Skip to content
Related Articles

Related Articles

Total number of Subsets of size at most K

Improve Article
Save Article
  • Last Updated : 21 Jan, 2022
Improve Article
Save Article

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:
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: 

  1. 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 
    \LARGE \sum _{i = 1}^{K} \text{ } ^{N}\textrm{C}_{i} =\text{ } ^{N}\textrm{C}_{1} + ^{N}\textrm{C}_{2} + ^{N}\textrm{C}_{3} + ...+ ^{N}\textrm{C}_{K}
     
  2. Inorder to calculate the value of NCK, Binomial Coefficient is used. Please refer this article to see how it works.
     
  3. 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++




// 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 k
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
int main()
{
    int n = 3, k = 2;
    cout << count(n, k) << endl;
 
    n = 5, k = 2;
    cout << count(n, k) << endl;
    return 0;
}

Java




// Java code to find total number of
// Subsets of size at most K
import 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 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(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

Python3




# 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 k
def 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 code
if __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#




// C# code to find total number of
// Subsets of size at most K
using 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

Javascript




<script>
// Javascript implementation of the
// above approach
 
// Function for the binomial coefficient
function 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 k
function 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 code
var 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>

Output

6
15

Time Complexity: O(n2 * k)

Auxiliary Space: O(n + k)


My Personal Notes arrow_drop_up
Related Articles

Start Your Coding Journey Now!