Given an array **arr[]** and a number **K** which is present in the array at least once, the task is to find the number of subsets in the array such that each subset contains only the given value **K**.

**Examples:**

Input:arr[] = {1, 0, 0, 1, 0, 1, 2, 5, 2, 1}, K = 0

Output:4

Explanation:

From the two 0’s present in the array at the index 2 and 3, 3 subsequences can be formed: {0}, {0}, {0, 0}

From the 0 present in the array at the index 5, 1 subsequence can be formed: {0}

Therefore, a total of 4 subsequences are formed.

Input:arr[] = {1, 0, 0, 1, 1, 0, 0, 2, 3, 5}, K = 1

Output:4

**Approach:** In order to find the number of subsets, one observation needs to be made on the number of subsets formed for the different number of elements in the given set.

So, let N be the number of elements for which we need to find the subsets.

Then, if:

N = 1: Only one subset can be formed. N = 2: Three subsets can be formed. N = 3: Six subsets can be formed. N = 4: Ten subsets can be formed. . . .N = K: (K * (K + 1))/2subsets can be formed.

Since we are calculating the number of subsets formed by the continuous occurrence of the value K, the idea is to find the count of continuous K’s present in the given array and find the count by using the given formula.

Below is the implementation of the above approach:

## C++

`// C++ implementation to find the ` `// number of subsets formed by ` `// the given value K ` `#include <iostream> ` `using` `namespace` `std; ` ` ` `// Function to find the number ` `// of subsets formed by the ` `// given value K ` `int` `count(` `int` `arr[], ` `int` `N, ` `int` `K) ` `{ ` ` ` `// Count is used to maintain the ` ` ` `// number of continuous K's ` ` ` `int` `count = 0, ans = 0; ` ` ` ` ` `// Iterating through the array ` ` ` `for` `(` `int` `i = 0; i < N; i++) { ` ` ` ` ` `// If the element in the array ` ` ` `// is equal to K ` ` ` `if` `(arr[i] == K) { ` ` ` `count = count + 1; ` ` ` `} ` ` ` `else` `{ ` ` ` ` ` `// count*(count+1)/2 is the ` ` ` `// total number of subsets ` ` ` `// with only K as their element ` ` ` `ans += (count * (count + 1)) / 2; ` ` ` ` ` `// Change count to 0 because ` ` ` `// other element apart from ` ` ` `// K has been found ` ` ` `count = 0; ` ` ` `} ` ` ` `} ` ` ` ` ` `// To handle the last set of K's ` ` ` `ans = ans + (count * (count + 1)) / 2; ` ` ` `return` `ans; ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `int` `arr[] = { 1, 0, 0, 1, 1, 0, 0 }; ` ` ` `int` `N = ` `sizeof` `(arr) / ` `sizeof` `(` `int` `); ` ` ` `int` `K = 0; ` ` ` ` ` `cout << count(arr, N, K); ` `} ` |

*chevron_right*

*filter_none*

## Java

`// Java implementation to find the ` `// number of subsets formed by ` `// the given value K ` `class` `GFG{ ` ` ` `// Function to find the number ` `// of subsets formed by the ` `// given value K ` `static` `int` `count(` `int` `arr[], ` `int` `N, ` `int` `K) ` `{ ` ` ` `// Count is used to maintain the ` ` ` `// number of continuous K's ` ` ` `int` `count = ` `0` `, ans = ` `0` `; ` ` ` ` ` `// Iterating through the array ` ` ` `for` `(` `int` `i = ` `0` `; i < N; i++) { ` ` ` ` ` `// If the element in the array ` ` ` `// is equal to K ` ` ` `if` `(arr[i] == K) { ` ` ` `count = count + ` `1` `; ` ` ` `} ` ` ` `else` `{ ` ` ` ` ` `// count*(count+1)/2 is the ` ` ` `// total number of subsets ` ` ` `// with only K as their element ` ` ` `ans += (count * (count + ` `1` `)) / ` `2` `; ` ` ` ` ` `// Change count to 0 because ` ` ` `// other element apart from ` ` ` `// K has been found ` ` ` `count = ` `0` `; ` ` ` `} ` ` ` `} ` ` ` ` ` `// To handle the last set of K's ` ` ` `ans = ans + (count * (count + ` `1` `)) / ` `2` `; ` ` ` `return` `ans; ` `} ` ` ` `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` ` ` `int` `arr[] = { ` `1` `, ` `0` `, ` `0` `, ` `1` `, ` `1` `, ` `0` `, ` `0` `}; ` ` ` `int` `N = arr.length; ` ` ` `int` `K = ` `0` `; ` ` ` ` ` `System.out.print(count(arr, N, K)); ` `} ` `} ` ` ` `// This code is contributed by 29AjayKumar ` |

*chevron_right*

*filter_none*

## Python3

` ` `# Python 3 implementation to find the ` `# number of subsets formed by ` `# the given value K ` ` ` `# Function to find the number ` `# of subsets formed by the ` `# given value K ` `def` `count(arr, N, K): ` ` ` `# Count is used to maintain the ` ` ` `# number of continuous K's ` ` ` `count ` `=` `0` ` ` `ans ` `=` `0` ` ` ` ` `# Iterating through the array ` ` ` `for` `i ` `in` `range` `(N): ` ` ` `# If the element in the array ` ` ` `# is equal to K ` ` ` `if` `(arr[i] ` `=` `=` `K): ` ` ` `count ` `=` `count ` `+` `1` ` ` ` ` `else` `: ` ` ` `# count*(count+1)/2 is the ` ` ` `# total number of subsets ` ` ` `# with only K as their element ` ` ` `ans ` `+` `=` `(count ` `*` `(count ` `+` `1` `)) ` `/` `/` `2` ` ` ` ` `# Change count to 0 because ` ` ` `# other element apart from ` ` ` `# K has been found ` ` ` `count ` `=` `0` ` ` ` ` `# To handle the last set of K's ` ` ` `ans ` `=` `ans ` `+` `(count ` `*` `(count ` `+` `1` `)) ` `/` `/` `2` ` ` `return` `ans ` ` ` `# Driver code ` `if` `__name__ ` `=` `=` `'__main__'` `: ` ` ` `arr ` `=` `[` `1` `, ` `0` `, ` `0` `, ` `1` `, ` `1` `, ` `0` `, ` `0` `] ` ` ` `N ` `=` `len` `(arr) ` ` ` `K ` `=` `0` ` ` ` ` `print` `(count(arr, N, K)) ` ` ` `# This code is contributed by Surendra_Gangwar ` |

*chevron_right*

*filter_none*

## C#

`// C# implementation to find the ` `// number of subsets formed by ` `// the given value K ` `using` `System; ` ` ` `class` `GFG{ ` ` ` `// Function to find the number ` `// of subsets formed by the ` `// given value K ` `static` `int` `count(` `int` `[]arr, ` `int` `N, ` `int` `K) ` `{ ` ` ` `// Count is used to maintain the ` ` ` `// number of continuous K's ` ` ` `int` `count = 0, ans = 0; ` ` ` ` ` `// Iterating through the array ` ` ` `for` `(` `int` `i = 0; i < N; i++) ` ` ` `{ ` ` ` ` ` `// If the element in the array ` ` ` `// is equal to K ` ` ` `if` `(arr[i] == K) ` ` ` `{ ` ` ` `count = count + 1; ` ` ` ` ` `} ` ` ` `else` ` ` `{ ` ` ` `// count*(count+1)/2 is the ` ` ` `// total number of subsets ` ` ` `// with only K as their element ` ` ` `ans += (count * (count + 1)) / 2; ` ` ` ` ` `// Change count to 0 because ` ` ` `// other element apart from ` ` ` `// K has been found ` ` ` `count = 0; ` ` ` ` ` `} ` ` ` `} ` ` ` ` ` `// To handle the last set of K's ` ` ` `ans = ans + (count * (count + 1)) / 2; ` ` ` `return` `ans; ` `} ` ` ` `// Driver code ` `public` `static` `void` `Main(String[] args) ` `{ ` ` ` `int` `[]arr = { 1, 0, 0, 1, 1, 0, 0 }; ` ` ` `int` `N = arr.Length; ` ` ` `int` `K = 0; ` ` ` ` ` `Console.Write(count(arr, N, K)); ` `} ` `} ` ` ` `//This is contributed by shivanisinghss2110 ` |

*chevron_right*

*filter_none*

**Output:**

6

**Time Complexity:** *O(N)*, where N is the size of the array.

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.