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 = 1Input: 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 <cmath> #include <iostream> 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
<script> // JavaScript implementation to find the // previous perfect square and cube // smaller than the given number // Function to find the previous // perfect square of the number N function previousPerfectSquare(N)
{ let 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 function previousPerfectCube(N)
{ let prevN = 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 let n = 30;
document.write(previousPerfectSquare(n) + "<br>" );
document.write(previousPerfectCube(n) + "<br>" );
// This code is contributed by Manoj. </script> |
Output:
25 27
Time Complexity: O(log(n))
Auxiliary Space: O(1)