Count of subsets not containing adjacent elements
Last Updated : 29 Aug, 2023
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: Using bit masking
Idea: 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 it AND with the mask. If the result of the AND operation is 0, then the mask does not have consecutive sets 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 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 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# 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 <script> // Javascript implementation of the approach // Function to return the count // of possible subsets function cntSubsets(arr, n) { // Total possible subsets of n // sized array is (2^n - 1) var max = Math.pow(2, n); // To store the required // count of subsets var result = 0; // Run from i 000..0 to 111..1 for (var i = 0; i < max; i++) { var counter = i; // If current subset has consecutive // elements from the array if (counter & (counter >> 1)) continue; result++; } return result; } // Driver code var arr = [3, 5, 7]; var n = arr.length; document.write( cntSubsets(arr, n)); </script> Time Complexity: O(2n), where n is the size of the given array.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
Method 2: The above approach takes exponential time. In the above code, the number of bit-masks without consecutive 1s was 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 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 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# 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 <script> // Javascript implementation of the approach // Function to return the count // of possible subsets function cntSubsets(arr, n) { var a = Array(n); var b = Array(n); a[0] = b[0] = 1; for (var 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 var result = a[n - 1] + b[n - 1]; return result; } // Driver code var arr = [3, 5, 7 ]; var n = arr.length; document.write( cntSubsets(arr, n)); </script> Time Complexity: O(n)
Auxiliary Space: O(n), where n is the size of the given array
Another approach : Space optimized
we can space optimize previous approach by using only two variables to store the previous values instead of two arrays.
Implementation Steps:
- Initialize two variables a and b to 1.
- Use a loop to iterate over the array elements from index 1 to n-1.
- For each element at index i, update a and b using the following formulas:
a = a + b
b = previous value of a - After the loop, compute the total number of possible subsets as the sum of a and b.
- Return the total number of possible subsets.
Implementation:
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 = 1, b = 1; for (int i = 1; i < n; i++) { // If previous element was 0 then 0 // as well as 1 can be appended int temp = a; a = a + b; // If previous element was 1 then // only 0 can be appended b = temp; } // Store the count of all possible subsets int result = a + b; 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 implementation of the approach import java.io.*; class Main { // Function to return the count // of possible subsets static int cntSubsets(int[] arr, int n) { int a = 1, b = 1; for (int i = 1; i < n; i++) { // If previous element was 0 then 0 // as well as 1 can be appended int temp = a; a = a + b; // If previous element was 1 then // only 0 can be appended b = temp; } // Store the count of all possible subsets int result = a + b; 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)); } } # Function to return the count # of possible subsets def cntSubsets(arr, n): a = 1 b = 1 for i in range(1, n): # If previous element was 0 then 0 # as well as 1 can be appended temp = a a = a + b # If previous element was 1 then # only 0 can be appended b = temp # Store the count of all possible subsets result = a + b return result # Driver code arr = [3, 5, 7] n = len(arr) print(cntSubsets(arr, n))
using System; public class Program { public static int CountSubsets(int[] arr, int n) { int a = 1, b = 1; for (int i = 1; i < n; i++) { // If previous element was 0 then 0 // as well as 1 can be appended int temp = a; a = a + b; // If previous element was 1 then // only 0 can be appended b = temp; } // Store the count of all possible subsets int result = a + b; return result; } public static void Main() { int[] arr = { 3, 5, 7 }; int n = arr.Length; Console.WriteLine(CountSubsets(arr, n)); } } // Function to return the count of possible subsets function cntSubsets(arr) { let a = 1, b = 1; for (let i = 1; i < arr.length; i++) { // If previous element was 0 then 0 // as well as 1 can be appended let temp = a; a = a + b; // If previous element was 1 then // only 0 can be appended b = temp; } // Store the count of all possible subsets let result = a + b; return result; } // Driver code const arr = [3, 5, 7]; console.log(cntSubsets(arr)); Time Complexity: O(n)
Auxiliary Space: O(1)
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 method 5 of this article.
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