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Longest subsequence with a given AND value
• Last Updated : 14 May, 2021

Given an array arr[] and an integer M, 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:
Longest sub-sequence with AND value 3 is {3, 7, 3}.
Input: arr[] = {2, 2}, M = 3
Output:

Approach: A simple way to solve this will be to generate all the possible sub-sequences and then find the largest among them with required AND value.
However, for smaller values of M, an approach based on dynamic programming can be used.
Let’s look at the recurrence relation first.

dp[i][curr_and] = max(dp[i + 1][curr_and], dp[i + 1][curr_and & arr[i]] + 1)

Let’s understand the states of DP now. Here, dp[i][curr_and] stores the longest subsequence of subarray arr[i…N-1] such that curr_and & the AND of this subsequence is equal to M. At each step, either the index i can be chosen updating curr_and or it can be rejected.
Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;``#define maxN 20``#define maxM 64` `// To store the states of DP``int` `dp[maxN][maxM];``bool` `v[maxN][maxM];` `// Function to return the required length``int` `findLen(``int``* arr, ``int` `i, ``int` `curr,``            ``int` `n, ``int` `m)``{``    ``// Base case``    ``if` `(i == n) {``        ``if` `(curr == m)``            ``return` `0;``        ``else``            ``return` `-1;``    ``}` `    ``// If the state has been solved before``    ``// return the value of the state``    ``if` `(v[i][curr])``        ``return` `dp[i][curr];` `    ``// Setting the state as solved``    ``v[i][curr] = 1;` `    ``// Recurrence relation``    ``int` `l = findLen(arr, i + 1, curr, n, m);``    ``int` `r = findLen(arr, i + 1, curr & arr[i],``                    ``n, m);``    ``dp[i][curr] = l;``    ``if` `(r != -1)``        ``dp[i][curr] = max(dp[i][curr], r + 1);``    ``return` `dp[i][curr];``}` `// Driver code``int` `main()``{``    ``int` `arr[] = { 3, 7, 2, 3 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(``int``);``    ``int` `m = 3;` `    ``int` `ans = findLen(arr, 0, ((1 << 8) - 1),``                      ``n, m);``    ``if` `(ans == -1)``        ``cout << 0;``    ``else``        ``cout << ans;` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``import` `java.util.*;` `class` `GFG``{``static` `int` `maxN = ``300``;``static` `int` `maxM = ``300``;` `// To store the states of DP``static` `int` `dp[][] = ``new` `int``[maxN][maxM];``static` `boolean` `v[][] = ``new` `boolean``[maxN][maxM];` `// Function to return the required length``static` `int` `findLen(``int``[] arr, ``int` `i,``                   ``int` `curr, ``int` `n, ``int` `m)``{``    ``// Base case``    ``if` `(i == n)``    ``{``        ``if` `(curr == m)``            ``return` `0``;``        ``else``            ``return` `-``1``;``    ``}` `    ``// If the state has been solved before``    ``// return the value of the state``    ``if` `(v[i][curr])``        ``return` `dp[i][curr];` `    ``// Setting the state as solved``    ``v[i][curr] = ``true``;` `    ``// Recurrence relation``    ``int` `l = findLen(arr, i + ``1``, curr, n, m);``    ``int` `r = findLen(arr, i + ``1``, curr & arr[i], n, m);``    ``dp[i][curr] = l;``    ``if` `(r != -``1``)``        ``dp[i][curr] = Math.max(dp[i][curr], r + ``1``);``    ``return` `dp[i][curr];``}` `// Driver code``public` `static` `void` `main(String args[])``{``    ``int` `arr[] = { ``3``, ``7``, ``2``, ``3` `};``    ``int` `n = arr.length;``    ``int` `m = ``3``;` `    ``int` `ans = findLen(arr, ``0``, ((``1` `<< ``8``) - ``1``), n, m);``    ``if` `(ans == -``1``)``        ``System.out.print( ``0``);``    ``else``        ``System.out.print( ans);``}``}` `// This code is contributed by Arnab Kundu`

## Python3

 `# Python3 implementation of the approach``import` `numpy as np` `maxN ``=` `20``maxM ``=` `256` `# To store the states of DP``dp ``=` `np.zeros((maxN, maxM));``v ``=` `np.zeros((maxN, maxM));` `# Function to return the required length``def` `findLen(arr, i, curr, n, m) :` `    ``# Base case``    ``if` `(i ``=``=` `n) :``        ``if` `(curr ``=``=` `m) :``            ``return` `0``;``        ``else` `:``            ``return` `-``1``;` `    ``# If the state has been solved before``    ``# return the value of the state``    ``if` `(v[i][curr]) :``        ``return` `dp[i][curr];` `    ``# Setting the state as solved``    ``v[i][curr] ``=` `1``;` `    ``# Recurrence relation``    ``l ``=` `findLen(arr, i ``+` `1``, curr, n, m);``    ``r ``=` `findLen(arr, i ``+` `1``, curr & arr[i], n, m);``    ` `    ``dp[i][curr] ``=` `l;``    ` `    ``if` `(r !``=` `-``1``) :``        ``dp[i][curr] ``=` `max``(dp[i][curr], r ``+` `1``);``        ` `    ``return` `dp[i][curr];` `# Driver code``if` `__name__ ``=``=` `"__main__"` `:` `    ``arr ``=` `[ ``3``, ``7``, ``2``, ``3` `];``    ``n ``=` `len``(arr);``    ``m ``=` `3``;` `    ``ans ``=` `findLen(arr, ``0``, ((``1` `<< ``8``) ``-` `1``), n, m);``    ` `    ``if` `(ans ``=``=` `-``1``) :``        ``print``(``0``);``    ``else` `:``        ``print``(ans);` `# This code is contributed by AnkitRai01`

## C#

 `// C# implementation of the approach``using` `System;` `class` `GFG``{``static` `int` `maxN = 300;``static` `int` `maxM = 300;` `// To store the states of DP``static` `int``[,] dp = ``new` `int``[maxN, maxM];``static` `bool``[,] v = ``new` `bool``[maxN, maxM];` `// Function to return the required length``static` `int` `findLen(``int``[] arr, ``int` `i,``                   ``int` `curr, ``int` `n, ``int` `m)``{``    ``// Base case``    ``if` `(i == n)``    ``{``        ``if` `(curr == m)``            ``return` `0;``        ``else``            ``return` `-1;``    ``}` `    ``// If the state has been solved before``    ``// return the value of the state``    ``if` `(v[i, curr])``        ``return` `dp[i, curr];` `    ``// Setting the state as solved``    ``v[i, curr] = ``true``;` `    ``// Recurrence relation``    ``int` `l = findLen(arr, i + 1, curr, n, m);``    ``int` `r = findLen(arr, i + 1, curr & arr[i], n, m);``    ``dp[i, curr] = l;``    ``if` `(r != -1)``        ``dp[i, curr] = Math.Max(dp[i, curr], r + 1);``    ``return` `dp[i, curr];``}` `// Driver code``public` `static` `void` `Main(String[] args)``{``    ``int``[] arr = { 3, 7, 2, 3 };``    ``int` `n = arr.Length;``    ``int` `m = 3;` `    ``int` `ans = findLen(arr, 0, ((1 << 8) - 1), n, m);``    ``if` `(ans == -1)``        ``Console.WriteLine(0);``    ``else``        ``Console.WriteLine(ans);``}``}` `// This code is contributed by``// sanjeev2552`

## Javascript

 ``
Output:
`3`

Time Complexity: O(N * maxVal) where maxVal is the maximum element from the given array.

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