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Number of subsets with a given AND value
• Last Updated : 14 May, 2021

Given an array arr of length N and an integer X, the task is to find the number of subsets whose AND value is X.
Examples:

Input: arr[] = {2, 3, 2} X = 2
Output:
All possible subsets and there AND values are:
{2} = 2
{3} = 3
{2} = 2
{2, 3} = 2 & 3 = 2
{3, 2} = 3 & 2 = 2
{2, 2} = 2 & 2 = 2
{2, 3, 2} = 2 & 3 & 2 = 2
Input: arr[] = {0, 0, 0}, X = 0
Output:

Approach: A simple approach is to solve the problem by generating all the possible subsets and then by counting the number of subsets with the given AND value. However, for smaller values of array elements, it can be solved using dynamic programming
Let’s look at the recurrence relation first.

dp[i][curr_and] = dp[i + 1][curr_and] + dp[i + 1][curr_and & arr[i]]

The above recurrence relation can be defined as the number of subsets of sub-array arr[i…N-1] such that ANDing them with curr_and will yield the required AND value.
The recurrence relation is justified as there are only paths. Either, take the current element and AND it with curr_and or ignore it and move forward.
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 states of DP``int` `dp1[maxN][maxM];``bool` `v1[maxN][maxM];` `// Function to return the required count``int` `findCnt(``int``* arr, ``int` `i, ``int` `curr, ``int` `n, ``int` `m)``{``    ``// Base case``    ``if` `(i == n) {``        ``return` `(curr == m);``    ``}` `    ``// If the state has been solved before``    ``// return the value of the state``    ``if` `(v1[i][curr])``        ``return` `dp1[i][curr];` `    ``// Setting the state as solved``    ``v1[i][curr] = 1;` `    ``// Recurrence relation``    ``dp1[i][curr]``        ``= findCnt(arr, i + 1, curr, n, m)``          ``+ findCnt(arr, i + 1, (curr & arr[i]), n, m);` `    ``return` `dp1[i][curr];``}` `// Driver code``int` `main()``{``    ``int` `arr[] = { 0, 0, 0 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(``int``);``    ``int` `m = 0;` `    ``cout << findCnt(arr, 0, ((1 << 6) - 1), n, m);` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``class` `GFG``{``static` `int` `maxN = ``20``;``static` `int` `maxM = ``64``;` `// To store states of DP``static` `int` `[][]dp1 = ``new` `int``[maxN][maxM];``static` `boolean` `[][]v1 = ``new` `boolean``[maxN][maxM];` `// Function to return the required count``static` `int` `findCnt(``int` `[]arr, ``int` `i,``                   ``int` `curr, ``int` `n, ``int` `m)``{``    ``// Base case``    ``if` `(i == n)``    ``{``        ``return` `(curr == m ? ``1` `: ``0``);``    ``}` `    ``// If the state has been solved before``    ``// return the value of the state``    ``if` `(v1[i][curr])``        ``return` `dp1[i][curr];` `    ``// Setting the state as solved``    ``v1[i][curr] = ``true``;` `    ``// Recurrence relation``    ``dp1[i][curr] = findCnt(arr, i + ``1``, curr, n, m) +``                   ``findCnt(arr, i + ``1``, (curr & arr[i]), n, m);` `    ``return` `dp1[i][curr];``}` `// Driver code``public` `static` `void` `main(String []args)``{``    ``int` `arr[] = { ``0``, ``0``, ``0` `};``    ``int` `n = arr.length;``    ``int` `m = ``0``;` `    ``System.out.println(findCnt(arr, ``0``, ((``1` `<< ``6``) - ``1``), n, m));``}``}` `// This code is contributed by 29AjayKumar`

## Python3

 `# Python3 implementation of the approach``import` `numpy as np` `maxN ``=` `20``maxM ``=` `64` `# To store states of DP``dp1 ``=` `np.zeros((maxN, maxM));``v1 ``=` `np.zeros((maxN, maxM));` `# Function to return the required count``def` `findCnt(arr, i, curr, n, m) :` `    ``# Base case``    ``if` `(i ``=``=` `n) :``        ``return` `(curr ``=``=` `m);` `    ``# If the state has been solved before``    ``# return the value of the state``    ``if` `(v1[i][curr]) :``        ``return` `dp1[i][curr];` `    ``# Setting the state as solved``    ``v1[i][curr] ``=` `1``;` `    ``# Recurrence relation``    ``dp1[i][curr] ``=` `findCnt(arr, i ``+` `1``, curr, n, m) ``+` `\``                   ``findCnt(arr, i ``+` `1``, (curr & arr[i]), n, m);` `    ``return` `dp1[i][curr];` `# Driver code``if` `__name__ ``=``=` `"__main__"` `:``    ` `    ``arr ``=` `[ ``0``, ``0``, ``0` `];``    ``n ``=` `len``(arr);``    ``m ``=` `0``;` `    ``print``(findCnt(arr, ``0``, ((``1` `<< ``6``) ``-` `1``), n, m));` `# This code is contributed by AnkitRai01`

## C#

 `// C# implementation of the approach``using` `System;` `class` `GFG``{``static` `int` `maxN = 20;``static` `int` `maxM = 64;` `// To store states of DP``static` `int` `[,]dp1 = ``new` `int``[maxN, maxM];``static` `bool` `[,]v1 = ``new` `bool``[maxN, maxM];` `// Function to return the required count``static` `int` `findCnt(``int` `[]arr, ``int` `i,``                   ``int` `curr, ``int` `n, ``int` `m)``{``    ``// Base case``    ``if` `(i == n)``    ``{``        ``return` `(curr == m ? 1 : 0);``    ``}` `    ``// If the state has been solved before``    ``// return the value of the state``    ``if` `(v1[i, curr])``        ``return` `dp1[i, curr];` `    ``// Setting the state as solved``    ``v1[i, curr] = ``true``;` `    ``// Recurrence relation``    ``dp1[i, curr] = findCnt(arr, i + 1, curr, n, m) +``                   ``findCnt(arr, i + 1, (curr & arr[i]), n, m);` `    ``return` `dp1[i, curr];``}` `// Driver code``public` `static` `void` `Main(String []args)``{``    ``int` `[]arr = { 0, 0, 0 };``    ``int` `n = arr.Length;``    ``int` `m = 0;` `    ``Console.WriteLine(findCnt(arr, 0, ((1 << 6) - 1), n, m));``}``}` `// This code is contributed by Rajput-Ji`

## Javascript

 ``
Output:
`7`

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