Check if the String has a Reversible Equal Substring at the Ends

In this problem, we need to find the reversible equal substring of maximum length from the start and end of the string.

The problem is very similar to finding the palindromic string. We can start traversing the string and traverse the string until characters from the start and end match.

Problem statement − We have given string str containing N characters. We need to check whether the string contains the reversible equal substring at the start and end of the string. If we find the substring according to the given condition, print the longest substring. Otherwise, print ‘false’ in the output.

Sample examples

Input

str = "tutorialsotut"

Output

‘tut’

Explanation − The ‘tut’ is present at the start, and the reverse of ‘tut’ is present at the end.

Input

 str = ‘abcytrfghgyuyjcba’

Output

‘abc’

Explanation − In the given string, ‘abc’ is present at the start, and ‘cab’ is present at the end.

Input

 str = ‘vdfdvcf’

Output

False

Explanation − The string doesn’t contain any reversible substring. So, it prints false.

Approach 1

In this approach, we will start traversing the string from the start. We will match the characters of the string from start and end. When we find unmatching characters, we will stop iterations.

Algorithm

Step 1 − Initialize the ‘p’ variable with 0 and the ‘q’ variable with the string length − 1.

Step 2 − Also, initialize the ‘res’ variable with an empty string to store the required string.

Step 3 − Use the loop to make iterations until ‘q’ is greater than or equal to zero.

Step 4 − If s[p] and s[q] characters are equal, append them to the ‘res’ string. Also, increase the p−value by 1 and decrease q by 1.

Step 5 − If s[p] and s[q] are not the same, break the loop using the ‘break’ keyword.

Step 6 − If the size of the ‘res’ string equals zero, print false. Otherwise, print the ‘res’ string.

Example

#include <bits/stdc++.h>
using namespace std;
// Check if the string has reversible equal substring at the start and end of a given string
void ReversibleEqual(string s) {
    // Initialize the length and pointer variables
    int len = s.size();
    int p = 0;
    int q = len - 1;
    string res = "";
    // Iterate over the string
    while (q >= 0) {
        // If characters at index p and q are the same
        if (s[p] == s[q]) {
            // append a character to 'res', and modify p and q value
            res += s[p];
            p++;
            q--;
        } else {
            break;
        }
    }
    // If the substring is not present, print False
    if (res.size() == 0)
        cout << "False";
    // print true
    else {
        cout << "True \n"
             << res;
    }
}
int main() {
    string str = "tutorialsotut";
    ReversibleEqual(str);
    return 0;
}

Output

True 
tuto

Time complexity− O(N) as we use a loop to iterate the string

Space complexity − O(N) as we store the reversible string.

Programmers can optimize the above solution by printing each character directly rather than appending it to the ‘res’ string. The solution approach is very similar to checking whether a string is a palindrome.