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Sum of width (max and min diff) of all Subsequences
• Difficulty Level : Hard
• Last Updated : 20 Nov, 2018

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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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:      Below is the implementation of above approach:

## C++

 // CPP implementation of above approach#include using namespace std;  #define MOD 1000000007  // Function to return sum of width of all subsetsint SubseqWidths(int A[], int n){    // Sort the array    sort(A, A + n);      int pow2[n];    pow2 = 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 programint main(){    int A[] = { 5, 6, 4, 3, 8 };      int n = sizeof(A) / sizeof(A);      cout << SubseqWidths(A, n);      return 0;}

## Java

 // Java implementation of above approachimport java.util.Arrays;   class GFG{static int MOD=1000000007;  // Function to return sum of width of all subsetsstatic int SubseqWidths(int[] A, int n){    // Sort the array    Arrays.sort(A);      int[] pow2=new int[n];    pow2 = 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 programpublic 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

## Python

 # Python3 implementation of above approach  # Function to return sum of width of all subsetsdef SubseqWidths(A):    MOD = 10**9 + 7    N = len(A)    A.sort()      pow2 =     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 programA = [5, 6, 4, 3, 8]  print(SubseqWidths(A))

## C#

 // C# implementation of above approachusing System;  class GFG{static int MOD = 1000000007;  // Function to return sum of // width of all subsetsstatic int SubseqWidths(int[] A, int n){    // Sort the array    Array.Sort(A);      int[] pow2 = new int[n];    pow2 = 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 Codestatic 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

## PHP

 
Output:
87


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

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