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:
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
. Similarly, the number of sets of size K having arr[i] as the minimum element are 
. - The value of
can be calculated efficiently by using the approach discussed in this article.
. Similarly, the number of sets of size K having arr[i] as the minimum element are 
.
can be calculated efficiently by using the approach discussed in 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}](data:image/svg+xml,%3Csvg%20xmlns='http://www.w3.org/2000/svg'%20viewBox='0%200%20157%2030'%3E%3C/svg%3E "Rendered by QuickLaTeX.com")
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}](data:image/svg+xml,%3Csvg%20xmlns='http://www.w3.org/2000/svg'%20viewBox='0%200%20218%2030'%3E%3C/svg%3E "Rendered by QuickLaTeX.com")
into sumMin. - The value of sumMax – sumMin is the required answer.
![arr[i] * {}^{i}C_{K - 1}](https://neveropen.co.uk/wp-content/uploads/2023/11/rubhabha_quicklatex.com-d99c5617f5b2a3341557361a6a1e205a_l3.png "Rendered by QuickLaTeX.com")
into sumMax.![arr[i] * {}^{N - i - 1}C_{K - 1}](https://neveropen.co.uk/wp-content/uploads/2023/11/wardslaus_rubhabha_quicklatex.com-b74668ec5b392477ae9e97aed9dc3208_l3.png "Rendered by QuickLaTeX.com")
into sumMin.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 1000000007int 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 elementsint 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 Codesigned 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 approachimport 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 approachusing 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 approachmax1 = 100000mod = 1000000007inv = [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 elementsdef 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 Codeif __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 approachlet 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 elementsfunction 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 Codelet 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)
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