Sum of width (max and min diff) of all Subsequences

Given an array A[] of integers. The task is to return the sum of the width of all subsequences of A. For any sequence S, the width of S is the difference between the maximum and minimum elements of S.

Note: Since the answer can be large, print the answer modulo 10^9 + 7.

Examples:

Input : A[] = {1, 3, 2}
Output : 6
Subsequences are {1}, {2}, {3}, {1, 3}, {1, 2} {3, 2} and {1, 3, 2}. Widths are 0, 0, 0, 2, 1, 1 and 2 respectively. Sum of widths is 6.

Input : A[] = [5, 6, 4, 3, 8]
Output : 87



Input : A[] = [1, 2, 3, 4, 5, 6, 7]
Output : 522

The idea is to first, sort the array as sorting the array won’t affect the final answer. After sorting, this allows us to know that the number of subsequences with minimum A[i] and maximum A[j] will be 2j-i-1.

Hence our answer boils down to finding:

 
\sum_{j>i}(2^{j-i-1})(A_{j}-A_{i})

= (\sum_{i=0}^{n-2}\sum_{j=i+1}^{n-1}(2^{j-i-1})(A_{j}))-(\sum_{i=0}^{n-2}\sum_{j=i+1}^{n-1}(2^{j-i-1})(A_{i})) 

= ((2^{0}A_{1}+2^{1}A_{2}+2^{2}A_{3}+...)+(2^{0}A_{2}+2^{1}A_{3}+2^{2}A_{4}+...)+(2^{0}A_{3}+2^{1}A_{4}+2^{2}A_{5}+...)+...)-      (\sum_{i=0}^{n-2}(2^{0}+2^{1}+...+2^{N-i-2})A_{i}+...)

= (\sum_{j=1}^{n-1}(2^{j}-1)A_{j})-(\sum_{i=0}^{n-2}(2^{N-i-1}-1)A_{i})

= \sum_{i=0}^{n-1}((2^{i}-1)A_{i}-(2^{N-i-1}-1)A_{i})

= \sum_{i=0}^{n-1}(2^{i}-2^{N-i-1})A_{i}

Below is the implementation of above approach:

C++

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// CPP implementation of above approach
#include <bits/stdc++.h>
using namespace std;
  
#define MOD 1000000007
  
// Function to return sum of width of all subsets
int SubseqWidths(int A[], int n)
{
    // Sort the array
    sort(A, A + n);
  
    int pow2[n];
    pow2[0] = 1;
  
    for (int i = 1; i < n; ++i)
        pow2[i] = (pow2[i - 1] * 2) % MOD;
  
    int ans = 0;
  
    for (int i = 0; i < n; ++i)
        ans = (ans + (pow2[i] - pow2[n - 1 - i]) * A[i]) % MOD;
  
    return ans;
}
  
// Driver program
int main()
{
    int A[] = { 5, 6, 4, 3, 8 };
  
    int n = sizeof(A) / sizeof(A[0]);
  
    cout << SubseqWidths(A, n);
  
    return 0;
}

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// Java implementation of above approach 


import java.util.Arrays;  


  


class GFG{ 


static int MOD=1000000007; 


  


// Function to return sum of width of all subsets 


static int SubseqWidths(int[] A, int n) 


{ 


    // Sort the array 


    Arrays.sort(A); 


  


    int[] pow2=new int[n]; 


    pow2[0] = 1; 


  


    for (int i = 1; i < n; ++i) 


        pow2[i] = (pow2[i - 1] * 2) % MOD; 


  


    int ans = 0; 


  


    for (int i = 0; i < n; ++i) 


        ans = (ans + (pow2[i] -  


                pow2[n - 1 - i]) * A[i]) % MOD; 


  


    return ans; 


} 


  


// Driver program 


public static void main(String[] args) 


{ 


    int[] A = new int[]{ 5, 6, 4, 3, 8 }; 


  


    int n = A.length; 


  


    System.out.println(SubseqWidths(A, n)); 


} 


} 


// This code is contributed by mits 



















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Python














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# Python3 implementation of above approach 


  


# Function to return sum of width of all subsets 


def SubseqWidths(A): 


    MOD = 10**9 + 7


    N = len(A) 


    A.sort() 


  


    pow2 = [1] 


    for i in range(1, N): 


        pow2.append(pow2[-1] * 2 % MOD) 


  


    ans = 0


    for i, x in enumerate(A): 


        ans = (ans + (pow2[i] - pow2[N - 1 - i]) * x) % MOD 


    return ans 


  


  


# Driver program 


A = [5, 6, 4, 3, 8] 


  


print(SubseqWidths(A)) 



















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// C# implementation of above approach 


using System; 


  


class GFG 


{ 


static int MOD = 1000000007; 


  


// Function to return sum of  


// width of all subsets 


static int SubseqWidths(int[] A, int n) 


{ 


    // Sort the array 


    Array.Sort(A); 


  


    int[] pow2 = new int[n]; 


    pow2[0] = 1; 


  


    for (int i = 1; i < n; ++i) 


        pow2[i] = (pow2[i - 1] * 2) % MOD; 


  


    int ans = 0; 


  


    for (int i = 0; i < n; ++i) 


        ans = (ans + (pow2[i] -  


                       pow2[n - 1 - i]) *  


                       A[i]) % MOD; 


  


    return ans; 


} 


  


// Driver Code 


static void Main() 


{ 


    int[] A = new int[]{ 5, 6, 4, 3, 8 }; 


  


    int n = A.Length; 


      


    Console.WriteLine(SubseqWidths(A, n)); 


} 


} 


  


// This code is contributed by mits 



















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<?php  


// PHP implementation of above approach 


$MOD = 1000000007; 


  


// Function to return sum  


// of width of all subsets 


function SubseqWidths(&$A, $n) 


{ 


    global $MOD; 


      


    // Sort the array 


    sort($A); 


  


    $pow2 = array_fill(0, $n, NULL); 


    $pow2[0] = 1; 


  


    for ($i = 1; $i < $n; ++$i) 


        $pow2[$i] = ($pow2[$i - 1] * 2) % $MOD; 


  


    $ans = 0; 


  


    for ($i = 0; $i < $n; ++$i) 


        $ans = ($ans + ($pow2[$i] -  


                        $pow2[$n - 1 - $i]) *  


                              $A[$i]) % $MOD; 


  


    return $ans; 


} 


  


// Driver Code 


$A = array(5, 6, 4, 3, 8 ); 


  


$n = sizeof($A); 


  


echo SubseqWidths($A, $n); 


  


// This code is contributed  


// by ChitraNayal 


?> 



















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Output:


87





Time Complexity: O(N*log(N)), where N is the length of A.


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