Total number of Subsets of size at most K

21 Jan, 2022
Given a number N which is the size of the set and a number K, the task is to find the count of subsets, of the set of N elements, having at most K elements in it, i.e. the size of subset is less than or equal to K.Examples:  
Input: N = 3, K = 2 Output: 6 Subsets with 1 element in it = {1}, {2}, {3} Subsets with 2 elements in it = {1, 2}, {1, 3}, {1, 2} Since K = 2, therefore only the above subsets will be considered for length atmost K. Therefore the count is 6.Input: N = 5, K = 2 Output: 15  
 
Approach: 
Since the number of subsets of exactly K elements that can be made from N items is (NCK). Therefore for “at most”, the required count will be  Inorder to calculate the value of NCK, Binomial Coefficient is used. Please refer this article to see how it works. So to get the required subsets for length atmost K, run a loop from 1 to K and add the NCi for each value of i.
Since the number of subsets of exactly K elements that can be made from N items is (NCK). Therefore for “at most”, the required count will be  
Inorder to calculate the value of NCK, Binomial Coefficient is used. Please refer this article to see how it works. 
So to get the required subsets for length atmost K, run a loop from 1 to K and add the NCi for each value of i.
Below is the implementation of the above approach:  
C++
Java
Python3
C#
Javascript
// C++ code to find total number of// Subsets of size at most K #include <bits/stdc++.h>using namespace std; // Function to compute the value// of Binomial Coefficient C(n, k)int binomialCoeff(int n, int k){    int C[n + 1][k + 1];    int i, j;     // Calculate value of Binomial Coefficient    // in bottom up manner    for (i = 0; i <= n; i++) {        for (j = 0; j <= min(i, k); j++) {             // Base Cases            if (j == 0 || j == i)                C[i][j] = 1;             // Calculate value using previously            // stored values            else                C[i][j] = C[i - 1][j - 1] + C[i - 1][j];        }    }     return C[n][k];} // Function to calculate sum of// nCj from j = 1 to kint count(int n, int k){    int sum = 0;    for (int j = 1; j <= k; j++) {         // Calling the nCr function        // for each value of j        sum = sum + binomialCoeff(n, j);    }     return sum;} // Driver codeint main(){    int n = 3, k = 2;    cout << count(n, k) << endl;     n = 5, k = 2;    cout << count(n, k) << endl;    return 0;}
// Java code to find total number of// Subsets of size at most Kimport java.lang.*;class GFG{ // Function to compute the value// of Binomial Coefficient C(n, k)public static int binomialCoeff(int n, int k){    int[][] C = new int[n + 1][k + 1];    int i, j;     // Calculate value of Binomial Coefficient    // in bottom up manner    for (i = 0; i <= n; i++)    {        for (j = 0; j <= Math.min(i, k); j++)        {             // Base Cases            if (j == 0 || j == i)                C[i][j] = 1;             // Calculate value using previously            // stored values            else                C[i][j] = C[i - 1][j - 1] + C[i - 1][j];        }    }     return C[n][k];} // Function to calculate sum of// nCj from j = 1 to kpublic static int count(int n, int k){    int sum = 0;    for (int j = 1; j <= k; j++)    {         // Calling the nCr function        // for each value of j        sum = sum + binomialCoeff(n, j);    }     return sum;} // Driver codepublic static void main(String args[]){    GFG g = new GFG();    int n = 3, k = 2;    System.out.print(count(n, k));     int n1 = 5, k1 = 2;    System.out.print(count(n1, k1));}} // This code is contributed by SoumikMondal
# Python code to find total number of# Subsets of size at most K # Function to compute the value# of Binomial Coefficient C(n, k)def binomialCoeff(n, k):    C = [[0 for i in range(k + 1)] for j in range(n + 1)];    i, j = 0, 0;     # Calculate value of Binomial Coefficient    # in bottom up manner    for i in range(n + 1):        for j in range( min(i, k) + 1):             # Base Cases            if (j == 0 or j == i):                C[i][j] = 1;             # Calculate value using previously            # stored values            else:                C[i][j] = C[i - 1][j - 1] + C[i - 1][j];    return C[n][k]; # Function to calculate sum of# nCj from j = 1 to kdef count(n, k):    sum = 0;    for j in range(1, k+1):         # Calling the nCr function        # for each value of j        sum = sum + binomialCoeff(n, j);    return sum; # Driver codeif __name__ == '__main__':    n = 3;    k = 2;    print(count(n, k), end="");     n1 = 5;    k1 = 2;    print(count(n1, k1)); # This code is contributed by 29AjayKumar
// C# code to find total number of// Subsets of size at most Kusing System; class GFG{     // Function to compute the value    // of Binomial Coefficient C(n, k)    public static int binomialCoeff(int n, int k)    {        int[,] C = new int[n + 1, k + 1];        int i, j;             // Calculate value of Binomial Coefficient        // in bottom up manner        for (i = 0; i <= n; i++)        {            for (j = 0; j <= Math.Min(i, k); j++)            {                     // Base Cases                if (j == 0 || j == i)                    C[i, j] = 1;                     // Calculate value using previously                // stored values                else                    C[i, j] = C[i - 1, j - 1] + C[i - 1, j];            }        }             return C[n, k];    }         // Function to calculate sum of    // nCj from j = 1 to k    public static int count(int n, int k)    {        int sum = 0;        for (int j = 1; j <= k; j++)        {                 // Calling the nCr function            // for each value of j            sum = sum + binomialCoeff(n, j);        }             return sum;    }         // Driver code    public static void Main()    {         int n = 3, k = 2;        Console.Write(count(n, k));             int n1 = 5, k1 = 2;        Console.Write(count(n1, k1));    }} // This code is contributed by AnkitRai01
<script>// Javascript implementation of the// above approach // Function for the binomial coefficientfunction binomialCoeff(n, k){     var C = new Array(n + 1);    // Loop to create 2D array using 1D array    for (var i = 0; i < C.length; i++) {        C[i] = new Array(k + 1);    }    var i, j;       // Calculate value of Binomial Coefficient    // in bottom up manner    for (i = 0; i <= n; i++) {        for (j = 0; j <= Math.min(i, k); j++) {            // Base Cases            if (j == 0 || j == i)                C[i][j] = 1;               // Calculate value using previously            // stored values            else                C[i][j] = C[i - 1][j - 1] + C[i - 1][j];        }    }       return C[n][k];}   // Function to calculate sum of// nCj from j = 1 to kfunction count(n, k){    var sum = 0;    for (var j = 1; j <= k; j++) {           // Calling the nCr function        // for each value of j        sum = sum + binomialCoeff(n, j);    }       return sum;} // Driver codevar n = 3;var k = 2;document.write(count(n, k)); var n = 5;var k = 2;document.write(count(n, k)); // This code is contributed by ShubhamSingh10</script>
6
15
Time Complexity: O(n2 * k)
Auxiliary Space: O(n + k)
SoumikMondal
ankthon
29AjayKumar
nidhi_biet
SHUBHAMSINGH10
anikakapoor
hcetefil
subhammahato348
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