Count of Subsets containing only the given value K

• Last Updated : 22 Mar, 2021

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:
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:

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))/2 subsets 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 using namespace std; // Function to find the number// of subsets formed by the// given value Kint 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 codeint main(){    int arr[] = { 1, 0, 0, 1, 1, 0, 0 };    int N = sizeof(arr) / sizeof(int);    int K = 0;     cout << count(arr, N, K);}

Java

 // Java implementation to find the// number of subsets formed by// the given value Kclass GFG{  // Function to find the number// of subsets formed by the// given value Kstatic 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 codepublic 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

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 Kdef 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 codeif __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

C#

 // C# implementation to find the// number of subsets formed by// the given value Kusing System; class GFG{ // Function to find the number// of subsets formed by the// given value Kstatic 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 codepublic 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

Javascript


Output:
6

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

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