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 = 3Input: 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++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;#define maxN 20#define maxM 64// To store states of DPint dp[maxN][maxM];bool v[maxN][maxM];// Function to return the required countint 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 codeint 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 approachimport java.util.*;class GFG { static int maxN = 20;static int maxM = 64;// To store states of DPstatic int [][]dp = new int[maxN][maxM];static boolean [][]v = new boolean[maxN][maxM];// Function to return the required countstatic 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 codepublic 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 npmaxN = 20maxM = 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 approachusing System; class GFG { static int maxN = 20;static int maxM = 64;// To store states of DPstatic int [,]dp = new int[maxN, maxM];static Boolean [,]v = new Boolean[maxN, maxM];// Function to return the required countstatic 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 codepublic 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 |
Javascript
<script>// Javascript implementation of the approachvar maxN = 20var maxM = 64// To store states of DPvar dp = Array.from(Array(maxN), ()=> Array(maxM));var v = Array.from(Array(maxN), ()=> Array(maxM));// Function to return the required countfunction 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 codevar 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.
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