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Count perfect power of K in a range [L, R]
  • Last Updated : 17 May, 2021

Given three integers L, R, and K, the task is to find the count of the number of integers between L to R, which are perfect power of K. That is the number of integers that are in the form of aK, where a can be any number.
Examples: 
 

Input: L = 3, R = 16, K = 3 
Output:
Explanation: 
There is only one integer between 3 to 16 which is in the for aK which is 8.
Input: L = 7, R = 18, K = 2 
Output:
Explanation: 
There are two such numbers that are in the form of aK and are in the range of 7 to 18 which is 9 and 16. 
 

 

Approach: The idea is to find the Kth root of the L and R respectively, where Kth root of a number N is a real number that gives N, when we raise it to integer power N. Then the count of integers which are the power of K in the range L and R can be defined as – 
 

Count = ( floor(Kthroot(R)) - ceil(Kthroot(L)) + 1 )

Kth root of a number N can be calculated using Newton’s Formulae, where ith iteration can be calculated using the below formulae – 
 



x(i + 1) = (1 / K) * ((K – 1) * x(i) + N / x(i) ^ (N – 1)) 
 

Below is the implementation of the above approach:
 

C++




// C++ implementation to find the
// count of numbers those are
// powers of K in range L to R
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the
// Nth root of the number
double nthRoot(int A, int N)
{
    // initially guessing a random
    // number between 0 to 9
    double xPre = rand() % 10;
 
    // Smaller eps,
    // denotes more accuracy
    double eps = 1e-3;
 
    // Initializing difference between two
    // roots by INT_MAX
    double delX = INT_MAX;
 
    // xK denotes
    // current value of x
    double xK;
 
    // loop untill we reach desired accuracy
    while (delX > eps) {
        // calculating current value
        // from previous value
        xK = ((N - 1.0) * xPre +
        (double)A / pow(xPre, N - 1))
                        / (double)N;
        delX = abs(xK - xPre);
        xPre = xK;
    }
    return xK;
}
 
// Function to count the perfect
// powers of K in range L to R
int countPowers(int a, int b, int k)
{
    return (floor(nthRoot(b, k)) -
           ceil(nthRoot(a, k)) + 1);
}
 
// Driver code
int main()
{
    int a = 7, b = 28, k = 2;
    cout << "Count of Powers is "
        << countPowers(a, b, k);
    return 0;
}

Java




// Java implementation to find the
// count of numbers those are
// powers of K in range L to R
class GFG{
  
// Function to find the
// Nth root of the number
static double nthRoot(int A, int N)
{
    // initially guessing a random
    // number between 0 to 9
    double xPre = Math.random()*10 % 10;
  
    // Smaller eps,
    // denotes more accuracy
    double eps = 1e-3;
  
    // Initializing difference between two
    // roots by Integer.MAX_VALUE
    double delX = Integer.MAX_VALUE;
  
    // xK denotes
    // current value of x
    double xK = 0;
  
    // loop untill we reach desired accuracy
    while (delX > eps)
    {
        // calculating current value
        // from previous value
        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 count the perfect
// powers of K in range L to R
static int countPowers(int a, int b, int k)
{
    return (int) (Math.floor(nthRoot(b, k)) -
           Math.ceil(nthRoot(a, k)) + 1);
}
  
// Driver code
public static void main(String[] args)
{
    int a = 7, b = 28, k = 2;
    System.out.print("Count of Powers is "
        + countPowers(a, b, k));
}
}
 
// This code is contributed by 29AjayKumar

Python 3




# Python 3 implementation to find the
# count of numbers those are
# powers of K in range L to R
import sys
from math import pow,ceil,floor
import random
 
# Function to find the
# Nth root of the number
def nthRoot(A,N):
 
    # initially guessing a random
    # number between 0 to 9
    xPre = (random.randint(0, 9))%10
 
    # Smaller eps,
    # denotes more accuracy
    eps = 1e-3
 
    # Initializing difference between two
    # roots by INT_MAX
    delX = sys.maxsize
 
    # xK denotes
    # current value of x
 
    # loo3p untill we reach desired accuracy
    while (delX > eps):
         
        # calculating current value
        # from previous value
        xK = ((N - 1.0) * xPre + A / pow(xPre, N - 1))/ N
        delX = abs(xK - xPre)
        xPre = xK
    return xK
 
# Function to count the perfect
# powers of K in range L to R
def countPowers(a, b, k):
    return (floor(nthRoot(b, k)) - ceil(nthRoot(a, k)) + 1)
 
# Driver code
if __name__ == '__main__':
    a = 7
    b = 28
    k = 2
    print("Count of Powers is",countPowers(a, b, k))
     
# This code is contributed by Surendra_Gangwar

Javascript




<script>
 
// JavaScript implementation to find the
// count of numbers those are
// powers of K in range L to R
 
// Function to find the
// Nth root of the number
function nthRoot(A, N)
{
    // initially guessing a random
    // number between 0 to 9
    var xPre = Math.random() % 10;
 
    // Smaller eps,
    // denotes more accuracy
    var eps = 0.001;
 
    // Initializing difference between two
    // roots by INT_MAX
    var delX = 1000000000;
 
    // xK denotes
    // current value of x
    var xK;
 
    // loop untill we reach desired accuracy
    while (delX > eps) {
        // calculating current value
        // from previous value
        xK = ((N - 1.0) * xPre +
         A / Math.pow(xPre, N - 1))
                        / N;
        delX = Math.abs(xK - xPre);
        xPre = xK;
    }
    return xK;
}
 
// Function to count the perfect
// powers of K in range L to R
function countPowers(a, b, k)
{
    return (Math.floor(nthRoot(b, k)) -
           Math.ceil(nthRoot(a, k)) + 1);
}
 
// Driver code
var a = 7, b = 28, k = 2;
document.write("Count of Powers is "
     + countPowers(a, b, k));
 
</script>
Output
Count of Powers is 3

Performance Analysis: 
 

  • Time Complexity: As in the above approach, there are two function calls for finding Nth root of the number which takes O(logN) time, Hence the Time Complexity will be O(logN).
  • Space Complexity: As in the above approach, there is no extra space used, Hence the space complexity will be O(1).

 

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