# Count perfect power of K in a range [L, R]

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: 1
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: 2
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.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 `

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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