Elements of Array which can be expressed as power of some integer to given exponent K

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Given an array arr[] of size N, and an integer K, the task is to print all the elements of the Array which can be expressed as a power of some integer (X) to the exponent K, i.e. X^K.
Examples:

Input: arr[] = {46656, 64, 256, 729, 16, 1000}, K = 6
Output: 46656 64 729
Explanation:
Only numbers 46656, 64, 729 can be expressed as a power of 6.
46656 = 6⁶,
64 = 2⁶,
729 = 3⁶
Input: arr[] = {23, 81, 256, 125, 16, 1000}, K = 4
Output: 81 256 16
Explanation:
The number 81, 256, 16 can be expressed as a power of 4.

Approach: To solve the problem mentioned above the main idea is to for each number in the Array, find the N-th root of a number. Then check whether this number is an integer or not. If yes, then print it, else skip to the next number.
Below is the implementation of the above approach:

CPP

// C++ implementation to print elements of
// the Array which can be expressed as
// power of some integer to given exponent K
 
#include <bits/stdc++.h>
using namespace std;
#define ll long long
 
// Method returns Nth power of A
double nthRoot(ll A, ll N)
{
 
    double xPre = 7;
 
    // Smaller eps, denotes more accuracy
    double eps = 1e-3;
 
    // Initializing difference between two
    // roots by INT_MAX
    double delX = INT_MAX;
 
    // x^K denotes current value of x
    double xK;
 
    // loop until we reach desired accuracy
    while (delX > eps) {
 
        // calculating current value from previous
        // value by newton's method
        xK = ((N - 1.0) * xPre
              + (double)A / pow(xPre, N - 1))
             / (double)N;
 
        delX = abs(xK - xPre);
        xPre = xK;
    }
 
    return xK;
}
 
// Function to check
// whether its k root
// is an integer or not
bool check(ll no, int k)
{
    double kth_root = nthRoot(no, k);
    ll num = kth_root;
 
    if (abs(num - kth_root) < 1e-4)
        return true;
 
    return false;
}
 
// Function to find the numbers
void printExpo(ll arr[], int n, int k)
{
    for (int i = 0; i < n; i++) {
        if (check(arr[i], k))
            cout << arr[i] << " ";
    }
}
 
// Driver code
int main()
{
 
    int K = 6;
 
    ll arr[] = { 46656, 64, 256,
                 729, 16, 1000 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    printExpo(arr, n, K);
 
    return 0;
}

Java

// Java implementation to print elements of
// the Array which can be expressed as
// power of some integer to given exponent K
 
class GFG{
  
// Method returns Nth power of A
static double nthRoot(long A, long N)
{
  
    double xPre = 7;
  
    // Smaller eps, denotes more accuracy
    double eps = 1e-3;
  
    // Initializing difference between two
    // roots by Integer.MAX_VALUE
    double delX = Integer.MAX_VALUE;
  
    // x^K denotes current value of x
    double xK = 0;
  
    // loop until we reach desired accuracy
    while (delX > eps) {
  
        // calculating current value from previous
        // value by newton's method
        xK = ((N - 1.0) * xPre
              + (double)A / Math.pow(xPre, N - 1))
             / (double)N;
  
        delX = Math.abs(xK - xPre);
        xPre = xK;
    }
  
    return xK;
}
  
// Function to check
// whether its k root
// is an integer or not
static boolean check(long no, int k)
{
    double kth_root = nthRoot(no, k);
    long num = (long) kth_root;
  
    if (Math.abs(num - kth_root) < 1e-4)
        return true;
  
    return false;
}
  
// Function to find the numbers
static void printExpo(long arr[], int n, int k)
{
    for (int i = 0; i < n; i++) {
        if (check(arr[i], k))
            System.out.print(arr[i]+ " ");
    }
}
  
// Driver code
public static void main(String[] args)
{
  
    int K = 6;
  
    long arr[] = { 46656, 64, 256,
                 729, 16, 1000 };
    int n = arr.length;
  
    printExpo(arr, n, K);
  
}
}
 
// This code is contributed by sapnasingh4991

Python3

# Python3 implementation to print elements of
# the Array which can be expressed as
# power of some integer to given exponent K
 
# Method returns Nth power of A
def nthRoot(A, N):
 
    xPre = 7
 
    # Smaller eps, denotes more accuracy
    eps = 1e-3
 
    # Initializing difference between two
    # roots by INT_MAX
    delX = 10**9
 
    # x^K denotes current value of x
    xK = 0
 
    # loop until we reach desired accuracy
    while (delX > eps):
 
        # calculating current value from previous
        # value by newton's method
        xK = ((N - 1.0) * xPre+ A /pow(xPre, N - 1))/ N
 
        delX = abs(xK - xPre)
        xPre = xK
 
    return xK
 
# Function to check
# whether its k root
# is an integer or not
def check(no, k):
    kth_root = nthRoot(no, k)
    num = int(kth_root)
 
    if (abs(num - kth_root) < 1e-4):
        return True
 
    return False
 
# Function to find the numbers
def printExpo(arr, n, k):
    for i in range(n):
        if (check(arr[i], k)):
            print(arr[i],end=" ")
 
# Driver code
if __name__ == '__main__':
 
    K = 6
 
    arr = [46656, 64, 256,729, 16, 1000]
    n = len(arr)
 
    printExpo(arr, n, K)
 
# This code is contributed by mohit kumar 29

C#

// C# implementation to print elements of
// the Array which can be expressed as
// power of some integer to given exponent K
using System;
 
class GFG{
   
// Method returns Nth power of A
static double nthRoot(long A, long N)
{
   
    double xPre = 7;
   
    // Smaller eps, denotes more accuracy
    double eps = 1e-3;
   
    // Initializing difference between two
    // roots by int.MaxValue
    double delX = int.MaxValue;
   
    // x^K denotes current value of x
    double xK = 0;
   
    // loop until we reach desired accuracy
    while (delX > eps) {
   
        // calculating current value from previous
        // value by newton's method
        xK = ((N - 1.0) * xPre
              + (double)A / Math.Pow(xPre, N - 1))
             / (double)N;
   
        delX = Math.Abs(xK - xPre);
        xPre = xK;
    }
   
    return xK;
}
   
// Function to check
// whether its k root
// is an integer or not
static bool check(long no, int k)
{
    double kth_root = nthRoot(no, k);
    long num = (long) kth_root;
   
    if (Math.Abs(num - kth_root) < 1e-4)
        return true;
   
    return false;
}
   
// Function to find the numbers
static void printExpo(long []arr, int n, int k)
{
    for (int i = 0; i < n; i++) {
        if (check(arr[i], k))
            Console.Write(arr[i]+ " ");
    }
}
   
// Driver code
public static void Main(String[] args)
{
   
    int K = 6;
   
    long []arr = { 46656, 64, 256,
                 729, 16, 1000 };
    int n = arr.Length;
   
    printExpo(arr, n, K);
   
}
}
 
// This code is contributed by Princi Singh

Javascript

<script>
// Javascript implementation to print elements of
// the Array which can be expressed as
// power of some integer to given exponent K
   
 
// Method returns Nth power of A
function nthRoot(A,N)
{
    let xPre = 7;
    
    // Smaller eps, denotes more accuracy
    let eps = 1e-3;
    
    // Initializing difference between two
    // roots by Integer.MAX_VALUE
    let delX = Number.MAX_VALUE;
    
    // x^K denotes current value of x
    let xK = 0;
    
    // loop until we reach desired accuracy
    while (delX > eps) {
    
        // calculating current value from previous
        // value by newton's method
        xK = ((N - 1.0) * xPre
              + A / Math.pow(xPre, N - 1))
             / N;
    
        delX = Math.abs(xK - xPre);
        xPre = xK;
    }
    
    return xK;
}
 
// Function to check
// whether its k root
// is an integer or not
function check(no,k)
{
    let kth_root = nthRoot(no, k);
    let num =  Math.floor(kth_root);
    
    if (Math.abs(num - kth_root) < 1e-4)
        return true;
    
    return false;
}
 
// Function to find the numbers
function printExpo(arr,n,k)
{
    for (let i = 0; i < n; i++) {
        if (check(arr[i], k))
            document.write(arr[i]+ " ");
    }
}
 
// Driver code
let K = 6;
let arr=[46656, 64, 256,
                 729, 16, 1000];
let n = arr.length;
 printExpo(arr, n, K);
 
 
// This code is contributed by patel2127
</script>

Output

46656 64 729

Another Approach:

The code defines a function “isPower” to check if a number is equal to a specific power of a base.
The code defines a function “printPowerElements” to print elements in a vector that can be expressed as a specific power of a base.
The “printPowerElements” function iterates through each element in the input vector.
For each element, it checks if the element can be expressed as a power of a base value ranging from 2 to the square root of the element.
It uses the “isPower” function to perform the check.
If a matching power is found, the element is printed.
In the “main” function, an example vector “arr” and a specific power “K” are declared.
The “printPowerElements” function is called with “arr” and “K”.
The program terminates after the function execution.

Below is the implementation of the above approach:

C++

#include <iostream>
#include <vector>
#include <cmath>
 
// Function to check if a number is equal to a specific power of a base
bool isPower(int num, int base, int power) {
    int result = pow(base, power);  // Calculate the power of the base
    return result == num;  // Check if the calculated result is equal to the number
}
 
// Function to print elements in the vector that are a specific power of a base
void printPowerElements(const std::vector<int>& arr, int K) {
    for (int i = 0; i < arr.size(); i++) {
        bool found = false;  // Flag to track if a matching power is found
        for (int j = 2; j <= sqrt(arr[i]); j++) {
            // Check if the number in the array can be expressed as a power of 'j' with power 'K'
            if (isPower(arr[i], j, K)) {
                found = true;  // Found a matching power, set the flag to true
                break;  // No need to check for other 'j' values, exit the loop
            }
        }
        if (found)
            std::cout << arr[i] << " ";  // Print the number if a matching power is found
    }
}
 
int main() {
    std::vector<int> arr = {46656, 64, 256, 729, 16, 1000};
    int K = 6;
      
    printPowerElements(arr, K);
     
    return 0;
}

Java

import java.util.ArrayList;
import java.util.List;
 
class Main {
    // Function to check if a number is equal to a specific 
  // power of a base
    static boolean isPower(int num, int base, int power) {
        int result = (int) Math.pow(base, power);  // Calculate the power of the base
        return result == num;  // Check if the calculated result is equal to the number
    }
 
    // Function to print elements in the list that are a specific power of a base
    static void printPowerElements(List<Integer> arr, int K) {
        for (int num : arr) {
            boolean found = false;  // Flag to track if a matching power is found
            for (int j = 2; j <= Math.sqrt(num); j++) {
                // Check if the number in the list can be expressed as a power of 'j' with power 'K'
                if (isPower(num, j, K)) {
                    found = true;  // Found a matching power, set the flag to true
                    break;  // No need to check for other 'j' values, exit the loop
                }
            }
            if (found) {
                System.out.print(num + " ");  // Print the number if a matching power is found
            }
        }
    }
 
    public static void main(String[] args) {
        List<Integer> arr = new ArrayList<>();
        arr.add(46656);
        arr.add(64);
        arr.add(256);
        arr.add(729);
        arr.add(16);
        arr.add(1000);
        int K = 6;
 
        printPowerElements(arr, K);
 
        //This Code Is Contributed By Shubham Tiwari.
       
    }
}

Python3

import math
 
 
# Function to check if a number is equal to a specific power of a base
def is_power(num, base, power):
    result = pow(base, power)  # Calculate the power of the base
    return result == num  # Check if the calculated result is equal to the number
 
 
# Function to print elements in the list that are a specific power of a base
def print_power_elements(arr, K):
    for i in range(len(arr)):
        found = False  # Flag to track if a matching power is found
        for j in range(2, int(math.sqrt(arr[i])) + 1):
            # Check if the number in the list can be expressed as a power of 'j' with power 'K'
            if is_power(arr[i], j, K):
                found = True  # Found a matching power, set the flag to True
                break  # No need to check for other 'j' values, exit the loop
        if found:
            print(arr[i], end=" ")  # Print the number if a matching power is found
 
 
if __name__ == "__main__":
    arr = [46656, 64, 256, 729, 16, 1000]
    K = 6
 
    print_power_elements(arr, K)
    # This code is contributed by Shubham Tiwari.

C#

using System;
using System.Collections.Generic;
 
class GFG
{
    // Function to check if a number is equal to a specific power of a base
    static bool IsPower(int num, int baseNum, int power)
    {
        int result = (int)Math.Pow(baseNum, power);  // Calculate the power of the base
        return result == num;  // Check if the calculated result is equal to the number
    }
 
    // Function to print elements in the list that are a specific power of a base
    static void PrintPowerElements(List<int> arr, int K)
    {
        foreach (int num in arr)
        {
            bool found = false;  // Flag to track if a matching power is found
            for (int j = 2; j <= Math.Sqrt(num); j++)
            {
                // Check if the number in the list can be expressed as a power of 'j' with power 'K'
                if (IsPower(num, j, K))
                {
                    found = true;  // Found a matching power, set the flag to true
                    break;  // No need to check for other 'j' values, exit the loop
                }
            }
            if (found)
                Console.Write(num + " ");  // Print the number if a matching power is found
        }
    }
 
    static void Main(string[] args)
    {
        List<int> arr = new List<int> { 46656, 64, 256, 729, 16, 1000 };
        int K = 6;
 
        PrintPowerElements(arr, K);
         
    }
}
//This code is Contributed By Shubham Tiwari  

Javascript

// Function to check if a number is equal to a specific power of a base
function isPower(num, base, power) {
    let result = Math.pow(base, power); // Calculate the power of the base
    return result === num; // Check if the calculated result is equal to the number
}
 
// Function to print elements in the array that are a specific power of a base
function printPowerElements(arr, K) {
    for (let i = 0; i < arr.length; i++) {
        let found = false; // Flag to track if a matching power is found
        for (let j = 2; j <= Math.sqrt(arr[i]); j++) {
            // Check if the number in the array can be expressed as a power of 'j' with power 'K'
            if (isPower(arr[i], j, K)) {
                found = true; // Found a matching power, set the flag to true
                break; // No need to check for other 'j' values, exit the loop
            }
        }
        if (found) {
            console.log(arr[i] + " "); // Print the number if a matching power is found
        }
    }
}
 
const arr = [46656, 64, 256, 729, 16, 1000];
const K = 6;
 
printPowerElements(arr, K);
 
// This Code Is Contributed By Shubham Tiwari

Output

46656 64 729

Time Complexity: O(n * sqrt(arr[i]))

Auxiliary Space: O(N)

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