Given a string S of length N, the task is to find the length of the longest substring X of the string S such that:
- No non-empty substring of X is a prefix of S.
- No non-empty substring of X is a suffix of S.
- If no such string is possible, print −1.
Examples:
Input: S = “abcdefb”
Output: 4
Explanation: cdef is the substring satisfying the conditions.Input: S = “cccc”
Output: -1
Explanation: No substring can satisfy the conditions.
Approach: To solve the problem follow the below observation:
Observations:
Since a prefix starts from a starting of string S and a suffix starts from ending of a string S. Therefore maximum length of substring can be length of string S – 2 and it cannot have elements which are at the starting or ending of the string.
So select the substring not having any character which is the first or last character of the string. this will the largest substring satisfying the conditions.
Follow the steps mentioned to solve the problem:
- Traverse the string from i = 1 to end of the string.
- Check if any character is found which is not equal to first and last character of string.
- If found consider it a part of the resultant substring.
- Otherwise, it would not be included in the substring.
- The result would be maximum length of substring satisfying the conditions.
Below is the implementation of the above approach:
C++
// C++ code to implement the above approach#include <bits/stdc++.h>using namespace std;// Function to find the largest substringint findLargest(string s){ int maxi = -1; int sum = 0; // Loop to find the largest substring // satisfying the given conditions for (int i = 1; i < s.size() - 1; i++) { if (s[i] != s[0] && s[i] != s[s.length() - 1]) { sum++; } else { maxi = max(sum, maxi); sum = 0; } } maxi = max(sum, maxi); if (maxi == 0) { maxi = -1; } // Return the maximum length return maxi;}// Driver codeint main(){ string s = "abcdefb"; // Function call cout << (findLargest(s)); return 0;}// This code is contributed by rakeshsahni |
Java
// Java code to implement the above approachimport java.io.*;class GFG { // Function to find the largest substring public static int findLargest(String s) { int max = -1; int sum = 0; // Loop to find the largest substring // satisfying the given conditions for (int i = 1; i < s.length() - 1; i++) { if (s.charAt(i) != s.charAt(0) && s.charAt(i) != s.charAt(s.length() - 1)) { sum++; } else { max = Math.max(sum, max); sum = 0; } } max = Math.max(sum, max); if (max == 0) { max = -1; } // Return the maximum length return max; } // Driver code public static void main(String[] args) { String s = "abcdefb"; // Function call System.out.println(findLargest(s)); }} |
Python
# Python code to implement the above approach# Function to find the largest substringdef findLargest(s): maxi = -1 sum = 0 # Loop to find the largest substring # satisfying the given conditions for i in range(1, len(s) - 1): if (s[i] != s[0] and s[i] != s[len(s) - 1]): sum += 1 else: maxi = max(sum, maxi) sum = 0 maxi = max(sum, maxi) if (maxi == 0): maxi = -1 # Return the maximum length return maxi# Driver codeif __name__ == "__main__": s = "abcdefb" # Function call print(findLargest(s)) # This code is contributed by hrithikgarg03188. |
C#
// C# code to implement the above approachusing System;public class GFG{ // Function to find the largest substring public static int findLargest(string s) { int max = -1; int sum = 0; // Loop to find the largest substring // satisfying the given conditions for (int i = 1; i < s.Length - 1; i++) { if (s[i] != s[0] && s[i] != s[s.Length - 1]) { sum++; } else { max = Math.Max(sum, max); sum = 0; } } max = Math.Max(sum, max); if (max == 0) { max = -1; } // Return the maximum length return max; } // Driver code static public void Main (){ string s = "abcdefb"; // Function call Console.Write(findLargest(s)); }}// This code is contributed by hrithikgarg03188. |
Javascript
<script> // JavaScript code for the above approach // Function to find the largest substring function findLargest(s) { let maxi = -1; let sum = 0; // Loop to find the largest substring // satisfying the given conditions for (let i = 1; i < s.length - 1; i++) { if (s[i] != s[0] && s[i] != s[s.length - 1]) { sum++; } else { maxi = Math.max(sum, maxi); sum = 0; } } maxi = Math.max(sum, maxi); if (maxi == 0) { maxi = -1; } // Return the maximum length return maxi; } // Driver code let s = "abcdefb"; // Function call document.write(findLargest(s)); // This code is contributed by Potta Lokesh </script> |
Output
4
Time Complexity: O(N)
Auxiliary Space: O(1)
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