Skip to content
Related Articles

Related Articles

Improve Article
Save Article
Like Article

Find sum of difference of maximum and minimum over all possible subsets of size K

  • Last Updated : 04 Oct, 2021

Given an array arr[] of N integers and an integer K, the task is to find the sum of the difference between the maximum and minimum elements over all possible subsets of size K.

Examples:

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

Input: arr[] = {1, 1, 3, 4}, K = 2
Output: 11
Explanation:
There are 6 subsets of the given array of size K(= 2). They are {1, 1}, {1, 3}, {1, 4}, {1, 3}, {1, 4} and {3, 4}.
The values of maximum – minimum for each of the subsets respectively are 0, 2, 3, 2, 3, 1 and their sum is 11.



Input: arr[] = {1, 1, 1}, K = 1
Output: 0

 

Approach: The given problem can be solved based on the following observations:

  • The sum of the difference between maximum and minimum from all the sets is independent of each other, i.e, it can be calculated as the sum of maximum from all sets of size K – the sum of minimum from all sets of size K.
  • In a sorted array arr[], arr[i] is the maximum of all sets having elements from the array in the range [0, i – 1]. Therefore, the number of sets of size K having arr[i] as the maximum array element can be calculated as {}^{i}C_{K - 1}           . Similarly, the number of sets of size K having arr[i] as the minimum element are {}^{N - i - 1}C_{K - 1}           .
  • The value of {}^{N}C_{R}            can be calculated efficiently by using the approach discussed in this article.

Using the above observations, the given problem can be solved by following the below steps:

  • Sort the given array arr[] in non-decreasing order.
  • In order to calculate the sum of the maximum of all sets of size K, create a variable sumMax, and for each, the index i in the range [K – 1, N – 1], iterate through the array arr[] and add arr[i] * {}^{i}C_{K - 1}            into sumMax.
  • Similarly, In order to calculate the sum of the minimum of all sets of size K, create a variable sumMin and for each i in the range [0, N-K], iterate through the array arr[] and add arr[i] * {}^{N - i - 1}C_{K - 1}            into sumMin.
  • The value of sumMax – sumMin is the required answer.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
#define int long long int
#define max 100000
#define mod 1000000007
 
int inv[max], fact[max], facinv[max];
 
// Function to precompute factorial and
// the inverse of factorial values of all
// elements in the range [1, max] to find
// the value of nCr in O(1)
void ncrPrecomputation()
{
    inv[0] = inv[1] = 1;
    fact[0] = fact[1] = 1;
    facinv[0] = facinv[1] = 1;
 
    // Loop to iterate over all i in
    // the range [2, max]
    for (int i = 2; i < max; i++) {
 
        // Calculate Inverse of i
        inv[i] = inv[mod % i]
                 * (mod - mod / i) % mod;
 
        // Calculate Factorial of i
        fact[i] = (fact[i - 1] * i) % mod;
 
        // Calculate the Inverse of
        // factorial of i
        facinv[i] = (inv[i] * facinv[i - 1]) % mod;
    }
}
 
// Function to find nCr in O(1)
int nCr(int n, int r)
{
    return ((fact[n] * facinv[r]) % mod
            * facinv[n - r])
           % mod;
}
 
// Function to find the sum of difference
// between maximum and minimum over all
// sets of arr[] having K elements
int sumMaxMin(int arr[], int N, int K)
{
    // Sort the given array
    sort(arr, arr + N);
 
    // Stores the sum of maximum of
    // all the sets
    int sumMax = 0;
 
    // Loop to iterate arr[] in the
    // range [K-1, N-1]
    for (int i = K - 1; i < N; i++) {
 
        // Add sum of sets having arr[i]
        // as the maximum element
        sumMax += (arr[i] * nCr(i, K - 1));
    }
 
    // Stores the sum of the minimum of
    // all the sets
    int sumMin = 0;
 
    // Loop to iterate arr[] in the
    // range [0, N - K]
    for (int i = 0; i <= N - K; i++) {
 
        // Add sum of sets having arr[i]
        // as the minimum element
        sumMin += (arr[i] * nCr(N - i - 1, K - 1));
    }
 
    // Return answer
    return (sumMax - sumMin);
}
 
// Driver Code
signed main()
{
    int arr[] = { 1, 1, 3, 4 };
    int K = 2;
    int N = sizeof(arr) / sizeof(arr[0]);
 
    ncrPrecomputation();
 
    cout << sumMaxMin(arr, N, K);
 
    return 0;
}

Java




// Java program for the above approach
import java.io.*;
import java.util.*;
 
class GFG {
 
    static final int max = 100000;
    static final int mod = 1000000007;
 
    static long inv[] = new long[max], fact[] = new long[max],
               facinv[] = new long[max];
 
    // Function to precompute factorial and
    // the inverse of factorial values of all
    // elements in the range [1, max] to find
    // the value of nCr in O(1)
    static void ncrPrecomputation()
    {
        inv[0] = inv[1] = 1;
        fact[0] = fact[1] = 1;
        facinv[0] = facinv[1] = 1;
 
        // Loop to iterate over all i in
        // the range [2, max]
        for (int i = 2; i < max; i++) {
 
            // Calculate Inverse of i
            inv[i] = inv[mod % i] * (mod - mod / i) % mod;
 
            // Calculate Factorial of i
            fact[i] = (fact[i - 1] * i) % mod;
 
            // Calculate the Inverse of
            // factorial of i
            facinv[i] = (inv[i] * facinv[i - 1]) % mod;
        }
    }
 
    // Function to find nCr in O(1)
    static long nCr(long n, long r)
    {
        return ((fact[(int)n] * facinv[(int)r]) % mod * facinv[(int)(n - r)])
            % mod;
    }
 
    // Function to find the sum of difference
    // between maximum and minimum over all
    // sets of arr[] having K elements
    static long sumMaxMin(long arr[], long N, long K)
    {
        // Sort the given array
        Arrays.sort(arr);
 
        // Stores the sum of maximum of
        // all the sets
        long sumMax = 0;
 
        // Loop to iterate arr[] in the
        // range [K-1, N-1]
        for (int i = (int)K - 1; i < N; i++) {
 
            // Add sum of sets having arr[i]
            // as the maximum element
            sumMax += (arr[i] * nCr(i, K - 1));
        }
 
        // Stores the sum of the minimum of
        // all the sets
        long sumMin = 0;
 
        // Loop to iterate arr[] in the
        // range [0, N - K]
        for (int i = 0; i <= N - K; i++) {
 
            // Add sum of sets having arr[i]
            // as the minimum element
            sumMin += (arr[i] * nCr(N - i - 1, K - 1));
        }
 
        // Return answer
        return (sumMax - sumMin);
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        long arr[] = { 1, 1, 3, 4 };
        long K = 2;
        long N = arr.length;
 
        ncrPrecomputation();
 
        System.out.println(sumMaxMin(arr, N, K));
    }
}
 
// This code is contributed by Dharanendra L V.

C#




// C# program for the above approach
using System;
public class GFG
{
 
    static readonly int max = 100000;
    static readonly int mod = 1000000007;
 
    static long []inv = new long[max];
    static long []fact = new long[max];
    static long []facinv = new long[max];
 
    // Function to precompute factorial and
    // the inverse of factorial values of all
    // elements in the range [1, max] to find
    // the value of nCr in O(1)
    static void ncrPrecomputation()
    {
        inv[0] = inv[1] = 1;
        fact[0] = fact[1] = 1;
        facinv[0] = facinv[1] = 1;
 
        // Loop to iterate over all i in
        // the range [2, max]
        for (int i = 2; i < max; i++) {
 
            // Calculate Inverse of i
            inv[i] = inv[mod % i] * (mod - mod / i) % mod;
 
            // Calculate Factorial of i
            fact[i] = (fact[i - 1] * i) % mod;
 
            // Calculate the Inverse of
            // factorial of i
            facinv[i] = (inv[i] * facinv[i - 1]) % mod;
        }
    }
 
    // Function to find nCr in O(1)
    static long nCr(long n, long r)
    {
        return ((fact[(int)n] * facinv[(int)r]) % mod * facinv[(int)(n - r)])
            % mod;
    }
 
    // Function to find the sum of difference
    // between maximum and minimum over all
    // sets of []arr having K elements
    static long sumMaxMin(long []arr, long N, long K)
    {
        // Sort the given array
        Array.Sort(arr);
 
        // Stores the sum of maximum of
        // all the sets
        long sumMax = 0;
 
        // Loop to iterate []arr in the
        // range [K-1, N-1]
        for (int i = (int)K - 1; i < N; i++) {
 
            // Add sum of sets having arr[i]
            // as the maximum element
            sumMax += (arr[i] * nCr(i, K - 1));
        }
 
        // Stores the sum of the minimum of
        // all the sets
        long sumMin = 0;
 
        // Loop to iterate []arr in the
        // range [0, N - K]
        for (int i = 0; i <= N - K; i++) {
 
            // Add sum of sets having arr[i]
            // as the minimum element
            sumMin += (arr[i] * nCr(N - i - 1, K - 1));
        }
 
        // Return answer
        return (sumMax - sumMin);
    }
 
    // Driver Code
    public static void Main(String[] args)
    {
        long []arr = { 1, 1, 3, 4 };
        long K = 2;
        long N = arr.Length;
 
        ncrPrecomputation();
 
        Console.WriteLine(sumMaxMin(arr, N, K));
    }
}
 
// This code is contributed by 29AjayKumar

Python3




# Python 3 program for the above approach
 
max1 = 100000
mod = 1000000007
 
inv = [0 for i in range(max1)]
fact = [0 for i in range(max1)]
facinv = [0 for i in range(max1)]
 
# Function to precompute factorial and
# the inverse of factorial values of all
# elements in the range [1, max] to find
# the value of nCr in O(1)
def ncrPrecomputation():
    inv[0] = inv[1] = 1
    fact[0] = fact[1] = 1
    facinv[0] = facinv[1] = 1
 
    # Loop to iterate over all i in
    # the range [2, max]
    for i in range(2,max1,1):
        # Calculate Inverse of i
        inv[i] = inv[mod % i] * (mod - mod // i) % mod
 
        # Calculate Factorial of i
        fact[i] = (fact[i - 1] * i) % mod
 
        # Calculate the Inverse of
        # factorial of i
        facinv[i] = (inv[i] * facinv[i - 1]) % mod
 
# Function to find nCr in O(1)
def nCr(n,r):
    return ((fact[n] * facinv[r]) % mod * facinv[n - r]) % mod
 
# Function to find the sum of difference
# between maximum and minimum over all
# sets of arr[] having K elements
def sumMaxMin(arr, N, K):
    # Sort the given array
    arr.sort()
 
    # Stores the sum of maximum of
    # all the sets
    sumMax = 0
 
    # Loop to iterate arr[] in the
    # range [K-1, N-1]
    for i in range(K - 1,N,1):
        # Add sum of sets having arr[i]
        # as the maximum element
        sumMax += (arr[i] * nCr(i, K - 1))
 
    # Stores the sum of the minimum of
    # all the sets
    sumMin = 0
 
    # Loop to iterate arr[] in the
    # range [0, N - K]
    for i in range(N - K+1):
        # Add sum of sets having arr[i]
        # as the minimum element
        sumMin += (arr[i] * nCr(N - i - 1, K - 1))
 
    # Return answer
    return (sumMax - sumMin)
 
# Driver Code
if __name__ == '__main__':
    arr = [1, 1, 3, 4]
    K = 2
    N = len(arr)
    ncrPrecomputation()
    print(sumMaxMin(arr, N, K))
     
    # This code is contributed by SURENDRA_GANGWAR

Javascript




<script>
// JaVASCRIPT program for the above approach
 
let max = 100000;
let mod = 1000000007;
 
let inv = new Array(max).fill(0),
  fact = new Array(max).fill(0),
  facinv = new Array(max).fill(0);
 
// Function to precompute factorial and
// the inverse of factorial values of all
// elements in the range [1, max] to find
// the value of nCr in O(1)
function ncrPrecomputation() {
  inv[0] = inv[1] = 1;
  fact[0] = fact[1] = 1;
  facinv[0] = facinv[1] = 1;
 
  // Loop to iterate over all i in
  // the range [2, max]
  for (let i = 2; i < max; i++) {
    // Calculate Inverse of i
    inv[i] = (inv[mod % i] * Math.ceil(mod - mod / i)) % mod;
 
    // Calculate Factorial of i
    fact[i] = (fact[i - 1] * i) % mod;
 
    // Calculate the Inverse of
    // factorial of i
    facinv[i] = (inv[i] * facinv[i - 1]) % mod;
  }
}
 
// Function to find nCr in O(1)
function nCr(n, r) {
  return (((fact[n] * facinv[r]) % mod) * facinv[n - r]) % mod;
}
 
// Function to find the sum of difference
// between maximum and minimum over all
// sets of arr[] having K elements
function sumMaxMin(arr, N, K) {
  // Sort the given array
  arr.sort((a, b) => a - b);
 
  // Stores the sum of maximum of
  // all the sets
  let sumMax = 0;
 
  // Loop to iterate arr[] in the
  // range [K-1, N-1]
  for (let i = K - 1; i < N; i++) {
    // Add sum of sets having arr[i]
    // as the maximum element
    sumMax += arr[i] * nCr(i, K - 1);
  }
 
  // Stores the sum of the minimum of
  // all the sets
  let sumMin = 0;
 
  // Loop to iterate arr[] in the
  // range [0, N - K]
  for (let i = 0; i <= N - K; i++) {
    // Add sum of sets having arr[i]
    // as the minimum element
    sumMin += arr[i] * nCr(N - i - 1, K - 1);
  }
 
  // Return answer
  return sumMax - sumMin;
}
 
// Driver Code
 
let arr = [1, 1, 3, 4];
let K = 2;
let N = arr.length;
 
ncrPrecomputation();
 
document.write(sumMaxMin(arr, N, K));
 
// This code is contributed by saurabh_jaiswal.
</script>
Output: 
11

 

Time Complexity: O(N*log N)
Auxiliary Space: O(N)




My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!