# Check whether the number formed by concatenating two numbers is a perfect square or not

• Last Updated : 21 Sep, 2022

Given two numbers a and b and the task is to check whether the concatenation of a and b is a perfect square or not.
Examples:

Input: a = 1, b = 21
Output: Yes
121 = 11 × 11, is a perfect square.

Input: a = 100, b = 100
Output: No
100100 is not a perfect square.

Approach: Initialize the number as strings initially and concatenate them. Convert the string to a number using Integer.valueOf() function. Once the string has been converted to a number, check if the number is a perfect square or not.
Below is the implementation of the above approach.

## C++

 `// C++ program to check if the``// concatenation of two numbers``// is a perfect square or not``#include ``using` `namespace` `std;` `// Function to check if``// the concatenation is``// a perfect square``void` `checkSquare(string s1, string s2)``{` `    ``// Function to convert``    ``// concatenation of``    ``// strings to a number``    ``int` `c = stoi(s1 + s2);` `    ``// square root of number``    ``int` `d = ``sqrt``(c);` `    ``// check if it is a``    ``// perfect square``    ``if` `(d * d == c)``    ``{``        ``cout << ``"Yes"``;``    ``}``    ``else``    ``{``        ``cout << ``"No"``;``    ``}``}` `// Driver Code``int` `main()``{``    ``string s1 = ``"12"``;``    ``string s2 = ``"1"``;` `    ``checkSquare(s1, s2);``    ` `    ``return` `0;``}`

## Java

 `// Java program to check if the``// concatenation of two numbers``// is a perfect square or not``import` `java.lang.*;``class` `GFG {` `    ``// Function to check if the concatenation is``    ``// a perfect square``    ``static` `void` `checkSquare(String s1, String s2)``    ``{` `        ``// Function to convert concatenation``        ``// of strings to a number``        ``int` `c = Integer.valueOf(s1 + s2);` `        ``// square root of number``        ``int` `d = (``int``)Math.sqrt(c);` `        ``// check if it is a perfect square``        ``if` `(d * d == c) {``            ``System.out.println(``"Yes"``);``        ``}``        ``else` `{``            ``System.out.println(``"No"``);``        ``}``    ``}` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``String s1 = ``"12"``;``        ``String s2 = ``"1"``;` `        ``checkSquare(s1, s2);``    ``}``}`

## Python 3

 `# Python 3 program to check if the``# concatenation of two numbers``# is a perfect square or not``import` `math` `# Function to check if the concatenation``# is a perfect square``def` `checkSquare(s1, s2):` `    ``# Function to convert concatenation of``    ``# strings to a number``    ``c ``=` `int``(s1 ``+` `s2)` `    ``# square root of number``    ``d ``=` `math.sqrt(c)` `    ``# check if it is a perfect square``    ``if` `(d ``*` `d ``=``=` `c) :``        ``print``(``"Yes"``)``    ` `    ``else``:``        ``print``(``"No"``)` `# Driver Code``if` `__name__ ``=``=` `"__main__"``:``    ` `    ``s1 ``=` `"12"``    ``s2 ``=` `"1"` `    ``checkSquare(s1, s2)` `# This code is contributed by ita_c`

## C#

 `// C# program to check if the``// concatenation of two numbers``// is a perfect square or not``using` `System;``public` `class` `GFG {` `    ``// Function to check if the concatenation is``    ``// a perfect square``    ``static` `void` `checkSquare(String s1, String s2)``    ``{` `        ``// Function to convert concatenation``        ``// of strings to a number``        ``int` `c = Convert.ToInt32(s1 + s2 );``//int.ValueOf(s1 + s2);` `        ``// square root of number``        ``int` `d = (``int``)Math.Sqrt(c);` `        ``// check if it is a perfect square``        ``if` `(d * d == c) {``            ``Console.WriteLine(``"Yes"``);``        ``}``        ``else` `{``            ``Console.WriteLine(``"No"``);``        ``}``    ``}` `    ``// Driver Code``    ``public` `static` `void` `Main()``    ``{``        ``String s1 = ``"12"``;``        ``String s2 = ``"1"``;` `        ``checkSquare(s1, s2);``    ``}``}` `// This code is contributed by PrinciRaj1992`

## PHP

 ``

## Javascript

 ``

Output:

`Yes`

Time complexity: log(c) as using inbuilt sqrt function
Auxiliary space: O(1)

My Personal Notes arrow_drop_up