# Longest sub-sequence with a given OR value : O(N) Approach

• Last Updated : 09 Nov, 2021

Given an array arr[], the task is to find the longest subsequence with a given OR value M. If there is no such sub-sequence then print 0
Examples:

Input: arr[] = {3, 7, 2, 3}, M = 3
Output:
{3, 2, 3} is the required subsequence
3 | 2 | 3 = 3
Input: arr[] = {2, 2}, M = 3
Output :

Naive approach: A simple way to solve this problem is to generate all the possible sub-sequences and then find the largest among them with the required OR value.
Efficient approach: One key observation is that all of the numbers in the required sub-sequence should yield the value M when they get ORed with M. So filter out all of such elements whose OR with M equals to M
Now, the task is to find the longest sub-sequence among this filtered subset. It’s pretty obvious that all of these numbers will be ORed together. If the result of this OR is M then the answer will be equal to the size of this filtered set. Otherwise answer will be 0. This is because OR only sets the unset bits. So, the larger the numbers in the set, the more optimal it is.
Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;` `// Function to return the required length``int` `findLen(``int``* arr, ``int` `n, ``int` `m)``{``    ``// To store the filtered numbers``    ``vector<``int``> filter;` `    ``// Filtering the numbers``    ``for` `(``int` `i = 0; i < n; i++)``        ``if` `((arr[i] | m) == m)``            ``filter.push_back(arr[i]);` `    ``// If there are no elements to check``    ``if` `(filter.size() == 0)``        ``return` `0;` `    ``// Find the OR of all the``    ``// filtered elements``    ``int` `c_or = filter[0];``    ``for` `(``int` `i = 1; i < filter.size(); i++)``        ``c_or |= filter[i];` `    ``// Check if the OR is equal to m``    ``if` `(c_or == m)``        ``return` `filter.size();` `    ``return` `0;``}` `// Driver code``int` `main()``{``    ``int` `arr[] = { 7, 3, 3, 1, 3 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(``int``);``    ``int` `m = 3;` `    ``cout << findLen(arr, n, m);` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``import` `java.util.*;` `class` `GFG``{` `// Function to return the required length``static` `int` `findLen(``int` `arr[], ``int` `n, ``int` `m)``{``    ``// To store the filtered numbers``    ``Vector filter = ``new` `Vector();` `    ``// Filtering the numbers``    ``for` `(``int` `i = ``0``; i < n; i++)``        ``if` `((arr[i] | m) == m)``            ``filter.add(arr[i]);` `    ``// If there are no elements to check``    ``if` `(filter.size() == ``0``)``        ``return` `0``;` `    ``// Find the OR of all the``    ``// filtered elements``    ``int` `c_or = filter.get(``0``);``    ``for` `(``int` `i = ``1``; i < filter.size(); i++)``        ``c_or |= filter.get(i);` `    ``// Check if the OR is equal to m``    ``if` `(c_or == m)``        ``return` `filter.size();` `    ``return` `0``;``}` `// Driver code``public` `static` `void` `main(String args[])``{``    ``int` `arr[] = { ``7``, ``3``, ``3``, ``1``, ``3` `};``    ``int` `n = arr.length;``    ``int` `m = ``3``;` `    ``System.out.print(findLen(arr, n, m));``}``}` `// This code is contributed by Arnab Kundu`

## Python3

 `# Python3 implementation of the approach` `# Function to return the required length``def` `findLen(arr, n, m) :` `    ``# To store the filtered numbers``    ``filter` `=` `[];` `    ``# Filtering the numbers``    ``for` `i ``in` `range``(n) :``        ``if` `((arr[i] | m) ``=``=` `m) :``            ``filter``.append(arr[i]);` `    ``# If there are no elements to check``    ``if` `(``len``(``filter``) ``=``=` `0``) :``        ``return` `0``;` `    ``# Find the OR of all the``    ``# filtered elements``    ``c_or ``=` `filter``[``0``];``    ``for` `i ``in` `range``(``1``, ``len``(``filter``)) :``        ``c_or |``=` `filter``[i];` `    ``# Check if the OR is equal to m``    ``if` `(c_or ``=``=` `m) :``        ``return` `len``(``filter``);` `# Driver code``if` `__name__ ``=``=` `"__main__"` `:` `    ``arr ``=` `[ ``7``, ``3``, ``3``, ``1``, ``3` `];``    ``n ``=` `len``(arr);``    ``m ``=` `3``;` `    ``print``(findLen(arr, n, m));``    ` `# This code is contributed by AnkitRai01`

## C#

 `// C# implementation of the approach``using` `System;` `using` `System.Collections.Generic;` `class` `GFG``{` `// Function to return the required length``static` `int` `findLen(``int` `[] arr, ``int` `n, ``int` `m)``{``    ``// To store the filtered numbers``    ``List<``int``> filter = ``new` `List<``int``>();` `    ``// Filtering the numbers``    ``for` `(``int` `i = 0; i < n; i++)``        ``if` `((arr[i] | m) == m)``            ``filter.Add(arr[i]);` `    ``// If there are no elements to check``    ``if` `(filter.Count == 0)``        ``return` `0;` `    ``// Find the OR of all the``    ``// filtered elements``    ``int` `c_or = filter[0];``    ``for` `(``int` `i = 1; i < filter.Count; i++)``        ``c_or |= filter[i];` `    ``// Check if the OR is equal to m``    ``if` `(c_or == m)``        ``return` `filter.Count;` `    ``return` `0;``}` `// Driver code``public` `static` `void` `Main()``{``    ``int` `[]arr = { 7, 3, 3, 1, 3 };``    ``int` `n = arr.Length;``    ``int` `m = 3;` `    ``Console.Write(findLen(arr, n, m));``}``}` `// This code is contributed by Mohit kumar 29`

## Javascript

 ``
Output:
`4`

Time Complexity: O(n)

Auxiliary Space: O(n)

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