Given a number
N
, Find the minimum number that needs to be added to or subtracted from
N
, to make it a
. If the number is to be added, print it with a + sign, else if the number is to be subtracted, print it with a – sign.
Examples:
Input: N = 25 Output: 2 Nearest perfect cube before 25 = 8 Nearest perfect cube after 25 = 27 Therefore 2 needs to be added to 25 to get the closest perfect cube Input: N = 40 Output: -13 Nearest perfect cube before 40 = 25 Nearest perfect cube after 40 = 64 Therefore 13 needs to be subtracted from 40 to get the closest perfect cube
Approach
:
- Get the number.
- Find the cube root of the number and convert the result as an integer.
- After converting the double value to integer, this will contain the root of the perfect cube before N, i.e. floor(cube root(N)).
- Then find the cube of this number, which will be the perfect cube before N.
- Find the root of the perfect cube after N, i.e. the ceil(cube root(N)).
- Then find the cube of this number, which will be the perfect cube after N.
- Check whether the cube of floor value is nearest to N or the ceil value.
- If the cube of floor value is nearest to N, print the difference with a -sign. Else print the difference between the cube of the ceil value and N with a + sign.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to return the Least numberint nearest(int n){ // Get the perfect cube // before and after N int prevCube = cbrt(n); int nextCube = prevCube + 1; prevCube = prevCube * prevCube * prevCube; nextCube = nextCube * nextCube * nextCube; // Check which is nearest to N int ans = (n - prevCube) < (nextCube - n) ? (prevCube - n) : (nextCube - n); // return the result return ans;}// Driver codeint main(){ int n = 25; cout << nearest(n) << endl; n = 27; cout << nearest(n) << endl; n = 40; cout << nearest(n) << endl; return 0;} |
Java
// Java implementation of the approach class GFG { // Function to return the Least number static int nearest(int n) { // Get the perfect cube // before and after N int prevCube = (int)Math.cbrt(n); int nextCube = prevCube + 1; prevCube = prevCube * prevCube * prevCube; nextCube = nextCube * nextCube * nextCube; // Check which is nearest to N int ans = (n - prevCube) < (nextCube - n) ? (prevCube - n) : (nextCube - n); // return the result return ans; } // Driver code public static void main (String[] args) { int n = 25; System.out.println(nearest(n)); n = 27; System.out.println(nearest(n)) ; n = 40; System.out.println(nearest(n)) ; } }// This code is contributed by Yash_R |
Python3
# Python3 implementation of the approachimport math# Function to return the least numberdef nearest(n): # Get the perfect cube before and after N prev_cube = int(n ** (1/3)) next_cube = prev_cube + 1 prev_cube = prev_cube ** 3 next_cube = next_cube ** 3 # Check which is nearest to N ans = prev_cube - n if n - prev_cube < next_cube - n else next_cube - n # Return the result return ans# Driver codeif __name__ == "__main__": n = 25 print(nearest(n)) n = 27 print(nearest(n)) n = 40 print(nearest(n))# by phasing17 |
C#
using System;class GFG { // Function to return the least number static int Nearest(int n) { // Get the perfect cube before and after N int prevCube = (int)(Math.Pow(n, 1.0 / 3.0)); int nextCube = prevCube + 1; prevCube = prevCube * prevCube * prevCube; nextCube = nextCube * nextCube * nextCube; // Check which is nearest to N int ans = (n - prevCube) < (nextCube - n) ? (prevCube - n) : (nextCube - n); // return the result return ans; } // Driver code public static void Main(string[] args) { int n = 25; Console.WriteLine(Nearest(n)); n = 27; Console.WriteLine(Nearest(n)); n = 40; Console.WriteLine(Nearest(n)); }} |
Javascript
// Javascript implementation for the approachfunction nearest(n) { // Get the perfect cube before and after N const prevCube = Math.floor(Math.cbrt(n)); const nextCube = prevCube + 1; const prevCubeValue = prevCube * prevCube * prevCube; const nextCubeValue = nextCube * nextCube * nextCube; // Check which is nearest to N const ans = Math.abs(n - prevCubeValue) < Math.abs(nextCubeValue - n) ? (prevCubeValue - n) : (nextCubeValue - n); // Return the result return ans;}// Driver codeconst n1 = 25;console.log(nearest(n1));const n2 = 27;console.log(nearest(n2));const n3 = 40;console.log(nearest(n3));// This code is contributed by Taranpreet Singh. |
Output
2 0 -13
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