Find the previous perfect square and perfect cube of N

Given a number N, the task is to find the previous perfect square and perfect cube smaller than N. Examples:

Examples :

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

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

Approach :

1. Find the square root of given N.
2. Calculate its floor value using floor function in C++.
3. Then subtract 1 from it if N is already a perfect square.
4. Print square of that number.
5. Find the cube root of given N.
6. Calculate its floor value using floor function in C++.
7. Then subtract 1 from it if N is already a perfect cube.
8. Print cube of that number.

Below is the implementation of above approach:

// C++ implementation of above approach
#include <iostream>
#include<cmath>
using namespace std;

// Function to find the previous perfect square
int previousPerfectSquare(int N)
{
    int prevN = floor(sqrt(N)) ;
    //If N is alreay 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;
}

25
27