# Previous perfect square and cube number smaller than number N

Given an integer N, the task is to find the previous perfect square or perfect cube smaller than the number N.
Examples

Input: N = 6
Output:
Perfect Square = 4
Perfect Cube = 1

Input: N = 30
Output:
Perfect Square = 25
Perfect Cube = 27

Approach: Previous perfect square number less than N can be computed as follows:

• Find the square root of given number N.
• Calculate its floor value using floor function of the respective language.
• Then subtract 1 from it if N is already a perfect square.
• Print square of that number.

Previous perfect cube number less than N can be computed as follows:

• Find the cube root of given N.
• Calculate its floor value using floor function of the respective language.
• Then subtract 1 from it if N is already a perfect cube.
• Print cube of that number.

Below is the implementation of above approach:

## C++

 `// C++ implementation to find the` `// previous perfect square and cube` `// smaller than the given number`   `#include ` `#include `   `using` `namespace` `std;`   `// Function to find the previous` `// perfect square of the number N` `int` `previousPerfectSquare(``int` `N)` `{` `    ``int` `prevN = ``floor``(``sqrt``(N));` `    `  `    ``// If N is already a perfect square` `    ``// decrease prevN by 1.` `    ``if` `(prevN * prevN == N)` `        ``prevN -= 1;`   `    ``return` `prevN * prevN;` `}`   `// Function to find the ` `// previous perfect cube` `int` `previousPerfectCube(``int` `N)` `{` `    ``int` `prevN = ``floor``(cbrt(N));` `    `  `    ``// If N is already a perfect cube` `    ``// decrease prevN by 1.` `    ``if` `(prevN * prevN * prevN == N)` `        ``prevN -= 1;` `        `  `    ``return` `prevN * prevN * prevN;` `}`   `// Driver Code` `int` `main()` `{` `    ``int` `n = 30;` `    ``cout << previousPerfectSquare(n) << ``"\n"``;` `    ``cout << previousPerfectCube(n) << ``"\n"``;` `    ``return` `0;` `}`

## Java

 `// Java implementation to find the` `// previous perfect square and cube` `// smaller than the given number` `import` `java.util.*;`   `class` `GFG{`   `// Function to find the previous` `// perfect square of the number N` `static` `int` `previousPerfectSquare(``int` `N)` `{` `    ``int` `prevN = (``int``)Math.floor(Math.sqrt(N));` `    `  `    ``// If N is already a perfect square` `    ``// decrease prevN by 1.` `    ``if` `(prevN * prevN == N)` `        ``prevN -= ``1``;`   `    ``return` `prevN * prevN;` `}`   `// Function to find the ` `// previous perfect cube` `static` `int` `previousPerfectCube(``int` `N)` `{` `    ``int` `prevN = (``int``)Math.floor(Math.cbrt(N));` `    `  `    ``// If N is already a perfect cube` `    ``// decrease prevN by 1.` `    ``if` `(prevN * prevN * prevN == N)` `        ``prevN -= ``1``;` `        `  `    ``return` `prevN * prevN * prevN;` `}`   `// Driver Code` `public` `static` `void` `main(String[] args)` `{` `    ``int` `n = ``30``;` `    ``System.out.println(previousPerfectSquare(n));` `    ``System.out.println(previousPerfectCube(n));` `}` `}`   `// This code is contributed by Rohit_ranjan`

## Python3

 `# Python3 implementation to find the` `# previous perfect square and cube` `# smaller than the given number` `import` `math` `import` `numpy as np `   `# Function to find the previous` `# perfect square of the number N` `def` `previousPerfectSquare(N):`   `    ``prevN ``=` `math.floor(math.sqrt(N));` `    `  `    ``# If N is already a perfect square` `    ``# decrease prevN by 1.` `    ``if` `(prevN ``*` `prevN ``=``=` `N):` `        ``prevN ``-``=` `1``;`   `    ``return` `prevN ``*` `prevN;`   `# Function to find the ` `# previous perfect cube` `def` `previousPerfectCube(N):`   `    ``prevN ``=` `math.floor(np.cbrt(N));` `    `  `    ``# If N is already a perfect cube` `    ``# decrease prevN by 1.` `    ``if` `(prevN ``*` `prevN ``*` `prevN ``=``=` `N):` `        ``prevN ``-``=` `1``;` `        `  `    ``return` `prevN ``*` `prevN ``*` `prevN;`   `# Driver Code` `n ``=` `30``;`   `print``(previousPerfectSquare(n));` `print``(previousPerfectCube(n));`   `# This code is contributed by Code_Mech`

## C#

 `// C# implementation to find the` `// previous perfect square and cube` `// smaller than the given number` `using` `System;`   `class` `GFG{`   `// Function to find the previous` `// perfect square of the number N` `static` `int` `previousPerfectSquare(``int` `N)` `{` `    ``int` `prevN = (``int``)Math.Floor(Math.Sqrt(N));` `    `  `    ``// If N is already a perfect square` `    ``// decrease prevN by 1.` `    ``if` `(prevN * prevN == N)` `        ``prevN -= 1;`   `    ``return` `prevN * prevN;` `}`   `// Function to find the ` `// previous perfect cube` `static` `int` `previousPerfectCube(``int` `N)` `{` `    ``int` `prevN = (``int``)Math.Floor(Math.Cbrt(N));` `    `  `    ``// If N is already a perfect cube` `    ``// decrease prevN by 1.` `    ``if` `(prevN * prevN * prevN == N)` `        ``prevN -= 1;` `        `  `    ``return` `prevN * prevN * prevN;` `}`   `// Driver Code` `public` `static` `void` `Main(String[] args)` `{` `    ``int` `n = 30;` `    `  `    ``Console.WriteLine(previousPerfectSquare(n));` `    ``Console.WriteLine(previousPerfectCube(n));` `}` `}`   `// This code is contributed by sapnasingh4991`

## Javascript

 ``

Output:

```25
27```

Time Complexity: O(log(n))
Auxiliary Space: O(1)

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