Given an array arr[] of N integers, the task is to find the count of all the subsets which do not contain adjacent elements from the given array.
Examples:
Input: arr[] = {2, 7}
Output: 3
All possible subsets are {}, {2} and {7}.Input: arr[] = {3, 5, 7}
Output: 5
Method 1: The idea is to use a bit-mask pattern to generate all the combinations as discussed in this article. While considering a subset, we need to check if it contains adjacent elements or not. A subset will contain adjacent elements if two or more consecutive bits are set in its bit-mask. In order to check if the bit-mask has consecutive bits set or not, we can right shift the mask by one bit and take its AND with the mask. If the result of the AND operation is 0, then the mask does not have consecutive bits set and therefore, the corresponding subset will not have adjacent elements from the array.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <iostream> #include <math.h> using namespace std; // Function to return the count // of possible subsets int cntSubsets(int* arr, int n) { // Total possible subsets of n // sized array is (2^n - 1) unsigned int max = pow(2, n); // To store the required // count of subsets int result = 0; // Run from i 000..0 to 111..1 for (int i = 0; i < max; i++) { int counter = i; // If current subset has consecutive // elements from the array if (counter & (counter >> 1)) continue; result++; } return result; } // Driver code int main() { int arr[] = { 3, 5, 7 }; int n = sizeof(arr) / sizeof(arr[0]); cout << cntSubsets(arr, n); return 0; } |
Java
// Java implementation of the approach import java.util.*; class GFG { // Function to return the count // of possible subsets static int cntSubsets(int[] arr, int n) { // Total possible subsets of n // sized array is (2^n - 1) int max = (int) Math.pow(2, n); // To store the required // count of subsets int result = 0; // Run from i 000..0 to 111..1 for (int i = 0; i < max; i++) { int counter = i; // If current subset has consecutive if ((counter & (counter >> 1)) > 0) continue; result++; } return result; } // Driver code static public void main (String []arg) { int arr[] = { 3, 5, 7 }; int n = arr.length; System.out.println(cntSubsets(arr, n)); } } // This code is contributed by Rajput-Ji |
Python3
# Python3 implementation of the approach # Function to return the count # of possible subsets def cntSubsets(arr, n): # Total possible subsets of n # sized array is (2^n - 1) max = pow(2, n) # To store the required # count of subsets result = 0 # Run from i 000..0 to 111..1 for i in range(max): counter = i # If current subset has consecutive # elements from the array if (counter & (counter >> 1)): continue result += 1 return result # Driver code arr = [3, 5, 7] n = len(arr) print(cntSubsets(arr, n)) # This code is contributed by Mohit Kumar |
C#
// C# implementation of the approach using System; class GFG { // Function to return the count // of possible subsets static int cntSubsets(int[] arr, int n) { // Total possible subsets of n // sized array is (2^n - 1) int max = (int) Math.Pow(2, n); // To store the required // count of subsets int result = 0; // Run from i 000..0 to 111..1 for (int i = 0; i < max; i++) { int counter = i; // If current subset has consecutive if ((counter & (counter >> 1)) > 0) continue; result++; } return result; } // Driver code static public void Main (String []arg) { int []arr = { 3, 5, 7 }; int n = arr.Length; Console.WriteLine(cntSubsets(arr, n)); } } // This code is contributed by Rajput-Ji |
5
Method 2: The above approach takes exponential time. In the above code, the number of bit-masks without consecutive 1s were required. This count can be obtained in linear time using dynamic programming as discussed in this article.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <iostream> using namespace std; // Function to return the count // of possible subsets int cntSubsets(int* arr, int n) { int a[n], b[n]; a[0] = b[0] = 1; for (int i = 1; i < n; i++) { // If previous element was 0 then 0 // as well as 1 can be appended a[i] = a[i - 1] + b[i - 1]; // If previous element was 1 then // only 0 can be appended b[i] = a[i - 1]; } // Store the count of all possible subsets int result = a[n - 1] + b[n - 1]; return result; } // Driver code int main() { int arr[] = { 3, 5, 7 }; int n = sizeof(arr) / sizeof(arr[0]); cout << cntSubsets(arr, n); return 0; } |
Java
// Java implementation of the approach import java.util.*; class GFG { // Function to return the count // of possible subsets static int cntSubsets(int []arr, int n) { int []a = new int[n]; int []b = new int[n]; a[0] = b[0] = 1; for (int i = 1; i < n; i++) { // If previous element was 0 then 0 // as well as 1 can be appended a[i] = a[i - 1] + b[i - 1]; // If previous element was 1 then // only 0 can be appended b[i] = a[i - 1]; } // Store the count of all possible subsets int result = a[n - 1] + b[n - 1]; return result; } // Driver code public static void main(String[] args) { int arr[] = { 3, 5, 7 }; int n = arr.length; System.out.println(cntSubsets(arr, n)); } } // This code is contributed by Princi Singh |
Python3
# Python3 implementation of the approach # Function to return the count # of possible subsets def cntSubsets(arr, n) : a = [0] * n b = [0] * n; a[0] = b[0] = 1; for i in range(1, n) : # If previous element was 0 then 0 # as well as 1 can be appended a[i] = a[i - 1] + b[i - 1]; # If previous element was 1 then # only 0 can be appended b[i] = a[i - 1]; # Store the count of all possible subsets result = a[n - 1] + b[n - 1]; return result; # Driver code if __name__ == "__main__" : arr = [ 3, 5, 7 ]; n = len(arr); print(cntSubsets(arr, n)); # This code is contributed by AnkitRai01 |
C#
// C# implementation of the approach using System; class GFG { // Function to return the count // of possible subsets static int cntSubsets(int []arr, int n) { int []a = new int[n]; int []b = new int[n]; a[0] = b[0] = 1; for (int i = 1; i < n; i++) { // If previous element was 0 then 0 // as well as 1 can be appended a[i] = a[i - 1] + b[i - 1]; // If previous element was 1 then // only 0 can be appended b[i] = a[i - 1]; } // Store the count of all possible subsets int result = a[n - 1] + b[n - 1]; return result; } // Driver code public static void Main(String[] args) { int []arr = { 3, 5, 7 }; int n = arr.Length; Console.WriteLine(cntSubsets(arr, n)); } } // This code is contributed by 29AjayKumar |
5
Method 3; If we take a closer look at the pattern, we can observe that the count is actually (N + 2)th Fibonacci number for N ≥ 1.
n = 1, count = 2 = fib(3)
n = 2, count = 3 = fib(4)
n = 3, count = 5 = fib(5)
n = 4, count = 8 = fib(6)
n = 5, count = 13 = fib(7)
…………….
Therefore, the subsets can be counted in O(log n) time using the method 5 of this article.
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.
