Skip to content
Related Articles

Related Articles

Lexicographically all Shortest Palindromic Substrings from a given string
  • Difficulty Level : Medium
  • Last Updated : 18 May, 2020

Given a string s of size N. The task is to find lexicographically all shortest palindromic substrings from the given string.

Examples:

Input: s= “programming”
Output: a g i m n o p r
Explanation:
The Lexicographical shortest palindrome substring for the word “programming” will be the single characters from the given string. Hence, the output is : a g i m n o p r.

Input: s= “geeksforgeeks”
Output: e f g k o r s

Approach:



To solve the problem mentioned above the very first observation is that the shortest palindromic substring will be of size 1. So as per the problem statement, we have to find all distinct substring of size 1 lexicographically which means all the characters in the given string.

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ program to find Lexicographically all
// Shortest Palindromic Substrings from a given string
  
#include <bits/stdc++.h>
using namespace std;
  
// Function to find all lexicographically
// shortest palindromic substring
void shortestPalindrome(string s)
{
  
    // Array to keep track of alphabetic characters
    int abcd[26] = { 0 };
  
    for (int i = 0; i < s.length(); i++)
        abcd[s[i] - 97] = 1;
  
    // Iterate to print all lexicographically shortest substring
    for (int i = 0; i < 26; i++) {
        if (abcd[i] == 1)
            cout << char(i + 97) << " ";
    }
}
  
// Driver code
int main()
{
    string s = "geeksforgeeks";
  
    shortestPalindrome(s);
  
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java program to find Lexicographically all
// Shortest Palindromic Substrings from a given string
class Main
{
    // Function to find all lexicographically
    // shortest palindromic substring
    static void shortestPalindrome(String s)
    {
  
        // Array to keep track of 
        // alphabetic characters
        int[] abcd = new int[26];
  
        for (int i = 0; i < s.length(); i++)
            abcd[s.charAt(i) - 97] = 1;
  
        // Iterate to print all lexicographically
        // shortest substring
        for (int i = 0; i < 26; i++)
        {
            if (abcd[i] == 1
            {
                System.out.print((char)(i + 97) + " ");
            }
        }
    }
  
    // Driver code
    public static void main(String[] args)
    {
        String s = "geeksforgeeks";
        shortestPalindrome(s);
    }
}

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# C++ program to find Lexicographically all
# Shortest Palindromic Substrings from a given string
  
# Function to find all lexicographically 
# shortest palindromic substring
def shortestPalindrome (s) :
      
    # Array to keep track of alphabetic characters
    abcd = [0]*26
  
    for i in range(len(s)):
        abcd[ord(s[i])-97] = 1
      
    # Iterate to print all lexicographically shortest substring
    for i in range(26): 
        if abcd[i]== 1 :
            print( chr(i + 97), end =' ' )
  
# Driver code
s = "geeksforgeeks"
  
shortestPalindrome (s)

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# program to find Lexicographically 
// all shortest palindromic substrings
// from a given string
using System;
  
class GFG{
      
// Function to find all lexicographically 
// shortest palindromic substring 
static void shortestPalindrome(string s) 
  
    // Array to keep track of
    // alphabetic characters 
    int[] abcd = new int[26]; 
  
    for(int i = 0; i < s.Length; i++) 
       abcd[s[i] - 97] = 1; 
  
    // Iterate to print all lexicographically 
    // shortest substring 
    for(int i = 0; i < 26; i++)
    
       if (abcd[i] == 1)
       
           Console.Write((char)(i + 97) + " "); 
       
    
  
// Driver code 
static public void Main(string[] args) 
    string s = "geeksforgeeks"
    shortestPalindrome(s); 
  
// This code is contributed by AnkitRai01

chevron_right


Output:

e f g k o r s

Time Complexity: O(N), where N is the size of the string.

Space Complexity: O(1)

competitive-programming-img

My Personal Notes arrow_drop_up
Recommended Articles
Page :