Given two integers N and K, the task is to check whether N can be made a perfect cube after adding to or subtracting from K to or from it.
Examples:
Input: N = 7, K = 1
Output: Yes
7 + 1 = 8 which is a perfect cube (23 = 8)Input: N = 5, K = 4
Output: Yes
5 – 4 = 1 which is a perfect cube (13 = 1)
Approach: The simplest way to solve this problem is to check whether either (N + K) or (N – K) is a perfect cube or not.
- Check whether (N + K) is a perfect cube or not
- If not, then check whether (N – K) is a perfect cube or not.
- If both are not perfect cube, then print “No”, else print “Yes”.
- In order to check whether a number is a perfect cube or not, the easiest way is to find the cube of the floor value of cube root of the number, and then check whether this cube is same as the number or not.
if(N3 == (floor(?N))3) Then N is a perfect cube
Below is the implementation of the above approach:
C++
// C++ implementation of the above approach #include <bits/stdc++.h> using namespace std; // Function to check if a number is // a perfect Cube or not bool isPerfectCube( int x) { int cr = round(cbrt(x)); return (cr * cr * cr == x); } void canBePerfectCube( int N, int K) { if (isPerfectCube(N + K) || isPerfectCube(N - K)) cout << "Yes\n" ; else cout << "No\n" ; } // Driver code int main() { int N = 7, K = 1; canBePerfectCube(N, K); N = 5, K = 4; canBePerfectCube(N, K); N = 7, K = 2; canBePerfectCube(N, K); return 0; } |
Java
// Java implementation of the above approach class GFG { // Function to check if a number is // a perfect Cube or not static boolean isPerfectCube( int x) { int cr = ( int )Math.cbrt(x); return (cr * cr * cr == x); } static void canBePerfectCube( int N, int K) { if (isPerfectCube(N + K) || isPerfectCube(N - K) == true ) System.out.println( "Yes" ); else System.out.println( "No" ); } // Driver code public static void main (String[] args) { int N = 7 ; int K = 1 ; canBePerfectCube(N, K); N = 5 ; K = 4 ; canBePerfectCube(N, K); N = 7 ; K = 2 ; canBePerfectCube(N, K); } } // This code is contributed by Yash_R |
Python3
# Python3 implementation of the above approach # Function to check if a number is # a perfect Cube or not def isPerfectCube(x) : cr = int (x * * ( 1 / 3 )); return (cr * cr * cr = = x); def canBePerfectCube(N, K) : if (isPerfectCube(N + K) or isPerfectCube(N - K)) : print ( "Yes" ); else : print ( "No" ); # Driver code if __name__ = = "__main__" : N = 7 ; K = 1 ; canBePerfectCube(N, K); N = 5 ; K = 4 ; canBePerfectCube(N, K); N = 7 ; K = 2 ; canBePerfectCube(N, K); # This code is contributed by Yash_R |
C#
// C# implementation of the above approach using System; class GFG { // Function to check if a number is // a perfect Cube or not static bool isPerfectCube( int x) { int cr = ( int )Math.Cbrt(x); return (cr * cr * cr == x); } static void canBePerfectCube( int N, int K) { if (isPerfectCube(N + K) || isPerfectCube(N - K) == true ) Console.WriteLine( "Yes" ); else Console.WriteLine( "No" ); } // Driver code public static void Main ( string [] args) { int N = 7; int K = 1; canBePerfectCube(N, K); N = 5; K = 4; canBePerfectCube(N, K); N = 7; K = 2; canBePerfectCube(N, K); } } // This code is contributed by AnkitRai01 |
Javascript
<script> // Javascript implementation of the above approach // Function to check if a number is // a perfect Cube or not function isPerfectCube(x) { var cr = Math.round(Math.cbrt(x)); return (cr * cr * cr == x); } function canBePerfectCube(N, K) { if (isPerfectCube(N + K) || isPerfectCube(N - K)) document.write( "Yes<br>" ); else document.write( "No<br>" ); } // Driver code var N = 7, K = 1; canBePerfectCube(N, K); N = 5, K = 4; canBePerfectCube(N, K); N = 7, K = 2; canBePerfectCube(N, K); // This code is contributed by rrrtnx. </script> |
Yes Yes No
Time Complexity : O(log(N+K)) because it is using cbrt function
Auxiliary Space: O(1)