# Count square and non-square numbers before n

Given a number n, we need to count square numbers smaller than or equal to n.
Examples :

```Input : n = 5
Output : Square Number : 2
Non-square numbers : 3
Explanation : Square numbers are 1 and 4.
Non square numbers are 2, 3 and 5.

Input : n = 10
Output : Square Number : 3
Non-square numbers : 7
Explanation : Square numbers are 1, 4 and 9.
Non square numbers are 2, 3, 5, 6, 7, 8 and 10.```

A simple solution is to traverse through all numbers from 1 to n and for every number check if n is perfect square or not.
An efficient solution is based on below formula.
Count of square numbers that are greater than 0 and smaller than or equal to n are floor(sqrt(n)) or ??(n)?
Count of non-square numbers = n – ??(n)?

## C++

 `// CPP program to count squares and` `// non-squares before a number.` `#include ` `using` `namespace` `std;`   `void` `countSquaresNonSquares(``int` `n)` `{` `    ``int` `sc = ``floor``(``sqrt``(n));` `    ``cout << ``"Count of squares "` `         ``<< sc << endl;` `    ``cout << ``"Count of non-squares "` `         ``<< n - sc << endl;` `}`   `// Driver Code` `int` `main()` `{` `    ``int` `n = 10;` `    ``countSquaresNonSquares(n);` `    ``return` `0;` `}`

## Java

 `// Java program to count squares and` `// non-squares before a number.` `import` `java.io.*;` `import` `java.math.*;`   `class` `GFG ` `{` `    ``static` `void` `countSquaresNonSquares(``int` `n)` `    ``{` `        ``int` `sc = (``int``)(Math.floor(Math.sqrt(n)));` `        ``System.out.println(``"Count of"` `+ ` `                     ``" squares "` `+ sc);` `        ``System.out.println(``"Count of"` `+ ` `                      ``" non-squares "` `+ ` `                            ``(n - sc) );` `    ``}`   `    ``// Driver code` `    ``public` `static` `void` `main(String args[])` `    ``{` `        ``int` `n = ``10``;` `        ``countSquaresNonSquares(n);` `    ``}` `}`   `// This code is contributed ` `// by Nikita Tiwari.`

## Python3

 `# Python 3 program to count ` `# squares and non-squares` `# before a number.` `import` `math`   `def` `countSquaresNonSquares(n) :` `    ``sc ``=` `(math.floor(math.sqrt(n)))` `    ``print``(``"Count of squares "``, sc)` `    ``print``(``"Count of non-squares "``, (n ``-` `sc) )` `    `  `    `  `# Driver code` `n ``=` `10` `countSquaresNonSquares(n)`   `# This code is contributed` `# by Nikita Tiwari.`

## C#

 `// C# program to count squares and` `// non-squares before a number.` `using` `System;`   `class` `GFG` `{` `static` `void` `countSquaresNonSquares(``int` `n)` `{` `    ``int` `sc = (``int``)Math.Sqrt(n);` `    ``Console.WriteLine( ``"Count of "` `+ ` `                        ``"squares "` `+ ` `                               ``sc) ;` `    ``Console.WriteLine(``"Count of "` `+ ` `                   ``"non-squares "` `+` `                         ``(n - sc));` `}`   `    ``// Driver Code` `    ``static` `public` `void` `Main ()` `    ``{` `    ``int` `n = 10;` `    ``countSquaresNonSquares(n);` `    ``}` `}`   `// This code is contributed by anuj_67.`

## PHP

 ``

## Javascript

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Output :

```Count of squares 3
Count of non-squares 7```

Time Complexity: O(logn)

Auxiliary Space: O(1) as using only constant variables

Approach 2: Using Loops:

Another approach to count the number of squares and non-squares before a given number is to iterate over all the numbers from 1 to n and check if each number is a perfect square or not. If a number is a perfect square, we increment the count of squares, otherwise, we increment the count of non-squares.

Here’s the code implementing this approach:

## C++

 `#include ` `#include `   `void` `countSquaresNonSquares(``int` `n)` `{` `    ``int` `countSquares = 0;` `    ``int` `countNonSquares = 0;` `    `  `    ``for` `(``int` `i = 1; i <= n; i++) {` `        ``int` `sqrtI = std::``sqrt``(i);` `        ``if` `(sqrtI * sqrtI == i) {` `            ``countSquares++;` `        ``} ``else` `{` `            ``countNonSquares++;` `        ``}` `    ``}` `    `  `    ``std::cout << ``"Count of squares "` `<< countSquares << std::endl;` `    ``std::cout << ``"Count of non-squares "` `<< countNonSquares << std::endl;` `}`   `int` `main()` `{` `    ``int` `n = 10;` `    ``countSquaresNonSquares(n);` `    ``return` `0;` `}`

## Java

 `import` `java.util.*;`   `public` `class` `Main {` `    ``public` `static` `void` `countSquaresNonSquares(``int` `n) {` `        ``int` `countSquares = ``0``;` `        ``int` `countNonSquares = ``0``;` `        `  `        ``for` `(``int` `i = ``1``; i <= n; i++) {` `            ``int` `sqrtI = (``int``)Math.sqrt(i);` `            ``if` `(sqrtI * sqrtI == i) {` `                ``countSquares++;` `            ``} ``else` `{` `                ``countNonSquares++;` `            ``}` `        ``}` `        `  `        ``System.out.println(``"Count of squares "` `+ countSquares);` `        ``System.out.println(``"Count of non-squares "` `+ countNonSquares);` `    ``}` `    `  `    ``public` `static` `void` `main(String[] args) {` `        ``int` `n = ``10``;` `        ``countSquaresNonSquares(n);` `    ``}` `}`

## Python3

 `import` `math`   `def` `countSquaresNonSquares(n):` `    ``countSquares ``=` `0` `    ``countNonSquares ``=` `0` `    `  `    ``for` `i ``in` `range``(``1``, n``+``1``):` `        ``sqrtI ``=` `int``(math.sqrt(i))` `        ``if` `sqrtI ``*` `sqrtI ``=``=` `i:` `            ``countSquares ``+``=` `1` `        ``else``:` `            ``countNonSquares ``+``=` `1` `    `  `    ``print``(``"Count of squares"``, countSquares)` `    ``print``(``"Count of non-squares"``, countNonSquares)`   `n ``=` `10` `countSquaresNonSquares(n)`

## C#

 `using` `System;`   `class` `MainClass {` `    ``public` `static` `void` `countSquaresNonSquares(``int` `n) {` `        ``int` `countSquares = 0;` `        ``int` `countNonSquares = 0;`   `        ``for` `(``int` `i = 1; i <= n; i++) {` `            ``int` `sqrtI = (``int``)Math.Sqrt(i);` `            ``if` `(sqrtI * sqrtI == i) {` `                ``countSquares++;` `            ``} ``else` `{` `                ``countNonSquares++;` `            ``}` `        ``}`   `        ``Console.WriteLine(``"Count of squares: "` `+ countSquares);` `        ``Console.WriteLine(``"Count of non-squares: "` `+ countNonSquares);` `    ``}`   `    ``public` `static` `void` `Main() {` `        ``int` `n = 10;` `        ``countSquaresNonSquares(n);` `    ``}` `}`

## Javascript

 `function` `countSquaresNonSquares(n) {` `  ``let countSquares = 0;` `  ``let countNonSquares = 0;`   `  ``for` `(let i = 1; i <= n; i++) {` `    ``let sqrtI = Math.sqrt(i);` `    ``if` `(sqrtI * sqrtI === i) {` `      ``countSquares++;` `    ``} ``else` `{` `        ``countNonSquares++;` `    ``}` `  ``}`   `  ``console.log(``"Count of squares "` `+ countSquares);` `  ``console.log(``"Count of non-squares "` `+ countNonSquares);` `}`   `let n = 10;`   `// function Call  ` `countSquaresNonSquares(n);`     `// This code is contributed by shivhack999`

Output :

```Count of squares 3
Count of non-squares 7```

Time Complexity:  O(nsqrt(n)), where n is the input variable, as described in the problem statement

Auxiliary Space: O(1) as using only constant variables

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