Given a string of lowercase, find the length of the longest substring that does not contain any palindrome as a substring.
Examples:
Input : str = "daiict" Output : 3 dai, ict are longest substring that do not contain any palindrome as substring Input : str = "a" Output : 0 a is itself a palindrome
The idea is to observe that if any character forms a palindrome, it can not be included in any substring. So, in that case, the required substring will be picked from before or after that character which forms a palindrome.
Therefore, a simple solution is to traverse the string and, for each character, check whether it forms a palindrome of length 2 or 3 with its adjacent characters. If it does not, then increase the length of the substring, otherwise re-initialize the length of the substring to zero. Using this approach, find the length of the maximum substring.
Below is the implementation of the above approach:
C++
// C++ implementation of the above approach#include <bits/stdc++.h>using namespace std;// Function to find the length of the longest// substringint lenoflongestnonpalindrome(string s){ // initializing the variables int max1 = 1, len = 0; for (int i = 0; i < s.length() - 1; i++) { // checking palindrome of size 2 // example: aa if (s[i] == s[i + 1]) len = 0; // checking palindrome of size 3 // example: aba else if (s[i + 1] == s[i - 1] && i > 0) len = 1; else // incrementing length of substring len++; max1 = max(max1, len + 1); // finding maximum } // if there exists single character then // it is always palindrome if (max1 == 1) return 0; else return max1;}// Driver Codeint main(){ string s = "synapse"; cout << lenoflongestnonpalindrome(s) << "\n"; return 0;} |
Java
// Java implementation of the above approachimport java.util.Arrays;import java.lang.Math;class GFG { // Function to find the length of the longest // substring public static int lenoflongestnonpalindrome(String s) { // initializing the variables int max1 = 1, len = 0; char[] new_str = s.toCharArray(); for (int i = 0; i < new_str.length - 1; i++) { // checking palindrome of size 2 // example: aa if (new_str[i] == new_str[i + 1]) len = 0; // checking palindrome of size 3 // example: aba else if (i > 0 && (new_str[i + 1] == new_str[i - 1])) len = 1; else // incrementing length of substring len++; max1 = Math.max(max1, len + 1); // finding maximum } // if there exits single character then // it is always palindrome if (max1 == 1) return 0; else return max1; } // Driver Code public static void main(String[] args) { String s = "synapse"; System.out.println(lenoflongestnonpalindrome(s)); }}// This code is contributed by princiraj1992 |
Python3
# Python3 implementation of the above approach # Function to find the length # of the longest substring def lenoflongestnonpalindrome(s): # initializing the variables max1, length = 1, 0 for i in range(0, len(s) - 1): # checking palindrome of # size 2 example: aa if s[i] == s[i + 1]: length = 0 # checking palindrome of # size 3 example: aba elif s[i + 1] == s[i - 1] and i > 0: length = 1 else: # incrementing length of substring length += 1 max1 = max(max1, length + 1) # finding maximum # If there exits single character # then it is always palindrome if max1 == 1: return 0 else: return max1 # Driver Code if __name__ == "__main__": s = "synapse" print(lenoflongestnonpalindrome(s)) # This code is contributed by Rituraj Jain |
C#
// C# implementation of the above approach using System; class GFG { // Function to find the length of the longest // substring public static int lenoflongestnonpalindrome(String s) { // initializing the variables int max1 = 1, len = 0; char[] new_str = s.ToCharArray(); for (int i = 0; i < new_str.Length - 1; i++) { // checking palindrome of size 2 // example: aa if (new_str[i] == new_str[i + 1]) len = 0; // checking palindrome of size 3 // example: aba else if (i > 0 && (new_str[i + 1] == new_str[i - 1])) len = 1; else // incrementing length of substring len++; max1 = Math.Max(max1, len + 1); // finding maximum } // if there exits single character then // it is always palindrome if (max1 == 1) return 0; else return max1; } // Driver Code public static void Main(String[] args) { String s = "synapse"; Console.WriteLine(lenoflongestnonpalindrome(s)); }}// This code has been contributed by 29AjayKumar |
PHP
<?php// PHP implementation of the above approach // Function to find the length of the longest // substring function lenoflongestnonpalindrome($s) { // initializing the variables $max1 = 1; $len = 0; for ($i = 0; $i < strlen($s) - 1; $i++) { // checking palindrome of size 2 // example: aa if ($s[$i] == $s[$i + 1]) $len = 0; // checking palindrome of size 3 // example: aba else if ($s[$i + 1] == $s[$i - 1] && $i > 0) $len = 1; else // incrementing length of substring $len++; $max1 = max($max1, $len + 1); // finding maximum } // if there exits single character then // it is always palindrome if ($max1 == 1) return 0; else return $max1; } // Driver Code $s = "synapse"; echo lenoflongestnonpalindrome($s), "\n"; // This code is contributed by AnkitRai01?> |
Javascript
<script>// JavaScript implementation of the above approach// Function to find the length of the longest// substringfunction lenoflongestnonpalindrome(s){ // initializing the variables let max1 = 1, len = 0; for (let i = 0; i < s.length - 1; i++) { // checking palindrome of size 2 // example: aa if (s[i] == s[i + 1]) len = 0; // checking palindrome of size 3 // example: aba else if (s[i + 1] == s[i - 1] && i > 0) len = 1; else // incrementing length of substring len++; max1 = Math.max(max1, len + 1); // finding maximum } // if there exits single character then // it is always palindrome if (max1 == 1) return 0; else return max1;}// Driver Code let s = "synapse"; document.write(lenoflongestnonpalindrome(s) + "<br>"); //This code is contributed by Manoj</script> |
Output:
7
Time complexity: O(n) where n is length of input string
Auxiliary Space: O(1)
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