Given an array arr[] of n integers. The task is to find the smallest perfect cube from the array. Print -1 if there is no perfect cube in the array.
Examples:
Input: arr[] = {16, 8, 25, 2, 3, 10}
Output: 8
8 is the only perfect cube in the arrayInput: arr[] = {27, 8, 1, 64}
Output: 1
All elements are perfect cubes but 1 is the minimum of all.
A simple solution is to sort the elements and sort the numbers and start checking from start for a perfect cube number using cbrt() function. The first number from the beginning 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 in 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 minimum element and store the minimum of all perfect cubes.
Below is the implementation of the above approach:
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std;
// Function that returns true // if n is a perfect cube bool checkPerfectcube( int n)
{ // Takes the sqrt of the number
int d = cbrt(n);
// Checks if it is a perfect
// cube number
if (d * d * d == n)
return true ;
return false ;
} // Function to return the smallest perfect // cube from the array int smallestPerfectCube( int a[], int n)
{ // Stores the minimum of all the
// perfect cubes from the array
int mini = INT_MAX;
// Traverse all elements in the array
for ( int i = 0; i < n; i++) {
// Store the minimum if current
// element is a perfect cube
if (checkPerfectcube(a[i])) {
mini = min(a[i], mini);
}
}
return mini;
} // Driver code int main()
{ int a[] = { 16, 8, 25, 2, 3, 10 };
int n = sizeof (a) / sizeof (a[0]);
cout << smallestPerfectCube(a, n);
return 0;
} |
// Java implementation of the approach import java.io.*;
class GFG {
// Function that returns true
// if n is a perfect cube
static boolean checkPerfectcube( int n)
{
// Takes the sqrt of the number
int d = ( int )Math.cbrt(n);
// Checks if it is a perfect
// cube number
if (d * d * d == n)
return true ;
return false ;
}
// Function to return the smallest perfect
// cube from the array
static int smallestPerfectCube( int a[], int n)
{
// Stores the minimum of all the
// perfect cubes from the array
int mini = Integer.MAX_VALUE;
// Traverse all elements in the array
for ( int i = 0 ; i < n; i++) {
// Store the minimum if current
// element is a perfect cube
if (checkPerfectcube(a[i])) {
mini = Math.min(a[i], mini);
}
}
return mini;
}
// Driver code
public static void main(String[] args)
{
int a[] = { 16 , 8 , 25 , 2 , 3 , 10 };
int n = a.length;
System.out.print(smallestPerfectCube(a, n));
}
} // This code is contributed by anuj_67.. |
# Python3 implementation of the approach import sys
# Function that returns true # if n is a perfect cube def checkPerfectcube(n):
# Takes the sqrt of the number
d = int (n * * ( 1 / 3 ))
# Checks if it is a perfect
# cube number
if (d * d * d = = n):
return True
return False
# Function to return the smallest perfect # cube from the array def smallestPerfectCube(a, n):
# Stores the minimum of all the
# perfect cubes from the array
mini = sys.maxsize
# Traverse all elements in the array
for i in range (n):
# Store the minimum if current
# element is a perfect cube
if (checkPerfectcube(a[i])):
mini = min (a[i], mini)
return mini
# Driver code if __name__ = = "__main__" :
a = [ 16 , 8 , 25 , 2 , 3 , 10 ]
n = len (a)
print (smallestPerfectCube(a, n))
# This code is contributed by AnkitRai01 |
// C# implementation of the approach using System;
class GFG {
// Function that returns true
// if n is a perfect cube
static bool checkPerfectcube( int n)
{
// Takes the sqrt of the number
int d = ( int )Math.Sqrt(n);
// Checks if it is a perfect
// cube number
if (d * d * d == n)
return true ;
return false ;
}
// Function to return the smallest perfect
// cube from the array
static int smallestPerfectCube( int [] a, int n)
{
// Stores the minimum of all the
// perfect cubes from the array
int mini = int .MaxValue;
// Traverse all elements in the array
for ( int i = 0; i < n; i++) {
// Store the minimum if current
// element is a perfect cube
if (checkPerfectcube(a[i])) {
mini = Math.Min(a[i], mini);
}
}
return mini;
}
// Driver code
static public void Main()
{
int [] a = { 16, 8, 25, 2, 3, 10 };
int n = a.Length;
Console.Write(smallestPerfectCube(a, n));
}
} // This code is contributed by ajit.. |
<script> // Javascript implementation of the approach // Function that returns true // if n is a perfect cube function checkPerfectcube(n)
{ // Takes the sqrt of the number
let d = parseInt(Math.cbrt(n));
// Checks if it is a perfect
// cube number
if (d * d * d == n)
return true ;
return false ;
} // Function to return the smallest perfect // cube from the array function smallestPerfectCube(a, n)
{ // Stores the minimum of all the
// perfect cubes from the array
let mini = Number.MAX_VALUE;
// Traverse all elements in the array
for (let i = 0; i < n; i++) {
// Store the minimum if current
// element is a perfect cube
if (checkPerfectcube(a[i])) {
mini = Math.min(a[i], mini);
}
}
return mini;
} // Driver code let a = [ 16, 8, 25, 2, 3, 10 ]; let n = a.length; document.write(smallestPerfectCube(a, n)); </script> |
8
Time Complexity: O(n log n) because the inbuilt cbrt function is used
Auxiliary Space: O(1)