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 Nint 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 cubeint 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 Codeint 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 numberimport java.util.*;class GFG{// Function to find the previous// perfect square of the number Nstatic 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 cubestatic 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 Codepublic 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 numberimport mathimport numpy as np # Function to find the previous# perfect square of the number Ndef 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 cubedef 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 Coden = 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 numberusing System;class GFG{// Function to find the previous// perfect square of the number Nstatic 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 cubestatic 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 Codepublic 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 Nfunction 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 cubefunction 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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