FIRSTLAST-Editorial

editorial
1

PROBLEM LINK:

Practice
Contest

Author, Tester and Editorialist: Jitin

DIFFICULTY:

EASY

PREREQUISITES:

None

PROBLEM:

You are given a string Y and you need to check whether there is a valid string X such that
we get the same string on shifting all characters in X one place left and appending the first character to as well as shifting all characters in X one place right and adding the last character at the beginning.

EXPLANATION:

If there is a valid X then first assume Y is the string obtained by left-shifting X. Hence original string X is obtained by right shifting Y. Similarly, another original string is obtained by left-shifting Y. If both are equal then X is valid but if they are not equal, then there doesn’t exist any valid X.

SOLUTIONS:

C++ Solution 
#include <iostream>
using namespace std;

string first_to_last(string x)
{
	int n = x.length();
	char first = x[0];
	for (int i = 1; i < n; i++)
	{
		x[i - 1] = x[i];
	}
	x[n - 1] = first;
	return x;
}

string last_to_first(string x)
{
	int n = x.length();
	char last = x[n - 1];
	for (int i = n - 1; i > 0; i--)
	{
		x[i] = x[i - 1];
	}
	x[0] = last;
	return x;
}

void solve()
{
	int t;
	cin >> t;
	while (t--)
	{
		string y;
		cin >> y;
		// If there is a valid X then first assume Y is the left-shifted.
		// Hence original string is obtained by right shifting it.
		// Similarly another original string is obtained by left-shifting it
		// If both are equal then X is valid.
		string l = first_to_last(y);
		string r = last_to_first(y);
		if (l == r)
		{
			cout << "YES";
		}
		else
		{
			cout << "NO";
		}
	}
}

int main()
{
	solve();
	return 0;
}