Given an array **arr[]** of length **N**, the task is to find the number of subsets with a given **OR** value **M**.**Examples:**

Input:arr[] = {2, 3, 2} M = 3Output:4

All possible subsets and there OR values are:

{2} = 2

{3} =3

{2} = 2

{2, 3} =2 | 3 = 3

{3, 2} =3 | 2 = 3

{2, 2} = 2 | 2 = 2

{2, 3, 2} =2 | 3 | 2 = 3

Input:arr[] = {1, 3, 2, 2}, M = 5Output:0

**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 OR 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_or] = dp[i + 1][curr_or] + dp[i + 1][curr_or | arr[i]]

The above recurrence relation can be defined as the number of subsets of sub-array **arr[i…N-1]** such that ORing them with **curr_or** will yield the required OR value.

The recurrence relation is justified as there are only paths. Either, take the current element and OR it with **curr_or** or ignore it and move forward.

Below is the implementation of the above approach:

## C++

`// C++ implementation of the approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `#define maxN 20` `#define maxM 64` `// To store states of DP` `int` `dp[maxN][maxM];` `bool` `v[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` `(v[i][curr])` ` ` `return` `dp[i][curr];` ` ` `// Setting the state as solved` ` ` `v[i][curr] = 1;` ` ` `// Recurrence relation` ` ` `dp[i][curr]` ` ` `= findCnt(arr, i + 1, curr, n, m)` ` ` `+ findCnt(arr, i + 1, (curr | arr[i]), n, m);` ` ` `return` `dp[i][curr];` `}` `// Driver code` `int` `main()` `{` ` ` `int` `arr[] = { 2, 3, 2 };` ` ` `int` `n = ` `sizeof` `(arr) / ` `sizeof` `(` `int` `);` ` ` `int` `m = 3;` ` ` `cout << findCnt(arr, 0, 0, n, m);` ` ` `return` `0;` `}` |

## Java

`// Java implementation of the approach` `import` `java.util.*;` `class` `GFG` `{` ` ` `static` `int` `maxN = ` `20` `;` `static` `int` `maxM = ` `64` `;` `// To store states of DP` `static` `int` `[][]dp = ` `new` `int` `[maxN][maxM];` `static` `boolean` `[][]v = ` `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` `(v[i][curr])` ` ` `return` `dp[i][curr];` ` ` `// Setting the state as solved` ` ` `v[i][curr] = ` `true` `;` ` ` `// Recurrence relation` ` ` `dp[i][curr] = findCnt(arr, i + ` `1` `, curr, n, m) +` ` ` `findCnt(arr, i + ` `1` `, (curr | arr[i]), n, m);` ` ` `return` `dp[i][curr];` `}` `// Driver code` `public` `static` `void` `main(String []args)` `{` ` ` `int` `arr[] = { ` `2` `, ` `3` `, ` `2` `};` ` ` `int` `n = arr.length;` ` ` `int` `m = ` `3` `;` ` ` `System.out.println(findCnt(arr, ` `0` `, ` `0` `, 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` `dp ` `=` `np.zeros((maxN, maxM));` `v ` `=` `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` `(v[i][curr]) :` ` ` `return` `dp[i][curr];` ` ` `# Setting the state as solved` ` ` `v[i][curr] ` `=` `1` `;` ` ` `# Recurrence relation` ` ` `dp[i][curr] ` `=` `findCnt(arr, i ` `+` `1` `, curr, n, m) ` `+` `\` ` ` `findCnt(arr, i ` `+` `1` `, (curr | arr[i]), n, m);` ` ` `return` `dp[i][curr];` `# Driver code` `if` `__name__ ` `=` `=` `"__main__"` `:` ` ` `arr ` `=` `[ ` `2` `, ` `3` `, ` `2` `];` ` ` `n ` `=` `len` `(arr);` ` ` `m ` `=` `3` `;` ` ` `print` `(findCnt(arr, ` `0` `, ` `0` `, 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` `[,]dp = ` `new` `int` `[maxN, maxM];` `static` `Boolean [,]v = ` `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` `(v[i, curr])` ` ` `return` `dp[i, curr];` ` ` `// Setting the state as solved` ` ` `v[i, curr] = ` `true` `;` ` ` `// Recurrence relation` ` ` `dp[i, curr] = findCnt(arr, i + 1, curr, n, m) +` ` ` `findCnt(arr, i + 1, (curr | arr[i]), n, m);` ` ` `return` `dp[i, curr];` `}` `// Driver code` `public` `static` `void` `Main(String []args)` `{` ` ` `int` `[]arr = { 2, 3, 2 };` ` ` `int` `n = arr.Length;` ` ` `int` `m = 3;` ` ` `Console.WriteLine(findCnt(arr, 0, 0, n, m));` `}` `}` `// This code is contributed by Rajput-Ji` |

**Output**

4

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