Maximum possible difference of two subsets of an array
Last Updated :
24 Mar, 2023
Given an array of n-integers. The array may contain repetitive elements but the highest frequency of any element must not exceed two. You have to make two subsets such that the difference of the sum of their elements is maximum and both of them jointly contain all elements of the given array along with the most important condition, no subset should contain repetitive elements.
Examples:
Input : arr[] = {5, 8, -1, 4}
Output : Maximum Difference = 18
Explanation :
Let Subset A = {5, 8, 4} & Subset B = {-1}
Sum of elements of subset A = 17, of subset B = -1
Difference of Sum of Both subsets = 17 - (-1) = 18
Input : arr[] = {5, 8, 5, 4}
Output : Maximum Difference = 12
Explanation :
Let Subset A = {5, 8, 4} & Subset B = {5}
Sum of elements of subset A = 17, of subset B = 5
Difference of Sum of Both subsets = 17 - 5 = 12
Before solving this question we have to take care of some given conditions, and they are listed as:
- While building up the subsets, take care that no subset should contain repetitive elements. And for this, we can conclude that all such elements whose frequency are 2, going to be part of both subsets, and hence overall they don't have any impact on the difference of subset-sum. So, we can easily ignore them.
- For making the difference of the sum of elements of both subset maximum we have to make subset in such a way that all positive elements belong to one subset and negative ones to other subsets.
Algorithm with time complexity O(n2):
for i=0 to n-1
isSingleOccurrence = true;
for j= i+1 to n-1
// if frequency of any element is two
// make both equal to zero
if arr[i] equals arr[j]
arr[i] = arr[j] = 0
isSingleOccurrence = false;
break;
if isSingleOccurrence == true
if (arr[i] > 0)
SubsetSum_1 += arr[i];
else
SubsetSum_2 += arr[i];
return abs(SubsetSum_1 - SubsetSum2)
Implementation:
C++
// CPP find maximum difference of subset sum
#include <bits/stdc++.h>
using namespace std;
// function for maximum subset diff
int maxDiff(int arr[], int n)
{
int SubsetSum_1 = 0, SubsetSum_2 = 0;
for (int i = 0; i <= n - 1; i++) {
bool isSingleOccurrence = true;
for (int j = i + 1; j <= n - 1; j++) {
// if frequency of any element is two
// make both equal to zero
if (arr[i] == arr[j]) {
isSingleOccurrence = false;
arr[i] = arr[j] = 0;
break;
}
}
if (isSingleOccurrence) {
if (arr[i] > 0)
SubsetSum_1 += arr[i];
else
SubsetSum_2 += arr[i];
}
}
return abs(SubsetSum_1 - SubsetSum_2);
}
// driver program
int main()
{
int arr[] = { 4, 2, -3, 3, -2, -2, 8 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << "Maximum Difference = " << maxDiff(arr, n);
return 0;
}
// java find maximum difference
// of subset sum
import java .io.*;
public class GFG {
// function for maximum subset diff
static int maxDiff(int []arr, int n)
{
int SubsetSum_1 = 0, SubsetSum_2 = 0;
for (int i = 0; i <= n - 1; i++)
{
boolean isSingleOccurrence = true;
for (int j = i + 1; j <= n - 1; j++)
{
// if frequency of any element
// is two make both equal to
// zero
if (arr[i] == arr[j])
{
isSingleOccurrence = false;
arr[i] = arr[j] = 0;
break;
}
}
if (isSingleOccurrence)
{
if (arr[i] > 0)
SubsetSum_1 += arr[i];
else
SubsetSum_2 += arr[i];
}
}
return Math.abs(SubsetSum_1 - SubsetSum_2);
}
// driver program
static public void main (String[] args)
{
int []arr = { 4, 2, -3, 3, -2, -2, 8 };
int n = arr.length;
System.out.println("Maximum Difference = "
+ maxDiff(arr, n));
}
}
// This code is contributed by vt_m.
# Python3 find maximum difference
# of subset sum
import math
# function for maximum subset diff
def maxDiff(arr, n) :
SubsetSum_1 = 0
SubsetSum_2 = 0
for i in range(0, n) :
isSingleOccurrence = True
for j in range(i + 1, n) :
# if frequency of any element
# is two make both equal to
# zero
if (arr[i] == arr[j]) :
isSingleOccurrence = False
arr[i] = arr[j] = 0
break
if (isSingleOccurrence == True) :
if (arr[i] > 0) :
SubsetSum_1 += arr[i]
else :
SubsetSum_2 += arr[i]
return abs(SubsetSum_1 - SubsetSum_2)
# Driver Code
arr = [4, 2, -3, 3, -2, -2, 8]
n = len(arr)
print ("Maximum Difference = {}"
. format(maxDiff(arr, n)))
# This code is contributed by Manish Shaw
# (manishshaw1)
// C# find maximum difference of
// subset sum
using System;
public class GFG {
// function for maximum subset diff
static int maxDiff(int []arr, int n)
{
int SubsetSum_1 = 0, SubsetSum_2 = 0;
for (int i = 0; i <= n - 1; i++)
{
bool isSingleOccurrence = true;
for (int j = i + 1; j <= n - 1; j++)
{
// if frequency of any element
// is two make both equal to
// zero
if (arr[i] == arr[j])
{
isSingleOccurrence = false;
arr[i] = arr[j] = 0;
break;
}
}
if (isSingleOccurrence)
{
if (arr[i] > 0)
SubsetSum_1 += arr[i];
else
SubsetSum_2 += arr[i];
}
}
return Math.Abs(SubsetSum_1 - SubsetSum_2);
}
// driver program
static public void Main ()
{
int []arr = { 4, 2, -3, 3, -2, -2, 8 };
int n = arr.Length;
Console.WriteLine("Maximum Difference = "
+ maxDiff(arr, n));
}
}
// This code is contributed by vt_m.
<?php
// PHP find maximum difference
// of subset sum
// function for maximum subset diff
function maxDiff($arr, $n)
{
$SubsetSum_1 = 0;
$SubsetSum_2 = 0;
for ($i = 0; $i <= $n - 1; $i++)
{
$isSingleOccurrence = true;
for ($j = $i + 1; $j <= $n - 1; $j++)
{
// if frequency of any element is two
// make both equal to zero
if ($arr[$i] == $arr[$j])
{
$isSingleOccurrence = false;
$arr[$i] = $arr[$j] = 0;
break;
}
}
if ($isSingleOccurrence)
{
if ($arr[$i] > 0)
$SubsetSum_1 += $arr[$i];
else
$SubsetSum_2 += $arr[$i];
}
}
return abs($SubsetSum_1 - $SubsetSum_2);
}
// Driver Code
$arr = array(4, 2, -3, 3, -2, -2, 8);
$n = sizeof($arr);
echo "Maximum Difference = " , maxDiff($arr, $n);
// This code is contributed by nitin mittal
?>
<script>
// JavaScript Program to find maximum difference
// of subset sum
// function for maximum subset diff
function maxDiff(arr, n)
{
let SubsetSum_1 = 0, SubsetSum_2 = 0;
for (let i = 0; i <= n - 1; i++)
{
let isSingleOccurrence = true;
for (let j = i + 1; j <= n - 1; j++)
{
// if frequency of any element
// is two make both equal to
// zero
if (arr[i] == arr[j])
{
isSingleOccurrence = false;
arr[i] = arr[j] = 0;
break;
}
}
if (isSingleOccurrence)
{
if (arr[i] > 0)
SubsetSum_1 += arr[i];
else
SubsetSum_2 += arr[i];
}
}
return Math.abs(SubsetSum_1 - SubsetSum_2);
}
// Driver program
let arr = [ 4, 2, -3, 3, -2, -2, 8 ];
let n = arr.length;
document.write("Maximum Difference = "
+ maxDiff(arr, n));
// This code is contributed by susmitakundugoaldanga.
</script>
OutputMaximum Difference = 20
Time Complexity O(n2)
Auxiliary Space: O(1)
Algorithm with time complexity O(n log n):
-> sort the array
-> for i =0 to n-2
// consecutive two elements are not equal
// add absolute arr[i] to result
if arr[i] != arr[i+1]
result += abs(arr[i])
// else skip next element too
else
i++;
// special check for last two elements
-> if (arr[n-2] != arr[n-1])
result += arr[n-1]
-> return result;
Implementation:
C++
// CPP find maximum difference of subset sum
#include <bits/stdc++.h>
using namespace std;
// function for maximum subset diff
int maxDiff(int arr[], int n)
{
int result = 0;
// sort the array
sort(arr, arr + n);
// calculate the result
for (int i = 0; i < n - 1; i++) {
if (arr[i] != arr[i + 1])
result += abs(arr[i]);
else
i++;
}
// check for last element
if (arr[n - 2] != arr[n - 1])
result += abs(arr[n - 1]);
// return result
return result;
}
// driver program
int main()
{
int arr[] = { 4, 2, -3, 3, -2, -2, 8 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << "Maximum Difference = " << maxDiff(arr, n);
return 0;
}
// java find maximum difference of
// subset sum
import java. io.*;
import java .util.*;
public class GFG {
// function for maximum subset diff
static int maxDiff(int []arr, int n)
{
int result = 0;
// sort the array
Arrays.sort(arr);
// calculate the result
for (int i = 0; i < n - 1; i++)
{
if (arr[i] != arr[i + 1])
result += Math.abs(arr[i]);
else
i++;
}
// check for last element
if (arr[n - 2] != arr[n - 1])
result += Math.abs(arr[n - 1]);
// return result
return result;
}
// driver program
static public void main (String[] args)
{
int[] arr = { 4, 2, -3, 3, -2, -2, 8 };
int n = arr.length;
System.out.println("Maximum Difference = "
+ maxDiff(arr, n));
}
}
// This code is contributed by vt_m.
# Python 3 find maximum difference
# of subset sum
# function for maximum subset diff
def maxDiff(arr, n):
result = 0
# sort the array
arr.sort()
# calculate the result
for i in range(n - 1):
if (abs(arr[i]) != abs(arr[i + 1])):
result += abs(arr[i])
else:
pass
# check for last element
if (arr[n - 2] != arr[n - 1]):
result += abs(arr[n - 1])
# return result
return result
# Driver Code
if __name__ == "__main__":
arr = [ 4, 2, -3, 3, -2, -2, 8 ]
n = len(arr)
print("Maximum Difference = " ,
maxDiff(arr, n))
# This code is contributed by ita_c
// C# find maximum difference
// of subset sum
using System;
public class GFG {
// function for maximum subset diff
static int maxDiff(int []arr, int n)
{
int result = 0;
// sort the array
Array.Sort(arr);
// calculate the result
for (int i = 0; i < n - 1; i++)
{
if (arr[i] != arr[i + 1])
result += Math.Abs(arr[i]);
else
i++;
}
// check for last element
if (arr[n - 2] != arr[n - 1])
result += Math.Abs(arr[n - 1]);
// return result
return result;
}
// driver program
static public void Main ()
{
int[] arr = { 4, 2, -3, 3, -2, -2, 8 };
int n = arr.Length;
Console.WriteLine("Maximum Difference = "
+ maxDiff(arr, n));
}
}
// This code is contributed by vt_m.
<?php
// PHP find maximum difference of subset sum
// function for maximum subset diff
function maxDiff( $arr, $n)
{
$result = 0;
// sort the array
sort($arr);
// calculate the result
for ( $i = 0; $i < $n - 1; $i++)
{
if ($arr[$i] != $arr[$i + 1])
$result += abs($arr[$i]);
else
$i++;
}
// check for last element
if ($arr[$n - 2] != $arr[$n - 1])
$result += abs($arr[$n - 1]);
// return result
return $result;
}
// Driver Code
$arr = array( 4, 2, -3, 3, -2, -2, 8 );
$n = count($arr);
echo "Maximum Difference = "
, maxDiff($arr, $n);
// This code is contributed by anuj_67.
?>
<script>
// Javascript find maximum difference of subset sum
// function for maximum subset diff
function maxDiff(arr, n)
{
var result = 0;
// sort the array
arr.sort((a,b)=> a-b)
// calculate the result
for (var i = 0; i < n - 1; i++) {
if (arr[i] != arr[i + 1])
result += Math.abs(arr[i]);
else
i++;
}
// check for last element
if (arr[n - 2] != arr[n - 1])
result += Math.abs(arr[n - 1]);
// return result
return result;
}
// driver program
var arr = [ 4, 2, -3, 3, -2, -2, 8 ];
var n = arr.length;
document.write( "Maximum Difference = " + maxDiff(arr, n));
</script>
OutputMaximum Difference = 20
Time Complexity: O(n log n)
Auxiliary Space: O(1)
Algorithm with time complexity O(n):
make hash table for positive elements:
for all positive elements(arr[i])
if frequency == 1
SubsetSum_1 += arr[i];
make hash table for negative elements:
for all negative elements
if frequency == 1
SubsetSum_2 += arr[i];
return abs(SubsetSum_1 - SubsetSum2)
Implementation:
C++
// CPP find maximum difference of subset sum
#include <bits/stdc++.h>
using namespace std;
// function for maximum subset diff
int maxDiff(int arr[], int n)
{
unordered_map<int, int> hashPositive;
unordered_map<int, int> hashNegative;
int SubsetSum_1 = 0, SubsetSum_2 = 0;
// construct hash for positive elements
for (int i = 0; i <= n - 1; i++)
if (arr[i] > 0)
hashPositive[arr[i]]++;
// calculate subset sum for positive elements
for (int i = 0; i <= n - 1; i++)
if (arr[i] > 0 && hashPositive[arr[i]] == 1)
SubsetSum_1 += arr[i];
// construct hash for negative elements
for (int i = 0; i <= n - 1; i++)
if (arr[i] < 0)
hashNegative[abs(arr[i])]++;
// calculate subset sum for negative elements
for (int i = 0; i <= n - 1; i++)
if (arr[i] < 0 &&
hashNegative[abs(arr[i])] == 1)
SubsetSum_2 += arr[i];
return abs(SubsetSum_1 - SubsetSum_2);
}
// driver program
int main()
{
int arr[] = { 4, 2, -3, 3, -2, -2, 8 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << "Maximum Difference = " << maxDiff(arr, n);
return 0;
}
// Java find maximum
// difference of subset sum
import java.util.*;
class GFG{
// Function for maximum subset diff
public static int maxDiff(int arr[],
int n)
{
HashMap<Integer,
Integer> hashPositive = new HashMap<>();
HashMap<Integer,
Integer> hashNegative = new HashMap<>();
int SubsetSum_1 = 0,
SubsetSum_2 = 0;
// Construct hash for
// positive elements
for (int i = 0; i <= n - 1; i++)
{
if (arr[i] > 0)
{
if(hashPositive.containsKey(arr[i]))
{
hashPositive.replace(arr[i],
hashPositive.get(arr[i]) + 1);
}
else
{
hashPositive.put(arr[i], 1);
}
}
}
// Calculate subset sum
// for positive elements
for (int i = 0; i <= n - 1; i++)
{
if(arr[i] > 0 &&
hashPositive.containsKey(arr[i]))
{
if(hashPositive.get(arr[i]) == 1)
{
SubsetSum_1 += arr[i];
}
}
}
// Construct hash for
// negative elements
for (int i = 0; i <= n - 1; i++)
{
if (arr[i] < 0)
{
if(hashNegative.containsKey(Math.abs(arr[i])))
{
hashNegative.replace(Math.abs(arr[i]),
hashNegative.get(Math.abs(arr[i])) + 1);
}
else
{
hashNegative.put(Math.abs(arr[i]), 1);
}
}
}
// Calculate subset sum for
// negative elements
for (int i = 0; i <= n - 1; i++)
{
if (arr[i] < 0 &&
hashNegative.containsKey(Math.abs(arr[i])))
{
if(hashNegative.get(Math.abs(arr[i])) == 1)
{
SubsetSum_2 += arr[i];
}
}
}
return Math.abs(SubsetSum_1 - SubsetSum_2);
}
// Driver code
public static void main(String[] args)
{
int arr[] = {4, 2, -3, 3,
-2, -2, 8};
int n = arr.length;
System.out.print("Maximum Difference = " +
maxDiff(arr, n));
}
}
// This code is contributed by divyeshrabadiya07
# Python3 find maximum difference of subset sum
# function for maximum subset diff
def maxDiff(arr, n):
hashPositive = dict()
hashNegative = dict()
SubsetSum_1, SubsetSum_2 = 0, 0
# construct hash for positive elements
for i in range(n):
if (arr[i] > 0):
hashPositive[arr[i]] = \
hashPositive.get(arr[i], 0) + 1
# calculate subset sum for positive elements
for i in range(n):
if (arr[i] > 0 and arr[i] in
hashPositive.keys() and
hashPositive[arr[i]] == 1):
SubsetSum_1 += arr[i]
# construct hash for negative elements
for i in range(n):
if (arr[i] < 0):
hashNegative[abs(arr[i])] = \
hashNegative.get(abs(arr[i]), 0) + 1
# calculate subset sum for negative elements
for i in range(n):
if (arr[i] < 0 and abs(arr[i]) in
hashNegative.keys() and
hashNegative[abs(arr[i])] == 1):
SubsetSum_2 += arr[i]
return abs(SubsetSum_1 - SubsetSum_2)
# Driver Code
arr = [4, 2, -3, 3, -2, -2, 8]
n = len(arr)
print("Maximum Difference =", maxDiff(arr, n))
# This code is contributed by mohit kumar
// C# find maximum
// difference of subset sum
using System;
using System.Collections.Generic;
class GFG {
// Function for maximum subset diff
static int maxDiff(int[] arr, int n)
{
Dictionary<int, int> hashPositive =
new Dictionary<int, int>();
Dictionary<int, int> hashNegative =
new Dictionary<int, int>();
int SubsetSum_1 = 0, SubsetSum_2 = 0;
// Construct hash for
// positive elements
for (int i = 0; i <= n - 1; i++)
{
if (arr[i] > 0)
{
if(hashPositive.ContainsKey(arr[i]))
{
hashPositive[arr[i]] += 1;
}
else
{
hashPositive.Add(arr[i], 1);
}
}
}
// Calculate subset sum
// for positive elements
for (int i = 0; i <= n - 1; i++)
{
if(arr[i] > 0 && hashPositive.ContainsKey(arr[i]))
{
if(hashPositive[arr[i]] == 1)
{
SubsetSum_1 += arr[i];
}
}
}
// Construct hash for
// negative elements
for (int i = 0; i <= n - 1; i++)
{
if (arr[i] < 0)
{
if(hashNegative.ContainsKey(Math.Abs(arr[i])))
{
hashNegative[(Math.Abs(arr[i]))] += 1;
}
else
{
hashNegative.Add(Math.Abs(arr[i]), 1);
}
}
}
// Calculate subset sum for
// negative elements
for (int i = 0; i <= n - 1; i++)
{
if (arr[i] < 0 &&
hashNegative.ContainsKey(Math.Abs(arr[i])))
{
if(hashNegative[(Math.Abs(arr[i]))] == 1)
{
SubsetSum_2 += arr[i];
}
}
}
return Math.Abs(SubsetSum_1 - SubsetSum_2);
}
// Driver code
static void Main() {
int[] arr = {4, 2, -3, 3, -2, -2, 8};
int n = arr.Length;
Console.WriteLine("Maximum Difference = " +
maxDiff(arr, n));
}
}
// This code is contributed by divesh072019
<script>
// Javascript find maximum
// difference of subset sum
// Function for maximum subset diff
function maxDiff(arr,n)
{
let hashPositive = new Map();
let hashNegative = new Map();
let SubsetSum_1 = 0,
SubsetSum_2 = 0;
// Construct hash for
// positive elements
for (let i = 0; i <= n - 1; i++)
{
if (arr[i] > 0)
{
if(hashPositive.has(arr[i]))
{
hashPositive.set(arr[i],
hashPositive.get(arr[i]) + 1);
}
else
{
hashPositive.set(arr[i], 1);
}
}
}
// Calculate subset sum
// for positive elements
for (let i = 0; i <= n - 1; i++)
{
if(arr[i] > 0 &&
hashPositive.has(arr[i]))
{
if(hashPositive.get(arr[i]) == 1)
{
SubsetSum_1 += arr[i];
}
}
}
// Construct hash for
// negative elements
for (let i = 0; i <= n - 1; i++)
{
if (arr[i] < 0)
{
if(hashNegative.has(Math.abs(arr[i])))
{
hashNegative.set(Math.abs(arr[i]),
hashNegative.get(Math.abs(arr[i])) + 1);
}
else
{
hashNegative.set(Math.abs(arr[i]), 1);
}
}
}
// Calculate subset sum for
// negative elements
for (let i = 0; i <= n - 1; i++)
{
if (arr[i] < 0 &&
hashNegative.has(Math.abs(arr[i])))
{
if(hashNegative.get(Math.abs(arr[i])) == 1)
{
SubsetSum_2 += arr[i];
}
}
}
return Math.abs(SubsetSum_1 - SubsetSum_2);
}
// Driver code
let arr = [4, 2, -3, 3,
-2, -2, 8];
let n = arr.length;
document.write("Maximum Difference = " +
maxDiff(arr, n));
// This code is contributed by rag2127
</script>
OutputMaximum Difference = 20
Time Complexity: O(n)
Auxiliary Space: O(n)
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Greedy AlgorithmsGreedy algorithms are a class of algorithms that make locally optimal choices at each step with the hope of finding a global optimum solution. At every step of the algorithm, we make a choice that looks the best at the moment. To make the choice, we sometimes sort the array so that we can always get
3 min read
Graph AlgorithmsGraph is a non-linear data structure like tree data structure. The limitation of tree is, it can only represent hierarchical data. For situations where nodes or vertices are randomly connected with each other other, we use Graph. Example situations where we use graph data structure are, a social net
3 min read
Dynamic Programming or DPDynamic Programming is an algorithmic technique with the following properties.It is mainly an optimization over plain recursion. Wherever we see a recursive solution that has repeated calls for the same inputs, we can optimize it using Dynamic Programming. The idea is to simply store the results of
3 min read
Bitwise AlgorithmsBitwise algorithms in Data Structures and Algorithms (DSA) involve manipulating individual bits of binary representations of numbers to perform operations efficiently. These algorithms utilize bitwise operators like AND, OR, XOR, NOT, Left Shift, and Right Shift.BasicsIntroduction to Bitwise Algorit
4 min read
Advanced
Segment TreeSegment Tree is a data structure that allows efficient querying and updating of intervals or segments of an array. It is particularly useful for problems involving range queries, such as finding the sum, minimum, maximum, or any other operation over a specific range of elements in an array. The tree
3 min read
Pattern SearchingPattern searching algorithms are essential tools in computer science and data processing. These algorithms are designed to efficiently find a particular pattern within a larger set of data. Patten SearchingImportant Pattern Searching Algorithms:Naive String Matching : A Simple Algorithm that works i
2 min read
GeometryGeometry is a branch of mathematics that studies the properties, measurements, and relationships of points, lines, angles, surfaces, and solids. From basic lines and angles to complex structures, it helps us understand the world around us.Geometry for Students and BeginnersThis section covers key br
2 min read
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