Number of subsets with a given OR value

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 = 3 
Output: 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 = 5 
Output: 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++ 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 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 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# 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

<script>

// Javascript implementation of the approach
var maxN = 20
var maxM = 64

// To store states of DP
var dp = Array.from(Array(maxN), ()=> Array(maxM));
var v = Array.from(Array(maxN), ()=> Array(maxM));

// Function to return the required count
function 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
var arr = [2, 3, 2 ];
var n = arr.length;
var m = 3;
document.write( findCnt(arr, 0, 0, n, m));

</script>    
    

4

Time Complexity: O(maxN*maxM), where maxN and maxM are the constant defined.
Auxiliary Space: O(maxN*maxM), where maxN and maxM are the constant defined.