Sum of (maximum element - minimum element) for all the subsets of an array.
Last Updated :
11 Aug, 2021
Given an array arr[], the task is to compute the sum of (max{A} - min{A}) for every non-empty subset A of the array arr[].
Examples:
Input: arr[] = { 4, 7 }
Output: 3
There are three non-empty subsets: { 4 }, { 7 } and { 4, 7 }.
max({4}) - min({4}) = 0
max({7}) - min({7}) = 0
max({4, 7}) - min({4, 7}) = 7 - 4 = 3.
Sum = 0 + 0 + 3 = 3
Input: arr[] = { 4, 3, 1 }
Output: 9
A naive solution is to generate all subsets and traverse every subset to find the maximum and minimum element and add their difference to the current sum. The time complexity of this solution is O(n * 2n).
An efficient solution is based on a simple observation stated below.
For example, A = { 4, 3, 1 }
Let value to be added in the sum for every subset be V.
Subsets with max, min and V values:
{ 4 }, max = 4, min = 4 (V = 4 - 4)
{ 3 }, max = 3, min = 3 (V = 3 - 3)
{ 1 }, max = 1, min = 1 (V = 1 - 1)
{ 4, 3 }, max = 4, min = 3 (V = 4 - 3)
{ 4, 1 }, max = 4, min = 1 (V = 4 - 1)
{ 3, 1 }, max = 3, min = 1 (V = 3 - 1)
{ 4, 3, 1 }, max = 4, min = 1 (V = 4 - 1)
Sum of all V values
= (4 - 4) + (3 - 3) + (1 - 1) + (4 - 3) + (4 - 1) + (3 - 1) + (4 - 1)
= 0 + 0 + 0 + (4 - 3) + (4 - 1) + (3 - 1) + (4 - 1)
= (4 - 3) + (4 - 1) + (3 - 1) + (4 - 1)
First 3 'V' values can be ignored since they evaluate to 0
(because they result from 1-sized subsets).
Rearranging the sum, we get:
= (4 - 3) + (4 - 1) + (3 - 1) + (4 - 1)
= (1 * 0 - 1 * 3) + (3 * 1 - 3 * 1) + (4 * 3 - 4 * 0)
= (1 * A - 1 * B) + (3 * C - 3 * D) + (4 * E - 4 * F)
where A = 0, B = 3, C = 1, D = 1, E = 3 and F = 0
If we closely look at the expression, instead of analyzing every subset, here we analyze every element of how many times it occurs as a minimum or a maximum element.
A = 0 implies that 1 doesn't occur as a maximum element in any of the subsets.
B = 3 implies that 1 occurs as a minimum element in 3 subsets.
C = 1 implies that 3 occurs as a maximum element in 1 subset.
D = 1 implies that 3 occurs as a minimum element in 1 subset.
E = 3 implies that 4 occurs as a maximum element in 3 subsets.
F = 0 implies that 4 doesn't occur as a minimum element in any of the subsets.
If we somehow know the count of subsets for every element in which it occurs as a maximum element and a minimum element then we can solve the problem in linear time, since the computation above is linear in nature.
Let A = { 6, 3, 89, 21, 4, 2, 7, 9 }
sorted(A) = { 2, 3, 4, 6, 7, 9, 21, 89 }
For example, we analyze element with value 6 (marked in bold). 3 elements are smaller than 6 and 4 elements are larger than 6. Therefore, if we think of all subsets in which 6 occurs with the 3 smaller elements, then in all those subsets 6 will be the maximum element. No of those subsets will be 23. Similar argument holds for 6 being the minimum element when it occurs with the 4 elements greater than 6.
Hence,
No of occurrences for an element as the maximum in all subsets = 2pos - 1
No of occurrences for an element as the minimum in all subsets = 2n - 1 - pos - 1
where pos is the index of the element in the sorted array.
Below is the implementation of the above approach.
C++
// C++ implementation of the above approach
#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int mod = 1000000007;
// Function to return a^n % mod
ll power(ll a, ll n)
{
if (n == 0)
return 1;
ll p = power(a, n / 2) % mod;
p = (p * p) % mod;
if (n & 1) {
p = (p * a) % mod;
}
return p;
}
// Compute sum of max(A) - min(A) for all subsets
ll computeSum(int* arr, int n)
{
// Sort the array.
sort(arr, arr + n);
ll sum = 0;
for (int i = 0; i < n; i++) {
// Maxs = 2^i - 1
ll maxs = (power(2, i) - 1 + mod) % mod;
maxs = (maxs * arr[i]) % mod;
// Mins = 2^(n-1-i) - 1
ll mins = (power(2, n - 1 - i) - 1 + mod) % mod;
mins = (mins * arr[i]) % mod;
ll V = (maxs - mins + mod) % mod;
sum = (sum + V) % mod;
}
return sum;
}
// Driver code
int main()
{
int arr[] = { 4, 3, 1 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << computeSum(arr, n);
return 0;
}
// Java implementation of the above approach
import java.util.*;
class GFG
{
static int mod = 1000000007;
// Function to return a^n % mod
static long power(long a, long n)
{
if (n == 0)
{
return 1;
}
long p = power(a, n / 2) % mod;
p = (p * p) % mod;
if (n == 1)
{
p = (p * a) % mod;
}
return p;
}
// Compute sum of max(A) - min(A) for all subsets
static long computeSum(int[] arr, int n)
{
// Sort the array.
Arrays.sort(arr);
long sum = 0;
for (int i = 0; i < n; i++)
{
// Maxs = 2^i - 1
long maxs = (power(2, i) - 1 + mod) % mod;
maxs = (maxs * arr[i]) % mod;
// Mins = 2^(n-1-i) - 1
long mins = (power(2, n - 1 - i) - 1 + mod) % mod;
mins = (mins * arr[i]) % mod;
long V = (maxs - mins + mod) % mod;
sum = (sum + V) % mod;
}
return sum;
}
// Driver code
public static void main(String[] args)
{
int arr[] = {4, 3, 1};
int n = arr.length;
System.out.println(computeSum(arr, n));
}
}
// This code has been contributed by 29AjayKumar
# Python3 implementation of the
# above approach
# Function to return a^n % mod
def power(a, n):
if n == 0:
return 1
p = power(a, n // 2) % mod
p = (p * p) % mod
if n & 1 == 1:
p = (p * a) % mod
return p
# Compute sum of max(A) - min(A)
# for all subsets
def computeSum(arr, n):
# Sort the array.
arr.sort()
Sum = 0
for i in range(0, n):
# Maxs = 2^i - 1
maxs = (power(2, i) - 1 + mod) % mod
maxs = (maxs * arr[i]) % mod
# Mins = 2^(n-1-i) - 1
mins = (power(2, n - 1 - i) -
1 + mod) % mod
mins = (mins * arr[i]) % mod
V = (maxs - mins + mod) % mod
Sum = (Sum + V) % mod
return Sum
# Driver code
if __name__ =="__main__":
mod = 1000000007
arr = [4, 3, 1]
n = len(arr)
print(computeSum(arr, n))
# This code is contributed
# by Rituraj Jain
// C# implementation of the above approach
using System;
using System.Collections;
class GFG
{
static int mod = 1000000007;
// Function to return a^n % mod
static long power(long a, long n)
{
if (n == 0)
{
return 1;
}
long p = power(a, n / 2) % mod;
p = (p * p) % mod;
if (n == 1)
{
p = (p * a) % mod;
}
return p;
}
// Compute sum of max(A) - min(A) for all subsets
static long computeSum(int []arr, int n)
{
// Sort the array.
Array.Sort(arr);
long sum = 0;
for (int i = 0; i < n; i++)
{
// Maxs = 2^i - 1
long maxs = (power(2, i) - 1 + mod) % mod;
maxs = (maxs * arr[i]) % mod;
// Mins = 2^(n-1-i) - 1
long mins = (power(2, n - 1 - i) - 1 + mod) % mod;
mins = (mins * arr[i]) % mod;
long V = (maxs - mins + mod) % mod;
sum = (sum + V) % mod;
}
return sum;
}
// Driver code
public static void Main()
{
int []arr = {4, 3, 1};
int n = arr.Length;
Console.WriteLine(computeSum(arr, n));
}
}
// This code has been contributed by mits
<?php
// PHP implementation of the above approach
$mod = 1000000007;
// Function to return a^n % mod
function power($a, $n)
{
global $mod;
if ($n == 0)
return 1;
$p = power($a, $n / 2) % $mod;
$p = ($p * $p) % $mod;
if ($n & 1)
{
$p = ($p * $a) % $mod;
}
return $p;
}
// Compute sum of max(A) - min(A)
// for all subsets
function computeSum(&$arr, $n)
{
global $mod;
// Sort the array.
sort($arr);
$sum = 0;
for ($i = 0; $i < $n; $i++)
{
// Maxs = 2^i - 1
$maxs = (power(2, $i) - 1 + $mod) % $mod;
$maxs = ($maxs * $arr[$i]) % $mod;
// Mins = 2^(n-1-i) - 1
$mins = (power(2, $n - 1 - $i) - 1 + $mod) % $mod;
$mins = ($mins * $arr[$i]) % $mod;
$V = ($maxs - $mins + $mod) % $mod;
$sum = ($sum + $V) % $mod;
}
return $sum;
}
// Driver code
$arr = array( 4, 3, 1 );
$n = sizeof($arr);
echo computeSum($arr, $n);
// This code is contributed by ita_c
?>
<script>
// Javascript implementation of the above approach
let mod = 1000000007;
// Function to return a^n % mod
function power(a,n)
{
if (n == 0)
{
return 1;
}
let p = power(a, n / 2) % mod;
p = (p * p) % mod;
if (n == 1)
{
p = (p * a) % mod;
}
return p;
}
// Compute sum of max(A) - min(A) for all subsets
function computeSum(arr,n)
{
// Sort the array.
arr.sort(function(a,b){return a-b;});
let sum = 0;
for (let i = 0; i < n; i++)
{
// Maxs = 2^i - 1
let maxs = (power(2, i) - 1 + mod) % mod;
maxs = (maxs * arr[i]) % mod;
// Mins = 2^(n-1-i) - 1
let mins = (power(2, n - 1 - i) - 1 + mod) % mod;
mins = (mins * arr[i]) % mod;
let V = (maxs - mins + mod) % mod;
sum = (sum + V) % mod;
}
return sum;
}
// Driver code
let arr=[4, 3, 1];
let n = arr.length;
document.write(computeSum(arr, n));
// This code is contributed by rag2127
</script>
Time Complexity: O(N * log(N))
Auxiliary Space: O(1)
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