# Length of Longest Perfect 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 the **Perfect number** in the array.

A number is a perfect number if it is equal to the sum of its proper divisors, that is, the sum of its positive divisors excluding the number itself.

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the

DSA Self Paced Courseat a student-friendly price and become industry ready. To complete your preparation from learning a language to DS Algo and many more, please referComplete Interview Preparation Course.In case you wish to attend

live classeswith experts, please referDSA Live Classes for Working ProfessionalsandCompetitive Programming Live for Students.

**Examples:**

Input:arr[] = { 3, 6, 11, 2, 28, 21, 8128 }Output:3Explanation:

The longest perfect number subsequence is {6, 28, 8128} and hence the answer is 3.

Input:arr[] = { 6, 4, 10, 13, 9, 25 }Output:1Explanation:

The longest perfect number subsequence is {6} and hence the answer is 1.

**Approach:**

To solve the problem mentioned above, follow the steps given below:

- Traverse the given array and for each element in the array, check if it is a perfect number or not.
- If the element is a perfect number, it will be in the Longest Perfect number Subsequence. Hence, increment the required length of the Longest Perfect number Subsequence by 1

Below is the implementation of the above approach:

## C++

`// C++ program to find the length of` `// Longest Perfect number Subsequence in an Array` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to check if` `// the number is a Perfect number` `bool` `isPerfect(` `long` `long` `int` `n)` `{` ` ` `// To store sum of divisors` ` ` `long` `long` `int` `sum = 1;` ` ` `// Find all divisors and add them` ` ` `for` `(` `long` `long` `int` `i = 2; i * i <= n; i++) {` ` ` `if` `(n % i == 0) {` ` ` `if` `(i * i != n)` ` ` `sum = sum + i + n / i;` ` ` `else` ` ` `sum = sum + i;` ` ` `}` ` ` `}` ` ` `// Check if sum of divisors is equal to` ` ` `// n, then n is a perfect number` ` ` `if` `(sum == n && n != 1)` ` ` `return` `true` `;` ` ` `return` `false` `;` `}` `// Function to find the longest subsequence` `// which contain all Perfect numbers` `int` `longestPerfectSubsequence(` `int` `arr[], ` `int` `n)` `{` ` ` `int` `answer = 0;` ` ` `// Find the length of longest` ` ` `// Perfect number subsequence` ` ` `for` `(` `int` `i = 0; i < n; i++) {` ` ` `if` `(isPerfect(arr[i]))` ` ` `answer++;` ` ` `}` ` ` `return` `answer;` `}` `// Driver code` `int` `main()` `{` ` ` `int` `arr[] = { 3, 6, 11, 2, 28, 21, 8128 };` ` ` `int` `n = ` `sizeof` `(arr) / ` `sizeof` `(arr[0]);` ` ` `cout << longestPerfectSubsequence(arr, n) << endl;` ` ` `return` `0;` `}` |

## Java

`// Java program to find the length of` `// longest perfect number subsequence` `// in an array` `class` `GFG {` ` ` `// Function to check if the` `// number is a perfect number` `static` `boolean` `isPerfect(` `long` `n)` `{` ` ` ` ` `// To store sum of divisors` ` ` `long` `sum = ` `1` `;` ` ` ` ` `// Find all divisors and add them` ` ` `for` `(` `long` `i = ` `2` `; i * i <= n; i++)` ` ` `{` ` ` `if` `(n % i == ` `0` `)` ` ` `{` ` ` `if` `(i * i != n)` ` ` `sum = sum + i + n / i;` ` ` `else` ` ` `sum = sum + i;` ` ` `}` ` ` `}` ` ` ` ` `// Check if sum of divisors is equal ` ` ` `// to n, then n is a perfect number` ` ` `if` `(sum == n && n != ` `1` `)` ` ` `{` ` ` `return` `true` `;` ` ` `}` ` ` `return` `false` `;` `}` ` ` `// Function to find the longest subsequence` `// which contain all Perfect numbers` `static` `int` `longestPerfectSubsequence(` `int` `arr[],` ` ` `int` `n)` `{` ` ` `int` `answer = ` `0` `;` ` ` ` ` `// Find the length of longest` ` ` `// perfect number subsequence` ` ` `for` `(` `int` `i = ` `0` `; i < n; i++)` ` ` `{` ` ` `if` `(isPerfect(arr[i]) == ` `true` `)` ` ` `answer++;` ` ` `}` ` ` `return` `answer;` `}` ` ` `// Driver code` `public` `static` `void` `main (String[] args)` `{` ` ` `int` `arr[] = { ` `3` `, ` `6` `, ` `11` `, ` `2` `, ` `28` `, ` `21` `, ` `8128` `};` ` ` `int` `n = arr.length;` ` ` ` ` `System.out.println(longestPerfectSubsequence(arr, n));` `}` `}` `// This code is contributed by AnkitRai01` |

## Python3

`# Python3 program to find the length of` `# Longest Perfect number Subsequence in an Array` `# Function to check if` `# the number is Perfect number` `def` `isPerfect( n ):` ` ` ` ` `# To store sum of divisors` ` ` `sum` `=` `1` ` ` ` ` `# Find all divisors and add them` ` ` `i ` `=` `2` ` ` `while` `i ` `*` `i <` `=` `n:` ` ` `if` `n ` `%` `i ` `=` `=` `0` `:` ` ` `sum` `=` `sum` `+` `i ` `+` `n ` `/` `i` ` ` `i ` `+` `=` `1` ` ` ` ` `# Check if sum of divisors is equal to` ` ` `# n, then n is a perfect number` ` ` ` ` `return` `(` `True` `if` `sum` `=` `=` `n ` `and` `n !` `=` `1` `else` `False` `)` `# Function to find the longest subsequence` `# which contain all Perfect numbers` `def` `longestPerfectSubsequence( arr, n):` ` ` ` ` `answer ` `=` `0` ` ` ` ` `# Find the length of longest` ` ` `# Perfect number subsequence` ` ` `for` `i ` `in` `range` `(n):` ` ` `if` `(isPerfect(arr[i])):` ` ` `answer ` `+` `=` `1` ` ` ` ` `return` `answer` `# Driver code` `if` `__name__ ` `=` `=` `"__main__"` `:` ` ` `arr ` `=` `[ ` `3` `, ` `6` `, ` `11` `, ` `2` `, ` `28` `, ` `21` `, ` `8128` `]` ` ` `n ` `=` `len` `(arr)` ` ` ` ` `print` `(longestPerfectSubsequence(arr, n))` |

## C#

`// C# program to find the length of` `// longest perfect number subsequence` `// in an array` `using` `System;` `class` `GFG {` ` ` `// Function to check if the` `// number is a perfect number` `static` `bool` `isPerfect(` `long` `n)` `{` ` ` ` ` `// To store sum of divisors` ` ` `long` `sum = 1;` ` ` ` ` `// Find all divisors and add them` ` ` `for` `(` `long` `i = 2; i * i <= n; i++)` ` ` `{` ` ` `if` `(n % i == 0)` ` ` `{` ` ` `if` `(i * i != n)` ` ` `sum = sum + i + n / i;` ` ` `else` ` ` `sum = sum + i;` ` ` `}` ` ` `}` ` ` ` ` `// Check if sum of divisors is equal` ` ` `// to n, then n is a perfect number` ` ` `if` `(sum == n && n != 1)` ` ` `{` ` ` `return` `true` `;` ` ` `}` ` ` `return` `false` `;` `}` ` ` `// Function to find the longest subsequence` `// which contain all perfect numbers` `static` `int` `longestPerfectSubsequence(` `int` `[]arr,` ` ` `int` `n)` `{` ` ` `int` `answer = 0;` ` ` ` ` `// Find the length of longest` ` ` `// perfect number subsequence` ` ` `for` `(` `int` `i = 0; i < n; i++)` ` ` `{` ` ` `if` `(isPerfect(arr[i]) == ` `true` `)` ` ` `answer++;` ` ` `}` ` ` `return` `answer;` `}` ` ` `// Driver code` `public` `static` `void` `Main (` `string` `[] args)` `{` ` ` `int` `[]arr = { 3, 6, 11, 2, 28, 21, 8128 };` ` ` `int` `n = arr.Length;` ` ` ` ` `Console.WriteLine(longestPerfectSubsequence(arr, n));` `}` `}` `// This code is contributed by AnkitRai01` |

## Javascript

`<script>` `// Javascript program to find the length of` `// Longest Perfect number Subsequence in an Array` `// Function to check if` `// the number is a Perfect number` `function` `isPerfect(n)` `{` ` ` `// To store sum of divisors` ` ` `var` `sum = 1;` ` ` `// Find all divisors and add them` ` ` `for` `(` `var` `i = 2; i * i <= n; i++) {` ` ` `if` `(n % i == 0) {` ` ` `if` `(i * i != n)` ` ` `sum = sum + i + n / i;` ` ` `else` ` ` `sum = sum + i;` ` ` `}` ` ` `}` ` ` `// Check if sum of divisors is equal to` ` ` `// n, then n is a perfect number` ` ` `if` `(sum == n && n != 1)` ` ` `return` `true` `;` ` ` `return` `false` `;` `}` `// Function to find the longest subsequence` `// which contain all Perfect numbers` `function` `longestPerfectSubsequence(arr, n)` `{` ` ` `var` `answer = 0;` ` ` `// Find the length of longest` ` ` `// Perfect number subsequence` ` ` `for` `(` `var` `i = 0; i < n; i++) {` ` ` `if` `(isPerfect(arr[i]))` ` ` `answer++;` ` ` `}` ` ` `return` `answer;` `}` `// Driver code` `var` `arr = [3, 6, 11, 2, 28, 21, 8128];` `var` `n = arr.length;` `document.write( longestPerfectSubsequence(arr, n));` `</script>` |

**Output:**

3

**Time Complexity:** O(N×√N)**Auxiliary Space Complexity:** O(1)