# Count of subsets not containing adjacent elements

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:**

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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 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

`// 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` |

## Javascript

`<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>` |

**Output:**

5

**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

`// 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` |

## Javascript

`<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>` |

**Output:**

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 method 5 of this article.