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Count perfect power of K in a range [L, R]
• Last Updated : 08 Apr, 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 ``using` `namespace` `std;` `// Function to find the``// Nth root of the number``double` `nthRoot(``int` `A, ``int` `N)``{``    ``// intially 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)``{``    ``// intially 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):` `    ``# intially 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

 ``
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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