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))/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 <iostream>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
<script> // Javascript 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 function count(arr, N, K) { // Count is used to maintain the // number of continuous K's let count = 0, ans = 0; // Iterating through the array for (let 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; } let arr = [ 1, 0, 0, 1, 1, 0, 0 ]; let N = arr.length; let K = 0; document.write(count(arr, N, K)); //This code is contributed by divyeshrabadiya</script> |
Output:
6
Time Complexity: O(N), where N is the size of the array.
Auxiliary Space: O(1), As constant extra space is used.
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
