Given two positive integers N and K, the task is to find the mean of the minimum of all possible subsets of size K from the first N natural numbers.
Examples:
Input: N = 3, K = 2
Output: 1.33333
Explanation:
All possible subsets of size K are {1, 2}, {1, 3}, {2, 3}. The minimum values in all the subsets are 1, 1, and 2 respectively. The mean of all the minimum values is (1 + 1 + 2)/3 = 1.3333.Input: N = 3, K = 1
Output: 2.00000
Naive Approach: The simplest approach to solve the given problem is to find all the subsets formed from elements [1, N] and find the mean of all the minimum elements of the subsets of size K.
Time Complexity: O(N*2N)
Auxiliary Space: O(1)
Efficient Approach: The above approach can also be optimized by observing the fact that the total number of subsets formed of size K is given by NCK and each element say i occurs (N – i)C(K – 1) number of times as the minimum element.
Therefore, the idea is to find the sum of (N – iCK – 1))*i over all possible values of i and divide it by the total number of subsets formed i.e., NCK.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to find the value of nCrint nCr(int n, int r, int f[]){ // Base Case if (n < r) { return 0; } // Find nCr recursively return f[n] / (f[r] * f[n - r]);}// Function to find the expected minimum// values of all the subsets of size Kint findMean(int N, int X){ // Find the factorials that will // be used later int f[N + 1]; f[0] = 1; // Find the factorials for (int i = 1; i <= N; i++) { f[i] = f[i - 1] * i; } // Total number of subsets int total = nCr(N, X, f); // Stores the sum of minimum over // all possible subsets int count = 0; // Iterate over all possible minimum for (int i = 1; i <= N; i++) { count += nCr(N - i, X - 1, f) * i; } // Find the mean over all subsets double E_X = double(count) / double(total); cout << setprecision(10) << E_X; return 0;}// Driver Codeint main(){ int N = 3, X = 2; findMean(N, X); return 0;} |
Java
// Java program for the above approachimport java.util.*;class GFG { // Function to find the value of nCr static int nCr(int n, int r, int f[]) { // Base Case if (n < r) { return 0; } // Find nCr recursively return f[n] / (f[r] * f[n - r]); } // Function to find the expected minimum // values of all the subsets of size K static int findMean(int N, int X) { // Find the factorials that will // be used later int[] f = new int[N + 1]; f[0] = 1; // Find the factorials for (int i = 1; i <= N; i++) { f[i] = f[i - 1] * i; } // Total number of subsets int total = nCr(N, X, f); // Stores the sum of minimum over // all possible subsets int count = 0; // Iterate over all possible minimum for (int i = 1; i <= N; i++) { count += nCr(N - i, X - 1, f) * i; } // Find the mean over all subsets double E_X = (double) (count) / (double) (total); System.out.print(String.format("%.6f", E_X)); return 0; } // Driver Code public static void main(String[] args) { int N = 3, X = 2; findMean(N, X); }}// This code contributed by Princi Singh |
Python3
# Python 3 program for the above approach# Function to find the value of nCrdef nCr(n, r, f): # Base Case if (n < r): return 0 # Find nCr recursively return f[n] / (f[r] * f[n - r])# Function to find the expected minimum# values of all the subsets of size Kdef findMean(N, X): # Find the factorials that will # be used later f = [0 for i in range(N + 1)] f[0] = 1 # Find the factorials for i in range(1,N+1,1): f[i] = f[i - 1] * i # Total number of subsets total = nCr(N, X, f) # Stores the sum of minimum over # all possible subsets count = 0 # Iterate over all possible minimum for i in range(1,N+1,1): count += nCr(N - i, X - 1, f) * i # Find the mean over all subsets E_X = (count) / (total) print("{0:.9f}".format(E_X)) return 0# Driver Codeif __name__ == '__main__': N = 3 X = 2 findMean(N, X) # This code is contributed by ipg201607. |
C#
// C# code for the above approach using System;public class GFG{ // Function to find the value of nCr static int nCr(int n, int r, int[] f) { // Base Case if (n < r) { return 0; } // Find nCr recursively return f[n] / (f[r] * f[n - r]); } // Function to find the expected minimum // values of all the subsets of size K static int findMean(int N, int X) { // Find the factorials that will // be used later int[] f = new int[N + 1]; f[0] = 1; // Find the factorials for (int i = 1; i <= N; i++) { f[i] = f[i - 1] * i; } // Total number of subsets int total = nCr(N, X, f); // Stores the sum of minimum over // all possible subsets int count = 0; // Iterate over all possible minimum for (int i = 1; i <= N; i++) { count += nCr(N - i, X - 1, f) * i; } // Find the mean over all subsets double E_X = (double) (count) / (double) (total); Console.Write("{0:F9}", E_X); return 0; } // Driver Code static public void Main (){ // Code int N = 3, X = 2; findMean(N, X); }}// This code is contributed by Potta Lokesh |
Javascript
<script>// Javascript program for the above approach// Function to find the value of nCrfunction nCr(n, r, f) { // Base Case if (n < r) { return 0; } // Find nCr recursively return f[n] / (f[r] * f[n - r]);}// Function to find the expected minimum// values of all the subsets of size Kfunction findMean(N, X) { // Find the factorials that will // be used later let f = new Array(N + 1).fill(0); f[0] = 1; // Find the factorials for (let i = 1; i <= N; i++) { f[i] = f[i - 1] * i; } // Total number of subsets let total = nCr(N, X, f); // Stores the sum of minimum over // all possible subsets let count = 0; // Iterate over all possible minimum for (let i = 1; i <= N; i++) { count += nCr(N - i, X - 1, f) * i; } // Find the mean over all subsets let E_X = count / total; document.write(parseFloat(E_X).toFixed(9)); return 0;}// Driver Codelet N = 3, X = 2;findMean(N, X);// This code is contributed by gfgking.</script> |
Output:
1.333333333
Time Complexity: O(N)
Auxiliary Space: O(N)
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