# Check if string is right to left diagonal or not

• Difficulty Level : Basic
• Last Updated : 15 Jun, 2021

Given string str of perfect square length. The task is to check whether the given string is a right to left diagonal or not. If it is a right to left diagonal then print “Yes” else print “No”.

Let the string be “abcdefghi”. It can be broken as:
“abc”
“def”
“ghi”
if the character c, e, and g are equal then the given string is a right to left diagonal otherwise not.

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It means, first break the string into a square box and check if right to left diagonal’s all character are the same or not. If it is the same then print “Yes”, otherwise print “No”.

Examples:

Input: str = “abcxabxcaxbcxabc”
Output: Yes
Explanation: Break the string in square box, see below
abcx
abxc
axbc
xabc
So, right ot left diagonal have same character.

Input: str=”abcdxabcxdabxcdaxbcdaxbcd”
Output: No
Explanation: Break the string in square box, see below
abcdx
abcxd
abxcd
axbcd
axbcd
So, right to left diagonal haven’t same character.

Approach: Follow the steps given below to solve the problem

• Calculate the length of the string.
• Check whether the length perfect square of any number or not.
• If not perfect square then, Print No
• Else proceed below steps
• Let length is the perfect square of k
• Check the indexes k – 1, 2k – 1, 3k – 1…and so on.
• If character at all the indexes is same then
• Print YES
• Else
• Print NO

Below is the implementation of the above approach:

## C++

 `// C++ program to Check if the``// given string is right to``// left diagonal or not``#include ``using` `namespace` `std;` `// Function to check if the given``// string is right to left diagonal or not``int` `is_rtol(string s)``{``    ``int` `tmp = ``sqrt``(s.length()) - 1;` `    ``char` `first = s[tmp];` `    ``// Iterate over string``    ``for` `(``int` `pos = tmp;``         ``pos < s.length() - 1; pos += tmp) {` `        ``// If character is not same as``        ``// the first character then``        ``// return false``        ``if` `(s[pos] != first) {``            ``return` `false``;``        ``}``    ``}` `    ``return` `true``;``}` `// Driver Code``int` `main()``{``    ``// Given String str``    ``string str = ``"abcxabxcaxbcxabc"``;` `    ``// Function Call``    ``if` `(is_rtol(str)) {``        ``cout << ``"Yes"` `<< endl;``    ``}``    ``else` `{``        ``cout << ``"No"` `<< endl;``    ``}` `    ``return` `0;``}`

## Java

 `// Java program to check if the``// given string is right to``// left diagonal or not``import` `java.io.*;` `class` `GFG{``    ` `// Function to check if the given``// string is right to left diagonal or not``public` `static` `boolean` `is_rtol(String s)``{``    ``int` `tmp = (``int``)(Math.sqrt(s.length())) - ``1``;``    ``char` `first = s.charAt(tmp);``    ` `    ``// Iterate over string``    ``for``(``int` `pos = tmp; pos < s.length() - ``1``;``            ``pos += tmp)``    ``{``        ` `        ``// If character is not same as``        ``// the first character then``        ``// return false``        ``if` `(s.charAt(pos) != first)``        ``{``            ``return` `false``;``        ``}``    ``}``    ``return` `true``;``}` `// Driver Code``public` `static` `void` `main(String args[])``{``    ` `    ``// Given String str``    ``String str = ``"abcxabxcaxbcxabc"``;``    ` `    ``// Function call``    ``if` `(is_rtol(str))``    ``{``        ``System.out.print(``"Yes"``);``    ``}``    ``else``    ``{``        ``System.out.print(``"No"``);``    ``}``}``}` `// This code is contributed by grand_master`

## Python3

 `# Python3 program to Check if the``# given is right to``# left diagonal or not``from` `math ``import` `sqrt, floor, ceil` `# Function to check if the given``# is right to left diagonal or not``def` `is_rtol(s):` `    ``tmp ``=` `floor(sqrt(``len``(s))) ``-` `1` `    ``first ``=` `s[tmp]` `    ``# Iterate over string``    ``for` `pos ``in` `range``(tmp, ``len``(s) ``-` `1``, tmp):` `        ``# If character is not same as``        ``# the first character then``        ``# return false``        ``if` `(s[pos] !``=` `first):``            ``return` `False` `    ``return` `True` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ` `    ``# Given String str``    ``str` `=` `"abcxabxcaxbcxabc"` `    ``# Function Call``    ``if` `(is_rtol(``str``)):``        ``print``(``"Yes"``)``    ``else``:``        ``print``(``"No"``)``        ` `# This code is contributed by Mohit Kumar`

## C#

 `// C# program to check if the``// given string is right to``// left diagonal or not``using` `System;` `class` `GFG{``    ` `// Function to check if the given``// string is right to left diagonal or not``public` `static` `bool` `is_rtol(String s)``{``    ``int` `tmp = (``int``)(Math.Sqrt(s.Length)) - 1;``    ``char` `first = s[tmp];``    ` `    ``// Iterate over string``    ``for``(``int` `pos = tmp; pos < s.Length - 1;``            ``pos += tmp)``    ``{``        ` `        ``// If character is not same as``        ``// the first character then``        ``// return false``        ``if` `(s[pos] != first)``        ``{``            ``return` `false``;``        ``}``    ``}``    ``return` `true``;``}` `// Driver Code``public` `static` `void` `Main(String []args)``{``    ` `    ``// Given String str``    ``String str = ``"abcxabxcaxbcxabc"``;``    ` `    ``// Function call``    ``if` `(is_rtol(str))``    ``{``        ``Console.Write(``"Yes"``);``    ``}``    ``else``    ``{``        ``Console.Write(``"No"``);``    ``}``}``}` `// This code is contributed by amal kumar choubey`

## Javascript

 ``
Output:
`Yes`

Time complexity: O(N)

Auxiliary space: O(1)

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