Given a string 
. The task is to find the lexicographically largest subsequence of the string which is a palindrome.
Examples:
Input : str = "abrakadabra" Output : rr Input : str = "neveropen" Output : ss
The idea is to observe a character a is said to be lexicographically larger than a character b if it’s ASCII value is greater than that of b.
Since the string has to be palindromic, the string should contain the largest characters only, as if we place any other smaller character in between the first and last character then it will make the string lexicographically smaller.
To find the lexicographically largest subsequence, first find the largest characters in the given string and append all of its occurrences in the original string to form the resultant subsequence string.
Below is the implementation of the above approach:
C++
// CPP program to find the largest// palindromic subsequence#include <bits/stdc++.h>using namespace std;// Function to find the largest// palindromic subsequencestring largestPalinSub(string s){ string res; char mx = s[0]; // Find the largest character for (int i = 1; i < s.length(); i++) mx = max(mx, s[i]); // Append all occurrences of largest character // to the resultant string for (int i = 0; i < s.length(); i++) if (s[i] == mx) res += s[i]; return res;}// Driver Codeint main(){ string s = "neveropen"; cout << largestPalinSub(s);} |
Java
// Java program to find the largest // palindromic subsequence class GFG{// Function to find the largest // palindromic subsequence static String largestPalinSub(String s) { String res = ""; char mx = s.charAt(0); // Find the largest character for (int i = 1; i < s.length(); i++) mx = (char)Math.max((int)mx, (int)s.charAt(i)); // Append all occurrences of largest // character to the resultant string for (int i = 0; i < s.length(); i++) if (s.charAt(i) == mx) res += s.charAt(i); return res; } // Driver Codepublic static void main(String []args){ String s = "neveropen"; System.out.println(largestPalinSub(s));}}// This code is contributed by// Rituraj Jain |
Python3
# Python3 program to find the largest # palindromic subsequence # Function to find the largest # palindromic subsequence def largestPalinSub(s): res = "" mx = s[0] # Find the largest character for i in range(1, len(s)): mx = max(mx, s[i]) # Append all occurrences of largest # character to the resultant string for i in range(0, len(s)): if s[i] == mx: res += s[i] return res # Driver Code if __name__ == "__main__": s = "neveropen" print(largestPalinSub(s)) # This code is contributed by# Rituraj Jain |
C#
// C# program to find the largest // palindromic subsequence using System;class GFG{ // Function to find the largest // palindromic subsequence static string largestPalinSub(string s) { string res = ""; char mx = s[0]; // Find the largest character for (int i = 1; i < s.Length; i++) mx = (char)Math.Max((int)mx, (int)s[i]); // Append all occurrences of largest // character to the resultant string for (int i = 0; i < s.Length; i++) if (s[i] == mx) res += s[i]; return res; } // Driver Code public static void Main() { string s = "neveropen"; Console.WriteLine(largestPalinSub(s)); }}// This code is contributed by Ryuga |
PHP
<?php // PHP program to find the largest// palindromic subsequence// Function to find the largest// palindromic subsequencefunction largestPalinSub($s){ $res=""; $mx = $s[0]; // Find the largest character for ($i = 1; $i < strlen($s); $i++) { $mx = max($mx, $s[$i]); } // Append all occurrences of largest character // to the resultant string for ($i = 0; $i < strlen($s); $i++) { if ($s[$i] == $mx) { $res.=$s[$i]; } } return $res;}// Driver Code$s = "neveropen";echo(largestPalinSub($s));// This code is contributed by princiraj1992?> |
Javascript
<script> // JavaScript program to find the largest // palindromic subsequence // Function to find the largest // palindromic subsequence function largestPalinSub(s) { let res = ""; let mx = s[0]; // Find the largest character for (let i = 1; i < s.length; i++) mx = String.fromCharCode(Math.max(mx.charCodeAt(), s[i].charCodeAt())); // Append all occurrences of largest // character to the resultant string for (let i = 0; i < s.length; i++) if (s[i] == mx) res += s[i]; return res; } let s = "neveropen"; document.write(largestPalinSub(s)); </script> |
Output
ss
Complexity Analysis:
- Time Complexity: O(N), where N is the length of the string.
- Auxiliary Space: O(1)
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