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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Group words with same set of charactersGiven a list of words with lower cases. Implement a function to find all Words that have the same unique character set. Example: Input: words[] = { "may", "student", "students", "dog", "studentssess", "god", "cat", "act", "tab", "bat", "flow", "wolf", "lambs", "amy", "yam", "balms", "looped", "poodl
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k-th distinct (or non-repeating) element among unique elements in an array.Given an integer array arr[], print kth distinct element in this array. The given array may contain duplicates and the output should print the k-th element among all unique elements. If k is more than the number of distinct elements, print -1.Examples:Input: arr[] = {1, 2, 1, 3, 4, 2}, k = 2Output:
7 min read
Intermediate problems on Hashing
Find Itinerary from a given list of ticketsGiven a list of tickets, find the itinerary in order using the given list.Note: It may be assumed that the input list of tickets is not cyclic and there is one ticket from every city except the final destination.Examples:Input: "Chennai" -> "Bangalore" "Bombay" -> "Delhi" "Goa" -> "Chennai"
11 min read
Find number of Employees Under every ManagerGiven a 2d matrix of strings arr[][] of order n * 2, where each array arr[i] contains two strings, where the first string arr[i][0] is the employee and arr[i][1] is his manager. The task is to find the count of the number of employees under each manager in the hierarchy and not just their direct rep
9 min read
Longest Subarray With Sum Divisible By KGiven an arr[] containing n integers and a positive integer k, he problem is to find the longest subarray's length with the sum of the elements divisible by k.Examples:Input: arr[] = [2, 7, 6, 1, 4, 5], k = 3Output: 4Explanation: The subarray [7, 6, 1, 4] has sum = 18, which is divisible by 3.Input:
10 min read
Longest Subarray with 0 Sum Given an array arr[] of size n, the task is to find the length of the longest subarray with sum equal to 0.Examples:Input: arr[] = {15, -2, 2, -8, 1, 7, 10, 23}Output: 5Explanation: The longest subarray with sum equals to 0 is {-2, 2, -8, 1, 7}Input: arr[] = {1, 2, 3}Output: 0Explanation: There is n
10 min read
Longest Increasing consecutive subsequenceGiven N elements, write a program that prints the length of the longest increasing consecutive subsequence. Examples: Input : a[] = {3, 10, 3, 11, 4, 5, 6, 7, 8, 12} Output : 6 Explanation: 3, 4, 5, 6, 7, 8 is the longest increasing subsequence whose adjacent element differs by one. Input : a[] = {6
10 min read
Count Distinct Elements In Every Window of Size KGiven an array arr[] of size n and an integer k, return the count of distinct numbers in all windows of size k. Examples: Input: arr[] = [1, 2, 1, 3, 4, 2, 3], k = 4Output: [3, 4, 4, 3]Explanation: First window is [1, 2, 1, 3], count of distinct numbers is 3. Second window is [2, 1, 3, 4] count of d
10 min read
Design a data structure that supports insert, delete, search and getRandom in constant timeDesign a data structure that supports the following operations in O(1) time.insert(x): Inserts an item x to the data structure if not already present.remove(x): Removes item x from the data structure if present. search(x): Searches an item x in the data structure.getRandom(): Returns a random elemen
5 min read
Subarray with Given Sum - Handles Negative NumbersGiven an unsorted array of integers, find a subarray that adds to a given number. If there is more than one subarray with the sum of the given number, print any of them.Examples: Input: arr[] = {1, 4, 20, 3, 10, 5}, sum = 33Output: Sum found between indexes 2 and 4Explanation: Sum of elements betwee
13 min read
Implementing our Own Hash Table with Separate Chaining in JavaAll data structure has their own special characteristics, for example, a BST is used when quick searching of an element (in log(n)) is required. A heap or a priority queue is used when the minimum or maximum element needs to be fetched in constant time. Similarly, a hash table is used to fetch, add
10 min read
Implementing own Hash Table with Open Addressing Linear ProbingIn Open Addressing, all elements are stored in the hash table itself. So at any point, size of table must be greater than or equal to total number of keys (Note that we can increase table size by copying old data if needed).Insert(k) - Keep probing until an empty slot is found. Once an empty slot is
13 min read
Maximum possible difference of two subsets of an arrayGiven 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 w
15+ min read
Sorting using trivial hash functionWe have read about various sorting algorithms such as heap sort, bubble sort, merge sort and others. Here we will see how can we sort N elements using a hash array. But this algorithm has a limitation. We can sort only those N elements, where the value of elements is not large (typically not above 1
15+ min read
Smallest subarray with k distinct numbersWe are given an array consisting of n integers and an integer k. We need to find the smallest subarray [l, r] (both l and r are inclusive) such that there are exactly k different numbers. If no such subarray exists, print -1 and If multiple subarrays meet the criteria, return the one with the smalle
14 min read