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. XK.
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 Adouble 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 notbool 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 numbersvoid printExpo(ll arr[], int n, int k){ for (int i = 0; i < n; i++) { if (check(arr[i], k)) cout << arr[i] << " "; }}// Driver codeint 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 Kclass GFG{ // Method returns Nth power of Astatic 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 notstatic 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 numbersstatic 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 codepublic 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 Adef 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 notdef 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 numbersdef printExpo(arr, n, k): for i in range(n): if (check(arr[i], k)): print(arr[i],end=" ")# Driver codeif __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 Kusing System;class GFG{ // Method returns Nth power of Astatic 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 notstatic 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 numbersstatic 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 codepublic 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 Afunction 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 notfunction 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 numbersfunction printExpo(arr,n,k){ for (let i = 0; i < n; i++) { if (check(arr[i], k)) document.write(arr[i]+ " "); }}// Driver codelet 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 basebool 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 basevoid 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 basedef 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 basedef 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 foundif __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 basefunction 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 basefunction 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)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
