Given a string s of size N. The task is to find lexicographically all the 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= “neveropen”
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 substrings of size 1 lexicographically, which means all the characters in the given string.
Below is the implementation of the above approach:
C++
// 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 substringvoid 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 codeint main(){ string s = "neveropen"; shortestPalindrome(s); return 0;} |
Java
// Java program to find Lexicographically all// Shortest Palindromic Substrings from a given stringclass 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 = "neveropen"; shortestPalindrome(s); }} |
Python3
# C++ program to find Lexicographically all# Shortest Palindromic Substrings from a given string# Function to find all lexicographically # shortest palindromic substringdef 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 codes = "neveropen"shortestPalindrome (s) |
C#
// C# program to find Lexicographically // all shortest palindromic substrings// from a given stringusing 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 = "neveropen"; shortestPalindrome(s); } } // This code is contributed by AnkitRai01 |
Javascript
<script>// Javascript program to find Lexicographically all// Shortest Palindromic Substrings from a given string // Function to find all lexicographically // shortest palindromic substring function shortestPalindrome(s) { // Array to keep track of // alphabetic characters let abcd = Array.from({length: 26}, (_, i) => 0); for (let i = 0; i < s.length; i++) abcd[s[i].charCodeAt() - 97] = 1; // Iterate to print all lexicographically // shortest substring for (let i = 0; i < 26; i++) { if (abcd[i] == 1) { document.write(String.fromCharCode(i + 97) + " "); } } }// Driver Code let s = "neveropen"; shortestPalindrome(s.split('')); </script> |
Output:
e f g k o r s
Time Complexity: O(N), where N is the size of the string.
Space Complexity: O(1)
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