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 arr_i 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) = arr_i for each i, you can find ? max(S). The necessary and sufficient condition of max(S) = arr_i is S contains arr_i, and also contains K-1 elements less than arr_i, so such number can be directly calculated by using binomial coefficients. You can calculate ? minS similarly. If A_i 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(arr_i, i and count the number of elements satisfying max(S) = (A_i, i). Therefore, you can also process in the same way in this case. Below is the implementation of the above approach: 

CPP

// 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;
}

Java

// 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

Python3

# 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

C#

// 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

Javascript

// JS implementation of the approach
let N = 100005
let mod = BigInt(1e9 + 7)

// To store the factorial and the
// factorial mod inverse of a number
let factorial = new Array(N)
let modinverse = new Array(N);

// Function to find (a ^ m1) % mod
function power(a, m1)
{
    if (m1 == 0n)
        return 1n;
    else if (m1 == 1n)
        return a;
    else if (m1 == 2n)
        return (a * a) % mod;
    else if (m1 & 1n)
        return (a
                * power(power(a, m1 / 2n), 2n))
               % mod;
    else
        return power(power(a, m1 / 2n), 2n) % mod;
}

// Function to find factorial
// of all the numbers
function factorialfun()
{
    factorial[0] = 1n;
    for (let i = 1n; i < N; i++)
        factorial[i] = (factorial[i - 1n] * i)
                       % mod;
}

// Function to find the factorial
// mod inverse of all the numbers
function modinversefun()
{
    modinverse[N - 1]
        = power(factorial[N - 1], mod - 2n) % mod;

    N = BigInt(N)
    for (var i = N - 2n; i >= 0n; i--)
        modinverse[i] = (modinverse[i + 1n]
                         * (i + 1n))
                        % mod;
}

// Function to return nCr
function binomial(n, r)
{
    if (r > n)
        return 0n;

    let 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
function max_min(a, n, k)
{

    n = BigInt(n)
    // Sort the given array
    a.sort();

    // Calculate the factorial and
    // modinverse of all elements
    factorialfun();
    modinversefun();

    let ans = 0n;
    k--;

    // For all the possible sets
    // Calculate max(S) and min(S)
    for (var i = 0n; i < n; i++) {
        var x = n - i - 1n;
        if (x >= k)
            ans -= binomial(x, k) * a[i] % mod;

        var y = i;
        if (y >= k)
            ans += binomial(y, k) * a[i] % mod;

        ans = (ans + mod) % mod;
    }

    return (ans);
}

// Driver code
let a= [ 1n, 1n, 3n, 4n ]
let k = 2n;
let n = a.length
console.log(max_min(a, n, k))

// This code is contributed by phasing17

Output:

11