
    // Print the palindromic String 
    // of max_len
    System.out.print(temp.substring(0, lps[n - 1]));
}
  
static String reverse(String input)
{
    char[] a = input.toCharArray();
    int l, r = a.length - 1;
      
    for(l = 0; l < r; l++, r--)
    {
       char temp = a[l];
       a[l] = a[r];
       a[r] = temp;
    }
    return String.valueOf(a);
}
  
// Driver Code
public static void main(String[] args)
{
  
    // Given String
    String str = "abaab";
  
    // Function Call
    LongestPalindromicPrefix(str);
}
}
  
// This code is contributed by Rajput-Ji

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C#

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// C# program for the above approach
using System;
  
class GFG{
  
// Function to find the longest 
// prefix which is palindromic
static void longestPalindromicPrefix(String str)
{
      
    // Create temporary String
    String temp = str + '?';
  
    // Reverse the String str
    str = reverse(str);
  
    // Append String str to temp
    temp += str;
  
    // Find the length of String temp
    int n = temp.Length;
  
    // lps[] array for String temp
    int []lps = new int[n];
  
    // Iterate the String temp
    for(int i = 1; i < n; i++)
    {
         
       // Length of longest prefix
       // till less than i
       int len = lps[i - 1];
         
       // Calculate length for i+1
       while (len > 0 && temp[len] != temp[i]) 
       {
           len = lps[len - 1];
       }
         
       // If character at current index
       // len are same then increament
       // length by 1
       if (temp[i] == temp[len])
       {
           len++;
       }
         
       // Update the length at current
       // index to len
       lps[i] = len;
    }
      
    // Print the palindromic String 
    // of max_len
    Console.Write(temp.Substring(0, lps[n - 1]));
}
  
static String reverse(String input)
{
    char[] a = input.ToCharArray();
    int l, r = a.Length - 1;
      
    for(l = 0; l < r; l++, r--)
    {
       char temp = a[l];
       a[l] = a[r];
       a[r] = temp;
    }
    return String.Join("", a);
}
  
// Driver Code
public static void Main(String[] args)
{
  
    // Given String
    String str = "abaab";
  
    // Function Call
    longestPalindromicPrefix(str);
}
}
  
// This code is contributed by Rajput-Ji

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Output:

aba

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

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