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 = 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)
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