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# Longest subsequence with a given AND value | O(N)

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

Input: arr[] = {3, 7, 2, 3}, M = 3
Output:
{3, 7, 3} is the required subsequence.
3 & 7 & 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 AND 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 ANDed with M. So filter out all of such elements whose AND 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 ANDed together. If the result of this AND is M then the answer will be equal to the size of this filtered set. Otherwise answer will be 0. This is because AND only unsets the set 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 AND of all the``    ``// filtered elements``    ``int` `c_and = filter;``    ``for` `(``int` `i = 1; i < filter.size(); i++)``        ``c_and &= filter[i];` `    ``// Check if the AND is equal to m``    ``if` `(c_and == 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 AND of all the``    ``// filtered elements``    ``int` `c_and = filter.get(``0``);``    ``for` `(``int` `i = ``1``; i < filter.size(); i++)``        ``c_and &= filter.get(i);` `    ``// Check if the AND is equal to m``    ``if` `(c_and == 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.println(findLen(arr, n, m));``}``}` `// This code is contributed by PrinciRaj1992`

## 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_and ``=` `filter``[``0``];``    ``for` `i ``in` `range``(``1``, ``len``(``filter``)) :``        ``c_and &``=` `filter``[i];` `    ``# Check if the OR is equal to m``    ``if` `(c_and ``=``=` `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 AND of all the``    ``// filtered elements``    ``int` `c_and = filter;``    ``for` `(``int` `i = 1; i < filter.Count; i++)``        ``c_and &= filter[i];` `    ``// Check if the AND is equal to m``    ``if` `(c_and == m)``        ``return` `filter.Count;` `    ``return` `0;``}` `// Driver code``public` `static` `void` `Main(String []args)``{``    ``int` `[]arr = { 7, 3, 3, 1, 3 };``    ``int` `n = arr.Length;``    ``int` `m = 3;` `    ``Console.WriteLine(findLen(arr, n, m));``}``}` `// This code is contributed by 29AjayKumar`

## Javascript

 ``

Output:

`4`

Time Complexity: O(n)

Auxiliary Space: O(n)