Perfect cubes in a range
Given two given numbers a and b where 1 <= a <= b, find perfect cubes between a and b (a and b inclusive).
Examples:
Input : a = 1, b = 100 Output : 1 8 27 64 Perfect cubes in the given range are 1, 8, 27, 64 Input : a = 24, b = 576 Output : 27 64 125 216 343 512 Perfect cubes in the given range are 27, 64, 125, 216, 343, 512
This problem is similar to Perfect squares between two numbers.
Method 1 (Naive) : One naive approach is to check all the numbers between a and b (inclusive a and b)
and print the perfect cube. Following is the code for the above approach:
C++
// A Simple Method to count cubes between a and b #include <bits/stdc++.h> using namespace std; void printCubes( int a, int b) { // Traverse through all numbers in given range // and one by one check if number is prime for ( int i = a; i <= b; i++) { // Check if current number 'i' // is perfect cube for ( int j = 1; j * j * j <= i; j++) { if (j * j * j == i) { cout << j * j * j << " " ; break ; } } } } // Driver code int main() { int a = 1, b = 100; cout << "Perfect cubes in given range:\n " ; printCubes(a, b); return 0; } |
Java
// A Simple Method to count cubes between a and b class Test { static void printCubes( int a, int b) { // Traverse through all numbers in given range // and one by one check if number is prime for ( int i = a; i <= b; i++) { // Check if current number 'i' // is perfect cube for ( int j = 1 ; j * j * j <= i; j++) { if (j * j * j == i) { System.out.print(j * j * j + " " ); break ; } } } } // Driver method public static void main(String[] args) { int a = 1 , b = 100 ; System.out.println( "Perfect cubes in given range:" ); printCubes(a, b); } } |
Python3
# A Simple Method to count cubes between a and b def printCubes(a, b) : # Traverse through all numbers in given range # and one by one check if number is prime for i in range (a, b + 1 ) : # Check if current number 'i' # is perfect cube j = 1 for j in range (j * * 3 , i + 1 ) : if (j * * 3 = = i) : print ( j * * 3 , end = " " ) break # Driver code a = 1 ; b = 100 print ( "Perfect cubes in given range: " ) printCubes(a, b) # This code is contributed by Nikita Tiwari. |
C#
// A Simple Method to count cubes // between a and b using System; class GFG { static void printCubes( int a, int b) { // Traverse through all numbers // in given range and one by // one check if number is prime for ( int i = a; i <= b; i++) { // Check if current number 'i' // is perfect cube for ( int j = 1; j * j * j <= i; j++) { if (j * j * j == i) { Console.Write(j * j * j + " " ); break ; } } } } // Driver method public static void Main() { int a = 1, b = 100; Console.WriteLine( "Perfect cubes in" + " given range:" ); printCubes(a, b); } } // This code contribute by parashar. |
PHP
<?php // A Simple Method to count // cubes between a and b function printCubes( $a , $b ) { // Traverse through all // numbers in given range // and one by one check // if number is prime for ( $i = $a ; $i <= $b ; $i ++) { // Check if current number 'i' // is perfect cube for ( $j = 1; $j * $j * $j <= $i ; $j ++) { if ( $j * $j * $j == $i ) { echo $j * $j * $j , " " ; break ; } } } } // Driver Code $a = 1; $b = 100; echo "Perfect cubes in given range:\n " ; printCubes( $a , $b ); // This code is contributed by ajit ?> |
Javascript
<script> // A Simple Method to count // cubes between a and b function printCubes(a, b) { // Traverse through all // numbers in given range // and one by one check // if number is prime for (let i = a; i <= b; i++) { // Check if current number 'i' // is perfect cube for (let j = 1; j * j * j <= i; j++) { if (j * j * j == i) { document.write(j * j * j + " " ); break ; } } } } // Driver Code let a = 1; let b = 100; document.write( "Perfect cubes in given range: <br> " ); printCubes(a, b); // This code is contributed by gfgking. </script> |
Output :
Perfect cubes in given range: 1 8 27 64
Method 2 (Efficient):
We can simply take cube root of ‘a’ and cube root of ‘b’ and print the cubes of number between them.
1- Given a = 24 b = 576 2- acr = cbrt(a)) bcr = cbrt(b) acr = 3 and bcr = 8 3- Print cubes of 3 to 8 that comes under the range of a and b(including a and b both) 27, 64, 125, 216, 343, 512
Below is implementation of above steps.
C++
// Efficient method to print cubes // between a and b #include <cmath> #include <iostream> using namespace std; // An efficient solution to print perfect // cubes between a and b void printCubes( int a, int b) { // Find cube root of both a and b int acrt = cbrt(a); int bcrt = cbrt(b); // Print cubes between acrt and bcrt for ( int i = acrt; i <= bcrt; i++) if (i * i * i >= a && i * i * i <= b) cout << i * i * i << " " ; } // Driver code int main() { int a = 24, b = 576; cout << "Perfect cubes in given range:\n" ; printCubes(a, b); return 0; } // improved by prophet1999 |
Java
// Java progroam for Efficient method // to print cubes between a and b class Test { // An efficient solution to print perfect // cubes between a and b static void printCubes( int a, int b) { // Find cube root of both a and b int acrt = ( int )Math.cbrt(a); int bcrt = ( int )Math.cbrt(b); // Print cubes between acrt and bcrt for ( int i = acrt; i <= bcrt; i++) if (i * i * i >= a && i * i * i <= b) System.out.print(i * i * i + " " ); } // Driver method public static void main(String[] args) { int a = 24 , b = 576 ; System.out.println( "Perfect cubes in given range:" ); printCubes(a, b); } } |
Python3
# Python3 code for Efficient method # to print cubes between a and b def cbrt(n) : return ( int )( n * * ( 1. / 3 )) # An efficient solution to print # perfect cubes between a and b def printCubes(a, b) : # Find cube root of # both a and b acrt = cbrt(a) bcrt = cbrt(b) # Print cubes between acrt and bcrt for i in range (acrt, bcrt + 1 ) : if (i * i * i > = a and i * i * i < = b) : print (i * i * i, " " , end = "") # Driver code a = 24 b = 576 print ( "Perfect cubes in given range:" ) printCubes(a, b) # This code is contributed # by Nikita Tiwari. |
C#
// C# progroam for Efficient // method to print cubes // between a and b using System; class GFG { // An efficient solution // to print perfect // cubes between a and b static void printCubes( int a, int b) { // Find cube root of // both a and b int acrt = ( int )Math.Pow(a, ( double )1 / 3); int bcrt = ( int )Math.Pow(b, ( double )1 / 3); // Print cubes between // acrt and bcrt for ( int i = acrt; i <= bcrt; i++) if (i * i * i >= a && i * i * i <= b) Console.Write(i * i * i + " " ); } // Driver Code static public void Main () { int a = 24; int b = 576; Console.WriteLine( "Perfect cubes " + "in given range:" ); printCubes(a, b); } } // This code is contributed // by ajit |
PHP
<?php // Efficient method to print // cubes between a and b // An efficient solution // to print perfect // cubes between a and b function printCubes( $a , $b ) { // Find cube root // of both a and b $acrt = (int)pow( $a , 1 / 3); $bcrt = (int)pow( $b , 1 / 3); // Print cubes between // acrt and bcrt for ( $i = $acrt ; $i <= $bcrt ; $i ++) if ( $i * $i * $i >= $a && $i * $i * $i <= $b ) echo $i * $i * $i , " " ; } // Driver code $a = 24; $b = 576; echo "Perfect cubes in given range:\n" , printCubes( $a , $b ); // This code is contributed by ajit ?> |
Javascript
<script> // Javascript progroam for Efficient // method to print cubes // between a and b // An efficient solution // to print perfect // cubes between a and b function printCubes(a, b) { // Find cube root of // both a and b let acrt = parseInt(Math.pow(a, 1 / 3), 10); let bcrt = parseInt(Math.pow(b, 1 / 3), 10); // Print cubes between // acrt and bcrt for (let i = acrt; i <= bcrt; i++) if (i * i * i >= a && i * i * i <= b) document.write((i * i * i) + " " ); } let a = 24; let b = 576; document.write( "Perfect cubes " + "in given range:" + "</br>" ); printCubes(a, b); // This code is contributed by rameshtravel07. </script> |
Output:
Perfect cubes in given range: 27 64 125 216 343 512
This article is contributed by Sahil Chhabra and improved by prophet1999.