Largest perfect cube number in an Array
Last Updated :
19 Sep, 2022
Given an array of N integers. The task is to find the largest number which is a perfect cube. Print -1 if there is no number that is perfect cube.
Examples:
Input : arr[] = {16, 8, 25, 2, 3, 10}
Output : 25
Explanation: 25 is the largest number
that is a perfect cube.
Input : arr[] = {36, 64, 10, 16, 29, 25|
Output : 64
A Simple Solution is to sort the elements and sort the N numbers and start checking from back for a perfect cube number using cbrt() function. The first number from the end which is a perfect cube number is our answer. The complexity of sorting is O(n log n) and of cbrt() function is log n, so at the worst case the complexity is O(n log n).
An Efficient Solution is to iterate for all the elements in O(n) and compare every time with the maximum element, and store the maximum of all perfect cubes.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
bool checkPerfectcube( int n)
{
int d = cbrt(n);
if (d * d * d == n)
return true ;
return false ;
}
int largestPerfectcubeNumber( int a[], int n)
{
int maxi = -1;
for ( int i = 0; i < n; i++) {
if (checkPerfectcube(a[i]))
maxi = max(a[i], maxi);
}
return maxi;
}
int main()
{
int a[] = { 16, 64, 25, 2, 3, 10 };
int n = sizeof (a) / sizeof (a[0]);
cout << largestPerfectcubeNumber(a, n);
return 0;
}
|
C
#include <stdio.h>
#include <math.h>
#include <stdbool.h>
int max( int a, int b)
{
int max = a;
if (max < b)
max = b;
return max;
}
bool checkPerfectcube( int n)
{
int d = cbrt(n);
if (d * d * d == n)
return true ;
return false ;
}
int largestPerfectcubeNumber( int a[], int n)
{
int maxi = -1;
for ( int i = 0; i < n; i++) {
if (checkPerfectcube(a[i]))
maxi = max(a[i], maxi);
}
return maxi;
}
int main()
{
int a[] = { 16, 64, 25, 2, 3, 10 };
int n = sizeof (a) / sizeof (a[0]);
printf ( "%d" ,largestPerfectcubeNumber(a, n));
return 0;
}
|
Java
class Solution
{
static boolean checkPerfectcube( int n)
{
int d =( int ) Math.cbrt(n);
if (d * d * d == n)
return true ;
return false ;
}
static int largestPerfectcubeNumber( int a[], int n)
{
int maxi = - 1 ;
for ( int i = 0 ; i < n; i++) {
if (checkPerfectcube(a[i]))
maxi = Math.max(a[i], maxi);
}
return maxi;
}
public static void main(String args[])
{
int a[] = { 16 , 64 , 25 , 2 , 3 , 10 };
int n =a.length;
System.out.print(largestPerfectcubeNumber(a, n));
}
}
|
Python3
import math
def checkPerfectcube(n):
cube_root = n * * ( 1. / 3. )
if round (cube_root) * * 3 = = n:
return True
else :
return False
def largestPerfectcubeNumber(a, n):
maxi = - 1
for i in range ( 0 , n, 1 ):
if (checkPerfectcube(a[i])):
maxi = max (a[i], maxi)
return maxi;
if __name__ = = '__main__' :
a = [ 16 , 64 , 25 , 2 , 3 , 10 ]
n = len (a)
print (largestPerfectcubeNumber(a, n))
|
C#
using System;
public class Solution
{
static bool checkPerfectcube( int n)
{
int d = ( int )Math.Ceiling(Math.Pow(n, ( double )1 / 3));
if (d * d * d == n)
return true ;
return false ;
}
static int largestPerfectcubeNumber( int []a, int n)
{
int maxi = -1;
for ( int i = 0; i < n; i++) {
if (checkPerfectcube(a[i]))
maxi = Math.Max(a[i], maxi);
}
return maxi;
}
public static void Main()
{
int []a = { 16, 64, 25, 2, 3, 10 };
int n =a.Length;
Console.WriteLine(largestPerfectcubeNumber(a, n));
}
}
|
PHP
<?php
function checkPerfectcube( $n )
{
$d = pow( $n , (1 / 3));
$d = round ( $d );
if ( $d * $d * $d == $n )
return true;
return false;
}
function largestPerfectcubeNumber(& $a , $n )
{
$maxi = -1;
for ( $i = 0; $i < $n ; $i ++)
{
if (checkPerfectcube( $a [ $i ]))
$maxi = max( $a [ $i ], $maxi );
}
return $maxi ;
}
$a = array ( 16, 64, 25, 2, 3, 10 );
$n = sizeof( $a );
echo largestPerfectcubeNumber( $a , $n );
?>
|
Javascript
<script>
function checkPerfectcube(n)
{
let d = parseInt(Math.cbrt(n));
if (d * d * d == n)
return true ;
return false ;
}
function largestPerfectcubeNumber(a, n)
{
let maxi = -1;
for (let i = 0; i < n; i++) {
if (checkPerfectcube(a[i]))
maxi = Math.max(a[i], maxi);
}
return maxi;
}
let a = [ 16, 64, 25, 2, 3, 10 ];
let n = a.length;
document.write(largestPerfectcubeNumber(a, n));
</script>
|
Time Complexity: O(NlogN)
Auxiliary Space: O(1)
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