Given a string S consisting of N characters, the task is to check if this string has a reversible equal substring from the start and the end. If yes, print True and then the longest substring present following the given conditions, otherwise print False.
Example:
Input: S = “abca”
Output:
True
a
Explanation:
The substring “a” is only the longest substring that satisfy the given criteria. Therefore, print a.Input: S = “acdfbcdca”
Output:
True
acd
Explanation:
The substring “acd” is only the longest substring that satisfy the given criteria. Therefore, print acd.Input: S = “abcdcb”
Output: False
Approach: The given problem can be solved by using the Two Pointer Approach. Following the steps below to solve the given problem:
- Initialize a string, say ans as “” that stores the resultant string satisfying the given criteria.
- Initialize two variables, say i and j as 0 and (N – 1) respectively.
- Iterate a loop until j is non-negative and f the characters S[i] and S[j] are the same, then just add the character S[i] in the variable ans and increment the value of i by 1 and decrement the value of j by 1. Otherwise, break the loop.
- After completing the above steps, if the string ans is empty then print False. Otherwise, print True and then print the string ans as the result.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to print longest substring// that appears at beginning of string// and also at end in reverse ordervoid commonSubstring(string s){ int n = s.size(); int i = 0; int j = n - 1; // Stores the resultant string string ans = ""; while (j >= 0) { // If the characters are same if (s[i] == s[j]) { ans += s[i]; i++; j--; } // Otherwise, break else { break; } } // If the string can't be formed if (ans.size() == 0) cout << "False"; // Otherwise print resultant string else { cout << "True \n" << ans; }}// Driver Codeint main(){ string S = "abca"; commonSubstring(S); return 0;} |
Java
// Java program for the above approachpublic class GFG { // Function to print longest substring // that appears at beginning of string // and also at end in reverse order static void commonSubstring(String s) { int n = s.length(); int i = 0; int j = n - 1; // Stores the resultant string String ans = ""; while (j >= 0) { // If the characters are same if (s.charAt(i) == s.charAt(j)) { ans += s.charAt(i); i++; j--; } // Otherwise, break else { break; } } // If the string can't be formed if (ans.length() == 0) System.out.println("False"); // Otherwise print resultant string else { System.out.println("True "); System.out.println(ans); } } // Driver Code public static void main(String []args) { String S = "abca"; commonSubstring(S); }}// This code is contributed by AnkThon |
Python3
# python program for the above approach# Function to print longest substring# that appears at beginning of string# and also at end in reverse orderdef commonSubstring(s): n = len(s) i = 0 j = n - 1 # Stores the resultant string ans = "" while (j >= 0): # // If the characters are same if (s[i] == s[j]): ans += s[i] i = i + 1 j = j - 1 # Otherwise, break else: break # If the string can't be formed if (len(ans) == 0): print("False") # Otherwise print resultant string else: print("True") print(ans)# Driver Codeif __name__ == "__main__": S = "abca" commonSubstring(S) # This code is contributed by rakeshsahni |
C#
// C# program for the above approachusing System;class GFG { // Function to print longest substring // that appears at beginning of string // and also at end in reverse order static void commonSubstring(string s) { int n = s.Length; int i = 0; int j = n - 1; // Stores the resultant string string ans = ""; while (j >= 0) { // If the characters are same if (s[i] == s[j]) { ans += s[i]; i++; j--; } // Otherwise, break else { break; } } // If the string can't be formed if (ans.Length == 0) Console.WriteLine("False"); // Otherwise print resultant string else { Console.WriteLine("True "); Console.WriteLine(ans); } } // Driver Code public static void Main() { string S = "abca"; commonSubstring(S); }}// This code is contributed by ukasp. |
Javascript
<script> // JavaScript Program to implement // the above approach // Function to print longest substring // that appears at beginning of string // and also at end in reverse order function commonSubstring(s) { let n = s.length; let i = 0; let j = n - 1; // Stores the resultant string let ans = ""; while (j >= 0) { // If the characters are same if (s[i] == s[j]) { ans += s[i]; i++; j--; } // Otherwise, break else { break; } } // If the string can't be formed if (ans.length == 0) document.write("False"); // Otherwise print resultant string else { document.write("True" + "<br>" + ans); } } // Driver Code let S = "abca"; commonSubstring(S); // This code is contributed by Potta Lokesh </script> |
Output
a
Time Complexity: O(N)
Auxiliary Space: O(N) because using extra space for string ans
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