Given an array arr[] and an integer K, the task is to calculate the Kth root of the arithmetic mean of the Kth powers of all array elements.
Examples:
Input: arr[] = {1, 2, 3, 4, 5}, K = 2
Output: 3.31662
Explanation:
Sum of all Kth powers of array elements = 1 + 4 + 9 + 16 + 25 = 55
The value of root mean Kth power of array elements = ?(55 / 5) = ?11 = 3.31662Input: arr[] = {10, 4, 6, 8}, K = 3
Output: 7.34847
Approach: The root mean of Kth powers of the array elements is given by the equation:
![\sqrt[k]{\frac{arr_{1}^{k} + arr_{2}^{k} + arr_{2}^{k} + ..... arr_{n}^{k}}{n}}](https://neveropen.co.uk/wp-content/uploads/2023/12/dominic-wardslaus-rubhabha_quicklatex.com-9eb4e31bc1f33c3b21f0e3a8f9d6da16_l3.png "Rendered by QuickLaTeX.com")
Therefore, the idea is to calculate the Kth power of each array element. Then, find the arithmetic mean of those elements. Finally, compute the Kth root of the calculated mean.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to find the Nth rootdouble nthRoot(int A, int N){ // Initially guessing random // numberbetween 0 and 9 double xPre = rand() % 10; // Smaller eps for more accuracy double eps = 1e-3; // Initialize difference between // the two roots by INT_MAX double delX = INT_MAX; // xK denotes current value of x double xK; // Iterate until desired // accuracy is reached while (delX > eps) { // Find the 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 calculate the Root// Mean kth power of array elementsfloat RMNValue(int arr[], int n, int k){ int Nth = 0; float mean = 0.0, root = 0.0; // Calculate sum of kth power for (int i = 0; i < n; i++) { Nth += pow(arr[i], k); } // Calculate Mean mean = (Nth / (float)(n)); // Calculate kth Root of mean root = nthRoot(mean, k); return root;}// Driver Codeint main(){ int arr[] = { 10, 4, 6, 8 }; int N = sizeof(arr) / sizeof(arr[0]); int K = 3; // Function Call cout << RMNValue(arr, N, K); return 0;} |
Java
// Java program for // the above approachclass GFG{// Function to find the // Nth rootstatic double nthRoot(int A, int N){ // Initially guessing random // numberbetween 0 and 9 double xPre = (Math.random() * 10) % 10; // Smaller eps for more accuracy double eps = 1e-3; // Initialize difference between // the two roots by Integer.MAX_VALUE double delX = Integer.MAX_VALUE; // xK denotes current value of x double xK = 0; // Iterate until desired // accuracy is reached while (delX > eps) { // Find the 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 calculate the Root// Mean kth power of array elementsstatic float RMNValue(int arr[], int n, int k){ int Nth = 0; float mean = 0, root = 0; // Calculate sum of kth power for (int i = 0; i < n; i++) { Nth += Math.pow(arr[i], k); } // Calculate Mean mean = (Nth / (float)(n)); // Calculate kth Root of mean root = (float) nthRoot((int)mean, k); return root;}// Driver Codepublic static void main(String[] args){ int arr[] = {10, 4, 6, 8}; int N = arr.length; int K = 3; // Function Call System.out.print(RMNValue(arr, N, K));}}// This code is contributed by Rajput-Ji |
Python3
# Python3 program for # the above approachimport sysimport random# Function to find # the Nth rootdef nthRoot(A, N): # Initially guessing random # numberbetween 0 and 9 xPre = random.random() % 10 # Smaller eps for # more accuracy eps = 1e-3 # Initialize difference between # the two roots by INT_MAX delX = sys.maxsize # xK denotes current # value of x xK = 0 # Iterate until desired # accuracy is reached while (delX > eps): # Find the 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 calculate the Root# Mean kth power of array elementsdef RMNValue(arr, n, k): Nth = 0 mean = 0.0 root = 0.0 # Calculate sum of kth power for i in range (n): Nth += pow(arr[i], k) # Calculate Mean mean = (Nth // (n)) # Calculate kth Root of mean root = nthRoot(mean, k) return root# Driver Codeif __name__ == "__main__": arr = [10, 4, 6, 8 ] N = len(arr) K = 3 # Function Call print ( RMNValue(arr, N, K))# This code is contributed by Chitranayal |
C#
// C# program for the above approach using System;class GFG{ // Function to find the // Nth rootstatic double nthRoot(int A, int N){ // Instantiate random number generator Random rand = new Random(); // Initially guessing random // numberbetween 0 and 9 double xPre = (rand.Next() * 10) % 10; // Smaller eps for more accuracy double eps = 1e-3; // Initialize difference between // the two roots by Integer.MAX_VALUE double delX = Int32.MaxValue; // xK denotes current value of x double xK = 0; // Iterate until desired // accuracy is reached while (delX > eps) { // Find the 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 calculate the Root// Mean kth power of array elementsstatic float RMNValue(int[] arr, int n, int k){ int Nth = 0; float mean = 0, root = 0; // Calculate sum of kth power for(int i = 0; i < n; i++) { Nth += (int)Math.Pow(arr[i], k); } // Calculate Mean mean = (Nth / (float)(n)); // Calculate kth Root of mean root = (float)nthRoot((int)mean, k); return root;} // Driver Codepublic static void Main(){ int[] arr = { 10, 4, 6, 8 }; int N = arr.Length; int K = 3; // Function call Console.Write(RMNValue(arr, N, K));}}// This code is contributed by code_hunt |
Javascript
<script>// Javascript program for the above approach // Function to find the // Nth rootfunction nthRoot(A, N){ // Initially guessing random // numberbetween 0 and 9 var xPre = (Math.random() * 10) % 10; // Smaller eps for more accuracy var eps = 1e-3; // Initialize difference between // the two roots by Integer.MAX_VALUE var delX = Number.MAX_VALUE; // xK denotes current value of x var xK = 0; // Iterate until desired // accuracy is reached while (delX > eps) { // Find the 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 calculate the Root// Mean kth power of array elementsfunction RMNValue(arr, n, k){ var Nth = 0; var mean = 0, root = 0; // Calculate sum of kth power for(var i = 0; i < n; i++) { Nth += Math.pow(arr[i], k); } // Calculate Mean mean = (Nth / (n)); // Calculate kth Root of mean root = nthRoot(mean, k); return root;}// Driver Codevar arr = [ 10, 4, 6, 8 ];var N = arr.length;var K = 3;// Function Calldocument.write(RMNValue(arr, N, K));// This code is contributed by Ankita saini</script> |
Output
7.65172
Time Complexity: O(N*log(sum)), where the sum is the sum of the square of all the given numbers in the array.
Auxiliary Space: O(1)
Method: Algorithmic approach
Steps:
- initialize a variable sum with zero
- Iterate through the array and calculate the Kth power of each element.
- Add the Kth power of each element to the sum.
- Calculate the arithmetic mean of Kth powers by dividing the sum by the length of the array.
- Take the Kth root of the arithmetic mean calculated in step 4 to get the final answer.
Below is the code implementation:
C++
#include <iostream>#include <cmath>using namespace std;int main() { // Initialize the array and K int arr[] = {1, 2, 3, 4, 5}; int K = 2; int n = sizeof(arr)/sizeof(arr[0]); // Initialize the sum to 0 double sum = 0.0; for(int i=0; i<n; i++) { sum += pow(arr[i], K); } double mean = sum/n; double result = pow(mean, 1.0/K); // Output the result cout << result << endl; return 0;} |
Java
import java.util.Arrays;public class MeanPowerRoot { public static void main(String[] args) { // Initialize the array and K int[] arr = {1, 2, 3, 4, 5}; int K = 2; int n = arr.length; // Initialize the sum to 0 double sum = 0.0; // Calculate the sum of Kth powers of array elements for (int i = 0; i < n; i++) { sum += Math.pow(arr[i], K); } // Calculate the mean double mean = sum / n; // Calculate the Kth root of the mean double result = Math.pow(mean, 1.0 / K); // Output the result System.out.println(result); }} |
Python3
import math# Initialize the array and Karr = [1, 2, 3, 4, 5]K = 2n = len(arr)# Initialize the sum to 0sum = 0.0for i in range(n): sum += math.pow(arr[i], K)mean = sum / nresult = math.pow(mean, 1.0 / K)# Output the resultprint(result) |
C#
using System;class Program{ static void Main() { // Initialize the array and K int[] arr = { 1, 2, 3, 4, 5 }; int K = 2; int n = arr.Length; // Initialize the sum to 0 double sum = 0.0; for (int i = 0; i < n; i++) { sum += Math.Pow(arr[i], K); } double mean = sum / n; double result = Math.Pow(mean, 1.0 / K); // Output the result Console.WriteLine(result); }} |
Javascript
// Initialize the array and Kconst arr = [1, 2, 3, 4, 5];const K = 2;const n = arr.length;// Initialize the sum to 0let sum = 0.0;for (let i = 0; i < n; i++) { sum += Math.pow(arr[i], K);}const mean = sum / n;const result = Math.pow(mean, 1.0 / K);// Output the resultconsole.log(result); |
Output
3.31662
Time Complexity: O(n), where n is the number of elements in the input array.
Auxiliary Space: O(1)
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