# Length of longest Powerful number subsequence in an Array

Given an array arr[] containing non-negative integers of length N, the task is to print the length of the longest subsequence of Powerful numbers in the array.

A number n is said to be Powerful Number if, for every prime factor p of it, p2 also divides it.

Examples:

Input: arr[] = { 3, 4, 11, 2, 9, 21 }
Output: 2
Explanation:
Longest Powerful number Subsequence is {4, 9} and hence the answer is 2.

Input: arr[] = { 6, 4, 10, 13, 9, 25 }
Output: 3
Explanation:
Longest Powerful number Subsequence is {4, 9, 25} and hence the answer is 3.

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

Approach: To solve the problem mentioned above, we have to follow the steps given below:

• Traverse the given array and for each element in the array, check if it is Powerful number or not.
• If the element is a Powerful number, it will be in Longest Powerful number Subsequence.
• Hence increment the required length of Longest Powerful number Subsequence by 1
• Return the final length after reaching the size of the array.

Below is the implementation of the above approach:

## C++

 `// C++ program to find the length of ` `// Longest Powerful Subsequence in an Array ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Function to check if the number is powerful ` `bool` `isPowerful(``int` `n) ` `{ ` `    ``// First divide the number repeatedly by 2 ` `    ``while` `(n % 2 == 0) { ` `        ``int` `power = 0; ` `        ``while` `(n % 2 == 0) { ` `            ``n /= 2; ` `            ``power++; ` `        ``} ` ` `  `        ``// Check if only 2^1 divides n, ` `        ``// then return false ` `        ``if` `(power == 1) ` `            ``return` `false``; ` `    ``} ` ` `  `    ``// Check if n is not a power of 2 ` `    ``// then this loop will execute ` `    ``// repeat above process ` `    ``for` `(``int` `factor = 3; ` `         ``factor <= ``sqrt``(n); ` `         ``factor += 2) { ` ` `  `        ``// Find highest power of "factor" ` `        ``// that divides n ` `        ``int` `power = 0; ` `        ``while` `(n % factor == 0) { ` `            ``n = n / factor; ` `            ``power++; ` `        ``} ` ` `  `        ``// If only factor^1 divides n, ` `        ``// then return false ` `        ``if` `(power == 1) ` `            ``return` `false``; ` `    ``} ` ` `  `    ``// n must be 1 now ` `    ``// if it is not a prime number. ` `    ``// Since prime numbers ` `    ``// are not powerful, we return ` `    ``// false if n is not 1. ` `    ``return` `(n == 1); ` `} ` ` `  `// Function to find the longest subsequence ` `// which contain all powerful numbers ` `int` `longestPowerfulSubsequence( ` `    ``int` `arr[], ``int` `n) ` `{ ` `    ``int` `answer = 0; ` ` `  `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``if` `(isPowerful(arr[i])) ` `            ``answer++; ` `    ``} ` ` `  `    ``return` `answer; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 6, 4, 10, 13, 9, 25 }; ` ` `  `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` ` `  `    ``cout << longestPowerfulSubsequence(arr, n) ` `         ``<< endl; ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java program to find the length of ` `// Longest Powerful Subsequence in an Array ` `class` `GFG{ ` ` `  `// Function to check if the number is powerful ` `static` `boolean` `isPowerful(``int` `n) ` `{ ` ` `  `    ``// First divide the number repeatedly by 2 ` `    ``while` `(n % ``2` `== ``0``) ` `    ``{ ` `        ``int` `power = ``0``; ` `        ``while` `(n % ``2` `== ``0``)  ` `        ``{ ` `            ``n /= ``2``; ` `            ``power++; ` `        ``} ` ` `  `        ``// Check if only 2^1 divides n, ` `        ``// then return false ` `        ``if` `(power == ``1``) ` `            ``return` `false``; ` `    ``} ` ` `  `    ``// Check if n is not a power of 2 ` `    ``// then this loop will execute ` `    ``// repeat above process ` `    ``for``(``int` `factor = ``3``;  ` `            ``factor <= Math.sqrt(n);  ` `            ``factor += ``2``) ` `    ``{ ` ` `  `       ``// Find highest power of "factor" ` `       ``// that divides n ` `       ``int` `power = ``0``; ` `       ``while` `(n % factor == ``0``) ` `       ``{ ` `           ``n = n / factor; ` `           ``power++; ` `       ``} ` `        `  `       ``// If only factor^1 divides n, ` `       ``// then return false ` `       ``if` `(power == ``1``) ` `       ``return` `false``; ` `         `  `    ``} ` ` `  `    ``// n must be 1 now ` `    ``// if it is not a prime number. ` `    ``// Since prime numbers ` `    ``// are not powerful, we return ` `    ``// false if n is not 1. ` `    ``return` `(n == ``1``); ` `} ` ` `  `// Function to find the longest subsequence ` `// which contain all powerful numbers ` `static` `int` `longestPowerfulSubsequence(``int` `arr[], ` `                                      ``int` `n) ` `{ ` `    ``int` `answer = ``0``; ` ` `  `    ``for``(``int` `i = ``0``; i < n; i++) ` `    ``{ ` `       ``if` `(isPowerful(arr[i])) ` `       ``answer++; ` `    ``} ` `     `  `    ``return` `answer; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `arr[] = { ``6``, ``4``, ``10``, ``13``, ``9``, ``25` `}; ` `    ``int` `n = arr.length; ` ` `  `    ``System.out.print(longestPowerfulSubsequence(arr, ` `                                                ``n) + ``"\n"``); ` `} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

## C#

 `// C# program to find the length of ` `// longest Powerful Subsequence in ` `// an array ` `using` `System; ` `class` `GFG{ ` ` `  `// Function to check if the ` `// number is powerful ` `static` `bool` `isPowerful(``int` `n) ` `{ ` ` `  `    ``// First divide the number ` `    ``// repeatedly by 2 ` `    ``while` `(n % 2 == 0) ` `    ``{ ` `        ``int` `power = 0; ` `        ``while` `(n % 2 == 0)  ` `        ``{ ` `            ``n /= 2; ` `            ``power++; ` `        ``} ` ` `  `        ``// Check if only 2^1 divides  ` `        ``// n, then return false ` `        ``if` `(power == 1) ` `            ``return` `false``; ` `    ``} ` ` `  `    ``// Check if n is not a power of 2 ` `    ``// then this loop will execute ` `    ``// repeat above process ` `    ``for``(``int` `factor = 3;  ` `            ``factor <= Math.Sqrt(n);  ` `            ``factor += 2) ` `    ``{ ` `        `  `       ``// Find highest power of "factor" ` `       ``// that divides n ` `       ``int` `power = 0; ` `       ``while` `(n % factor == 0) ` `       ``{ ` `           ``n = n / factor; ` `           ``power++; ` `       ``} ` `        `  `       ``// If only factor^1 divides n, ` `       ``// then return false ` `       ``if` `(power == 1) ` `           ``return` `false``; ` `    ``} ` `     `  `    ``// n must be 1 now ` `    ``// if it is not a prime number. ` `    ``// Since prime numbers ` `    ``// are not powerful, we return ` `    ``// false if n is not 1. ` `    ``return` `(n == 1); ` `} ` ` `  `// Function to find the longest subsequence ` `// which contain all powerful numbers ` `static` `int` `longestPowerfulSubsequence(``int` `[]arr, ` `                                      ``int` `n) ` `{ ` `    ``int` `answer = 0; ` ` `  `    ``for``(``int` `i = 0; i < n; i++) ` `    ``{ ` `       ``if` `(isPowerful(arr[i])) ` `           ``answer++; ` `    ``} ` `    ``return` `answer; ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``int` `[]arr = { 6, 4, 10, 13, 9, 25 }; ` `    ``int` `n = arr.Length; ` ` `  `    ``Console.Write(longestPowerfulSubsequence(arr, ` `                                             ``n) + ``"\n"``); ` `} ` `} ` ` `  `// This code is contributed by gauravrajput1 `

Output:

```3
```

Time Complexity: O(N*√N)

Auxiliary Space Complexity: O(1) My Personal Notes arrow_drop_up Check out this Author's contributed articles.

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Improved By : 29AjayKumar, GauravRajput1