Given a set S consisting of n numbers, find the sum of difference between last and first element of each subset. We find first and last element of every subset by keeping them in same order as they appear in input set S.

i.e., sumSetDiff(S) = ∑ (last(s) – first(s)),

where sum goes over all subsets s of S.

**Note:** Elements in the subset should be in the same order as in the set S.

Examples:

S = {5, 2, 9, 6}, n = 4 Subsets are: {5}, last(s)-first(s) = 0. {2}, last(s)-first(s) = 0. {9}, last(s)-first(s) = 0. {6}, last(s)-first(s) = 0. {5,2}, last(s)-first(s) = -3. {5,9}, last(s)-first(s) = 4. {5,6}, last(s)-first(s) = 1. {2,9}, last(s)-first(s) = 7. {2,6}, last(s)-first(s) = 4. {9,6}, last(s)-first(s) = -3. {5,2,9}, last(s)-first(s) = 4. {5,2,6}, last(s)-first(s) = 1. {5,9,6}, last(s)-first(s) = 1. {2,9,6}, last(s)-first(s) = 4. {5,2,9,6}, last(s)-first(s) = 1. Output = -3+4+1+7+4-3+4+1+1+4+1 = 21.

**A simple solution** for this problem is to find the difference between the last and first element for each subset s of set S and output the sum of ll these differences. Time complexity for this approach is O(2^{n}).

**An efficient solution** to solve the problem in linear time complexity.

We are given a set S consisting of n numbers, and we need to compute the sum of difference between last and first element of each subset of S, i.e.,

sumSetDiff(S) = ∑ (last(s) – first(s)), where sum goes over all subsets s of S.

Equivalently,

sumSetDiff(S) = ∑ (last(s)) – ∑ (first(s)),

In other words, we can compute the sum of last element of each subset, and the sum of first element of each subset separately, and then compute their difference.

Let us say that the elements of S are {a1, a2, a3,…, an}. Note the following observation:

- Subsets containing element
**a1**as the first element can be obtained by taking any subset of {a2, a3,…, an} and then including a1 into it. Number of such subsets will be 2^{n-1}. - Subsets containing element a2 as the first element can be obtained by taking any subset of {a3, a4,…, an} and then including a2 into it. Number of such subsets will be 2
^{n-2}. - Subsets containing element ai as the first element can be obtained by taking any subset of {ai, a(i+1),…, an} and then including ai into it. Number of such subsets will be 2
^{n-i}. - Sum of maximum and minimum of Kth subset ordered by increasing subset sum
- Maximum Subset Sum possible by negating the entire sum after selecting the first Array element
- Largest subset having with sum less than equal to sum of respective indices
- Sum of absolute differences of all pairs in a given array
- Maximize sum of consecutive differences in a circular array
- Minimum sum of differences with an element in an array
- Minimum and Maximum sum of absolute differences of pairs
- Array element with minimum sum of absolute differences
- Sum of absolute differences of pairs from the given array that satisfy the given condition
- Maximize the sum of differences of consecutive elements after removing exactly K elements
- Minimize the sum of differences of consecutive elements after removing exactly K elements
- Sum of consecutive bit differences of first N non-negative integers
- Sum of all differences between Maximum and Minimum of increasing Subarrays
- Clustering/Partitioning an array such that sum of square differences is minimum
- Sum of bit differences among all pairs
- Array formed using sum of absolute differences of that element with all other elements
- Sum of absolute differences of indices of occurrences of each array element
- Find the smallest positive integer value that cannot be represented as sum of any subset of a given array
- Subset sum queries using bitset
- Subset with no pair sum divisible by K

Therefore, the sum of first element of all subsets will be:

SumF = a1.2^{n-1} + a2.2^{n-2} +…+ an.1

In a similar way we can compute the sum of last element of all subsets of S (Taking at every step ai as last element instead of first element and then obtaining all the subsets).

SumL = a1.1 + a2.2 +…+ an.2^{n-1}

Finally, the answer of our problem will be **SumL – SumF**.

**Implementation:**

## C++

`// A C++ program to find sum of difference between ` `// last and first element of each subset ` `#include<bits/stdc++.h> ` ` ` `// Returns the sum of first elements of all subsets ` `int` `SumF(` `int` `S[], ` `int` `n) ` `{ ` ` ` `int` `sum = 0; ` ` ` ` ` `// Compute the SumF as given in the above explanation ` ` ` `for` `(` `int` `i = 0; i < n; i++) ` ` ` `sum = sum + (S[i] * ` `pow` `(2, n-i-1)); ` ` ` `return` `sum; ` `} ` ` ` `// Returns the sum of last elements of all subsets ` `int` `SumL(` `int` `S[], ` `int` `n) ` `{ ` ` ` `int` `sum = 0; ` ` ` ` ` `// Compute the SumL as given in the above explanation ` ` ` `for` `(` `int` `i = 0; i < n; i++) ` ` ` `sum = sum + (S[i] * ` `pow` `(2, i)); ` ` ` `return` `sum; ` `} ` ` ` `// Returns the difference between sum of last elements of ` `// each subset and the sum of first elements of each subset ` `int` `sumSetDiff(` `int` `S[], ` `int` `n) ` `{ ` ` ` `return` `SumL(S, n) - SumF(S, n); ` `} ` ` ` `// Driver program to test above function ` `int` `main() ` `{ ` ` ` `int` `n = 4; ` ` ` `int` `S[] = {5, 2, 9, 6}; ` ` ` `printf` `(` `"%d\n"` `, sumSetDiff(S, n)); ` ` ` `return` `0; ` `} ` |

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## Java

`// A Java program to find sum of difference ` `// between last and first element of each ` `// subset ` `class` `GFG { ` ` ` ` ` `// Returns the sum of first elements ` ` ` `// of all subsets ` ` ` `static` `int` `SumF(` `int` `S[], ` `int` `n) ` ` ` `{ ` ` ` `int` `sum = ` `0` `; ` ` ` ` ` `// Compute the SumF as given in ` ` ` `// the above explanation ` ` ` `for` `(` `int` `i = ` `0` `; i < n; i++) ` ` ` `sum = sum + (` `int` `)(S[i] * ` ` ` `Math.pow(` `2` `, n - i - ` `1` `)); ` ` ` `return` `sum; ` ` ` `} ` ` ` ` ` `// Returns the sum of last elements ` ` ` `// of all subsets ` ` ` `static` `int` `SumL(` `int` `S[], ` `int` `n) ` ` ` `{ ` ` ` `int` `sum = ` `0` `; ` ` ` ` ` `// Compute the SumL as given in ` ` ` `// the above explanation ` ` ` `for` `(` `int` `i = ` `0` `; i < n; i++) ` ` ` `sum = sum + (` `int` `)(S[i] * ` ` ` `Math.pow(` `2` `, i)); ` ` ` ` ` `return` `sum; ` ` ` `} ` ` ` ` ` `// Returns the difference between sum ` ` ` `// of last elements of each subset and ` ` ` `// the sum of first elements of each ` ` ` `// subset ` ` ` `static` `int` `sumSetDiff(` `int` `S[], ` `int` `n) ` ` ` `{ ` ` ` `return` `SumL(S, n) - SumF(S, n); ` ` ` `} ` ` ` ` ` `// Driver program ` ` ` `public` `static` `void` `main(String arg[]) ` ` ` `{ ` ` ` `int` `n = ` `4` `; ` ` ` `int` `S[] = { ` `5` `, ` `2` `, ` `9` `, ` `6` `}; ` ` ` ` ` `System.out.println(sumSetDiff(S, n)); ` ` ` `} ` `} ` ` ` `// This code is contributed by Anant Agarwal. ` |

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## Python3

`# Python3 program to find sum of ` `# difference between last and ` `# first element of each subset ` ` ` `# Returns the sum of first ` `# elements of all subsets ` `def` `SumF(S, n): ` ` ` ` ` `sum` `=` `0` ` ` ` ` `# Compute the SumF as given ` ` ` `# in the above explanation ` ` ` `for` `i ` `in` `range` `(n): ` ` ` `sum` `=` `sum` `+` `(S[i] ` `*` `pow` `(` `2` `, n ` `-` `i ` `-` `1` `)) ` ` ` `return` `sum` ` ` `# Returns the sum of last ` `# elements of all subsets ` `def` `SumL(S, n): ` ` ` ` ` `sum` `=` `0` ` ` ` ` `# Compute the SumL as given ` ` ` `# in the above explanation ` ` ` `for` `i ` `in` `range` `(n): ` ` ` `sum` `=` `sum` `+` `(S[i] ` `*` `pow` `(` `2` `, i)) ` ` ` `return` `sum` ` ` ` ` `# Returns the difference between sum ` `# of last elements of each subset and ` `# the sum of first elements of each subset ` `def` `sumSetDiff(S, n): ` ` ` ` ` `return` `SumL(S, n) ` `-` `SumF(S, n) ` ` ` `# Driver program ` `n ` `=` `4` `S ` `=` `[` `5` `, ` `2` `, ` `9` `, ` `6` `] ` `print` `(sumSetDiff(S, n)) ` ` ` `# This code is contributed by Anant Agarwal. ` |

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## C#

`// A C# program to find sum of difference ` `// between last and first element of each ` `// subset ` `using` `System; ` `class` `GFG { ` ` ` ` ` `// Returns the sum of first elements ` ` ` `// of all subsets ` ` ` `static` `int` `SumF(` `int` `[]S, ` `int` `n) ` ` ` `{ ` ` ` `int` `sum = 0; ` ` ` ` ` `// Compute the SumF as given in ` ` ` `// the above explanation ` ` ` `for` `(` `int` `i = 0; i < n; i++) ` ` ` `sum = sum + (` `int` `)(S[i] * ` ` ` `Math.Pow(2, n - i - 1)); ` ` ` `return` `sum; ` ` ` `} ` ` ` ` ` `// Returns the sum of last elements ` ` ` `// of all subsets ` ` ` `static` `int` `SumL(` `int` `[]S, ` `int` `n) ` ` ` `{ ` ` ` `int` `sum = 0; ` ` ` ` ` `// Compute the SumL as given in ` ` ` `// the above explanation ` ` ` `for` `(` `int` `i = 0; i < n; i++) ` ` ` `sum = sum + (` `int` `)(S[i] * ` ` ` `Math.Pow(2, i)); ` ` ` ` ` `return` `sum; ` ` ` `} ` ` ` ` ` `// Returns the difference between sum ` ` ` `// of last elements of each subset and ` ` ` `// the sum of first elements of each ` ` ` `// subset ` ` ` `static` `int` `sumSetDiff(` `int` `[]S, ` `int` `n) ` ` ` `{ ` ` ` `return` `SumL(S, n) - SumF(S, n); ` ` ` `} ` ` ` ` ` `// Driver program ` ` ` `public` `static` `void` `Main() ` ` ` `{ ` ` ` `int` `n = 4; ` ` ` `int` `[]S = { 5, 2, 9, 6 }; ` ` ` ` ` `Console.Write(sumSetDiff(S, n)); ` ` ` `} ` `} ` ` ` `// This code is contributed by nitin mittal. ` |

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## PHP

`<?php ` `// A PHP program to find sum ` `// of difference between last ` `// and first element of each subset ` ` ` `// Returns the sum of first ` `// elements of all subsets ` `function` `SumF( ` `$S` `, ` `$n` `) ` `{ ` ` ` `$sum` `= 0; ` ` ` ` ` `// Compute the SumF as given ` ` ` `// in the above explanation ` ` ` `for` `(` `$i` `= 0; ` `$i` `< ` `$n` `; ` `$i` `++) ` ` ` `$sum` `= ` `$sum` `+ (` `$S` `[` `$i` `] * ` ` ` `pow(2, ` `$n` `- ` `$i` `- 1)); ` ` ` `return` `$sum` `; ` `} ` ` ` `// Returns the sum of last ` `// elements of all subsets ` `function` `SumL( ` `$S` `, ` `$n` `) ` `{ ` ` ` `$sum` `= 0; ` ` ` ` ` `// Compute the SumL as given ` ` ` `// in the above explanation ` ` ` `for` `(` `$i` `= 0; ` `$i` `< ` `$n` `; ` `$i` `++) ` ` ` `$sum` `= ` `$sum` `+ (` `$S` `[` `$i` `] * ` ` ` `pow(2, ` `$i` `)); ` ` ` `return` `$sum` `; ` `} ` ` ` `// Returns the difference between ` `// sum of last elements of ` `// each subset and the sum of ` `// first elements of each subset ` `function` `sumSetDiff( ` `$S` `, ` `$n` `) ` `{ ` ` ` `return` `SumL(` `$S` `, ` `$n` `) - SumF(` `$S` `, ` `$n` `); ` `} ` ` ` ` ` `// Driver Code ` ` ` `$n` `= 4; ` ` ` `$S` `= ` `array` `(5, 2, 9, 6); ` ` ` `echo` `sumSetDiff(` `$S` `, ` `$n` `); ` ` ` `// This code is contributed by anuj_67. ` `?> ` |

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Output:

21

Time Complexity : O(n)

This article is contributed by **Akash Aggarwal**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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