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: 6
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: 7
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 <bits/stdc++.h> 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
<script> // Javascript implementation of the approach var maxN = 20; var maxM = 64; // To store states of DP var dp1 = Array.from(Array(maxN), ()=> Array(maxM)); var v1 = 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 (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 var arr = [0, 0, 0]; var n = arr.length; var m = 0; document.write( findCnt(arr, 0, ((1 << 6) - 1), n, m)); </script> |
7
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!