Sum of subset differences
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 2n-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 2n-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 2n-i.
- 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;
}
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.
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.
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.
Javascript
// Returns the sum of first elements of all subsets
function
sumF(S, n) {
let sum = 0;
// Compute the SumF as given in the above explanation
for
(let i = 0; i < n; i++) {
sum += S[i] * Math.pow(2, n - i - 1);
}
return
sum;
}
// Returns the sum of last elements of all subsets
function
sumL(S, n) {
let sum = 0;
// Compute the SumL as given in the above explanation
for
(let i = 0; i < n; i++) {
sum += 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
function
sumSetDiff(S, n) {
return
sumL(S, n) - sumF(S, n);
}
// Driver program to test the above functions
function
main() {
const n = 4;
const S = [5, 2, 9, 6];
console.log(sumSetDiff(S, n));
}
main();
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.
?>
- Output:
-
21
- Time Complexity : O(n) This article is contributed by
- Akash Aggarwal
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