Generate an N-length string having longest palindromic substring of length K
Given two integers N and K (K ≤ N), the task is to obtain a string of length N such that maximum length of a palindromic substring of this string is K.
Examples:
Input: N = 5, K = 3
Output: “abacd”
Explanation: Palindromic substrings are “a”, “b”, “c”, “d” and “aba”. Therefore, the longest palindromic substring from the given string is of length 3.Input: N = 8, K = 4
Output: “abbacdef”
Explanation: Palindromic substrings are “a”, “b”, “c”, “d”, “e”, “f”, “bb”, “abba”. Therefore, the longest palindromic substring from the given string is of length 4.
Approach: The idea is based on the following observation that the string of any length made up of a single character is always palindromic, e.g. {‘a’, ‘bbbbb’, ‘ccc’}. So, in order to generate a string with required conditions, print ‘a’ K times such that it has a longest palindromic substring of length K fill the remaining N – K slots by a non-palindromic sequence.
Follow the steps below to solve the problem:
- Print ‘a’ exactly K times.
- Consider a non-palindromic sequence, say “bcd”.
- Print the string.
Below is the implementation of the above approach:
C++
// C++ program to implement the above approach #include <bits/stdc++.h> using namespace std; // Function to generate a string of // length N having longest palindromic // substring of length K void string_palindrome( int N, int K) { // Fill first K characters with 'a' for ( int i = 0; i < K; i++) cout << "a" ; // Stores a non-palindromic sequence // to be repeated for N - k slots string s = "bcd" ; // Print N - k remaining characters for ( int i = 0; i < N - K; i++) cout << s[i % 3]; } // Driver Code int main() { // Given N and K int N = 5, K = 3; string_palindrome(N, K); return 0; } |
Java
// Java program to implement the above approach import java.util.*; class GFG { // Function to generate a String of // length N having longest palindromic // subString of length K static void String_palindrome( int N, int K) { // Fill first K characters with 'a' for ( int i = 0 ; i < K; i++) System.out.print( "a" ); // Stores a non-palindromic sequence // to be repeated for N - k slots String s = "bcd" ; // Print N - k remaining characters for ( int i = 0 ; i < N - K; i++) System.out.print(s.charAt(i % 3 )); } // Driver Code public static void main(String[] args) { // Given N and K int N = 5 , K = 3 ; String_palindrome(N, K); } } // This code is contributed by 29AjayKumar |
Python3
# Python3 program to implement the above approach # Function to generate a string of # length N having longest palindromic # substring of length K def string_palindrome(N, K): # Fill first K characters with 'a' for i in range (K): print ( "a" , end = "") # Stores a non-palindromic sequence # to be repeated for N - k slots s = "bcd" # Print N - k remaining characters for i in range (N - K): print (s[i % 3 ], end = "") # Driver Code if __name__ = = '__main__' : # Given N and K N, K = 5 , 3 string_palindrome(N, K) # This code is contributed by mohit kumar 29 |
C#
// C# program to implement the above approach using System; class GFG { // Function to generate a String of // length N having longest palindromic // subString of length K static void String_palindrome( int N, int K) { // Fill first K characters with 'a' for ( int i = 0; i < K; i++) Console.Write( "a" ); // Stores a non-palindromic sequence // to be repeated for N - k slots string s = "bcd" ; // Print N - k remaining characters for ( int i = 0; i < N - K; i++) Console.Write(s[i % 3]); } // Driver Code public static void Main( string [] args) { // Given N and K int N = 5, K = 3; String_palindrome(N, K); } } // This code is contributed by AnkThon |
Javascript
<script> // JavaScript program for above approach // Function to generate a String of // length N having longest palindromic // subString of length K function String_palindrome(N, K) { // Fill first K characters with 'a' for (let i = 0; i < K; i++) document.write( "a" ); // Stores a non-palindromic sequence // to be repeated for N - k slots let s = "bcd" ; // Print N - k remaining characters for (let i = 0; i < N - K; i++) document.write(s[i % 3]); } // Driver Code // Given N and K let N = 5, K = 3; String_palindrome(N, K); </script> |
aaabc
Time complexity: O(N)
Auxiliary space: O(1)
Please Login to comment...