# Check if string is right to left diagonal or not

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.

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  `

Output:

```Yes
```

Time complexity: O(N)

Auxiliary space: O(1)

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