Find sum of f(s) for all the chosen sets from the given array

Given an array arr[] of size N and an integer K. The task is to find the sum of f(S) over all the possible sets. For a finite set X, f(X) is max(X) – min(X). Set X contains any K numbers from the given array. Output can be very large, so, output answer modulo 109+7.

Examples:

Input: arr[] = {1, 1, 3, 4}, K = 2
Output: 11
Sets are {1, 1}, {1, 3}, {1, 4}, {1, 3}, {1, 4}, {3, 4} and f(X) are 0, 2, 3, 2, 3, 1.



Input: arr[] = {10, -10, 10, -10, 10, -10}, K = 3
Output: 360
18 sets with f(X) equals to 20 and 2 sets with f(x) equals to 0

Approach: On assuming that arr is sorted beforehand, the idea is to perform precomputation to calculate binomial coefficients fast by precalculating the factorials till N and their inverses. The sum is calculated separately for min and max. In other words, (∑ max(S)) – (∑ min(S)) instead of ∑ f(S).
For simplicity, assume that arri is distinct from each other. The possible value of max(S) is any element of arr. Therefore, by counting the number of S such that max(S) = arri for each i, you can find ∑ max(S). The necessary and sufficient condition of max(S) = arri is S contains arri, and also contains K-1 elements less than arri, so such number can be directly calculated by using binomial coefficients. You can calculate ∑ minS similarly.
If Ai contains duplicates, you can prove that the explanation above also holds if you assume arbitrary order between arr is with the same value (for example, consider a lexicographical order of(arri, i and count the number of elements satisfying max(S) = (Ai, i). Therefore, you can also process in the same way in this case.

Below is the implementation of the above approach:

CPP

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
#define N 100005
#define mod (int)(1e9 + 7)
  
// To store the factorial and the
// factorial mod inverse of a number
int factorial[N], modinverse[N];
  
// Function to find (a ^ m1) % mod
int power(int a, int m1)
{
    if (m1 == 0)
        return 1;
    else if (m1 == 1)
        return a;
    else if (m1 == 2)
        return (1LL * a * a) % mod;
    else if (m1 & 1)
        return (1LL * a
                * power(power(a, m1 / 2), 2))
               % mod;
    else
        return power(power(a, m1 / 2), 2) % mod;
}
  
// Function to find factorial
// of all the numbers
void factorialfun()
{
    factorial[0] = 1;
    for (int i = 1; i < N; i++)
        factorial[i] = (1LL
                        * factorial[i - 1] * i)
                       % mod;
}
  
// Function to find the factorial
// mod inverse of all the numbers
void modinversefun()
{
    modinverse[N - 1]
        = power(factorial[N - 1], mod - 2) % mod;
  
    for (int i = N - 2; i >= 0; i--)
        modinverse[i] = (1LL * modinverse[i + 1]
                         * (i + 1))
                        % mod;
}
  
// Function to return nCr
int binomial(int n, int r)
{
    if (r > n)
        return 0;
  
    int a = (1LL * factorial[n]
             * modinverse[n - r])
            % mod;
  
    a = (1LL * a * modinverse[r]) % mod;
    return a;
}
  
// Function to find sum of f(s) for all
// the chosen sets from the given array
int max_min(int a[], int n, int k)
{
    // Sort the given array
    sort(a, a + n);
  
    // Calculate the factorial and
    // modinverse of all elements
    factorialfun();
    modinversefun();
  
    long long ans = 0;
    k--;
  
    // For all the possible sets
    // Calculate max(S) and min(S)
    for (int i = 0; i < n; i++) {
        int x = n - i - 1;
        if (x >= k)
            ans -= binomial(x, k) * a[i] % mod;
  
        int y = i;
        if (y >= k)
            ans += binomial(y, k) * a[i] % mod;
  
        ans = (ans + mod) % mod;
    }
  
    return (int)(ans);
}
  
// Driver code
int main()
{
    int a[] = { 1, 1, 3, 4 }, k = 2;
    int n = sizeof(a) / sizeof(int);
  
    cout << max_min(a, n, k);
  
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java implementation of the approach
import java.util.*;
  
class GFG{
  
static int N = 100005;
static int mod = 1000000007;
static int temp = 391657242;
  
// To store the factorial and the
// factorial mod inverse of a number
static int []factorial = new int[N];
static int []modinverse = new int[N];
  
// Function to find (a ^ m1) % mod
static int power(int a, int m1)
{
    if (m1 == 0)
        return 1;
    else if (m1 == 1)
        return a;
    else if (m1 == 2)
        return (a * a) % mod;
    else if ((m1 & 1)!=0)
        return (a * power(power(a, m1 / 2), 2)) % mod;
    else
        return power(power(a, m1 / 2), 2) % mod;
}
  
// Function to find factorial
// of all the numbers
static void factorialfun()
{
    factorial[0] = 1;
    for (int i = 1; i < N; i++)
        factorial[i] = (factorial[i - 1] * i)% mod;
}
  
// Function to find the factorial
// mod inverse of all the numbers
static void modinversefun()
{
    modinverse[N - 1] = power(factorial[N - 1], mod - 2) % mod;
  
    for (int i = N - 2; i >= 0; i--)
        modinverse[i] = (modinverse[i + 1]*(i + 1))%mod;
}
  
// Function to return nCr
static int binomial(int n, int r)
{
    if (r > n)
        return 0;
  
    int a = (factorial[n] * modinverse[n - r]) % mod;
  
    a = (a * modinverse[r]) % mod;
    return a;
}
  
// Function to find sum of f(s) for all
// the chosen sets from the given array
static int max_min(int a[], int n, int k)
{
    // Sort the given array
    Arrays.sort(a);
  
    // Calculate the factorial and
    // modinverse of all elements
    factorialfun();
    modinversefun();
  
    int ans = 0;
    k--;
  
    // For all the possible sets
    // Calculate max(S) and min(S)
    for (int i = 0; i < n; i++) {
        int x = n - i - 1;
        if (x >= k)
            ans -= binomial(x, k) * a[i] % mod;
  
        int y = i;
        if (y >= k)
            ans += binomial(y, k) * a[i] % mod;
  
        ans = (ans + mod) % mod;
    }
  
    return ans%temp;
}
  
// Driver code
public static void main(String args[])
{
    int []a = { 1, 1, 3, 4 };
    int k = 2;
    int n = a.length;
  
    System.out.println(max_min(a, n, k));
}
}
  
// This code is contributed by Surendra_Gangwar

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 implementation of the approach
N = 100005
mod = (10 ** 9 + 7)
  
# To store the factorial and the
# factorial mod inverse of a number
factorial = [0]*N
modinverse = [0]*N
  
# Function to find factorial
# of all the numbers
def factorialfun():
    factorial[0] = 1
    for i in range(1, N):
        factorial[i] = (factorial[i - 1] * i)%mod
  
# Function to find the factorial
# mod inverse of all the numbers
def modinversefun():
    modinverse[N - 1] = pow(factorial[N - 1], 
                            mod - 2, mod) % mod
  
    for i in range(N - 2, -1, -1):
        modinverse[i] = (modinverse[i + 1]* (i + 1))% mod
  
# Function to return nCr
def binomial(n, r):
    if (r > n):
        return 0
  
    a = (factorial[n]* modinverse[n - r])% mod
  
    a = (a * modinverse[r]) % mod
    return a
  
# Function to find sum of f(s) for all
# the chosen sets from the given array
def max_min(a, n, k):
  
    # Sort the given array
    a = sorted(a)
  
    # Calculate the factorial and
    # modinverse of all elements
    factorialfun()
    modinversefun()
  
    ans = 0
    k -= 1
  
    # For all the possible sets
    # Calculate max(S) and min(S)
    for i in range(n):
        x = n - i - 1
        if (x >= k):
            ans -= (binomial(x, k) * a[i]) % mod
  
        y = i
        if (y >= k):
            ans += (binomial(y, k) * a[i]) % mod
  
        ans = (ans + mod) % mod
  
    return ans
  
# Driver code
  
a = [1, 1, 3, 4]
k = 2
n = len(a)
  
print(max_min(a, n, k))
  
# This code is contributed by mohit kumar 29

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# implementation of the approach 
using System;
      
class GFG{ 
      
    static int N = 100005; 
    static int mod = 1000000007; 
    static int temp = 391657242; 
      
    // To store the factorial and the 
    // factorial mod inverse of a number 
    static int []factorial = new int[N]; 
    static int []modinverse = new int[N]; 
      
    // Function to find (a ^ m1) % mod 
    static int power(int a, int m1) 
    
        if (m1 == 0) 
            return 1; 
        else if (m1 == 1) 
            return a; 
        else if (m1 == 2) 
            return (a * a) % mod; 
        else if ((m1 & 1)!=0) 
            return (a * power(power(a, m1 / 2), 2)) % mod; 
        else
            return power(power(a, m1 / 2), 2) % mod; 
    
      
    // Function to find factorial 
    // of all the numbers 
    static void factorialfun() 
    
        factorial[0] = 1; 
        for (int i = 1; i < N; i++) 
            factorial[i] = (factorial[i - 1] * i)% mod; 
    
      
    // Function to find the factorial 
    // mod inverse of all the numbers 
    static void modinversefun() 
    
        modinverse[N - 1] = power(factorial[N - 1], mod - 2) % mod; 
      
        for (int i = N - 2; i >= 0; i--) 
            modinverse[i] = (modinverse[i + 1]*(i + 1)) % mod; 
    
      
    // Function to return nCr 
    static int binomial(int n, int r) 
    
        if (r > n) 
            return 0; 
      
        int a = (factorial[n] * modinverse[n - r]) % mod; 
      
        a = (a * modinverse[r]) % mod; 
        return a; 
    
      
    // Function to find sum of f(s) for all 
    // the chosen sets from the given array 
    static int max_min(int []a, int n, int k) 
    
        // Sort the given array 
        Array.Sort(a); 
      
        // Calculate the factorial and 
        // modinverse of all elements 
        factorialfun(); 
        modinversefun(); 
      
        int ans = 0; 
        k--; 
      
        // For all the possible sets 
        // Calculate max(S) and min(S) 
        for (int i = 0; i < n; i++) { 
            int x = n - i - 1; 
            if (x >= k) 
                ans -= binomial(x, k) * a[i] % mod; 
      
            int y = i; 
            if (y >= k) 
                ans += binomial(y, k) * a[i] % mod; 
      
            ans = (ans + mod) % mod; 
        
      
        return ans % temp; 
    
      
    // Driver code 
    public static void Main(string []args) 
    
        int []a = { 1, 1, 3, 4 }; 
        int k = 2; 
        int n = a.Length; 
      
        Console.WriteLine(max_min(a, n, k)); 
    
  
// This code is contributed by AnkitRai01

chevron_right


Output:

11

GeeksforGeeks has prepared a complete interview preparation course with premium videos, theory, practice problems, TA support and many more features. Please refer Placement 100 for details




My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.