Maximum difference between two subsets of m elements Last Updated : 04 Mar, 2025 Summarize Comments Improve Suggest changes Share Like Article Like Report Given an array of n integers and a number m, find the maximum possible difference between two sets of m elements chosen from given array.Examples: Input : arr[] = 1 2 3 4 5 m = 4Output : 4The maximum four elements are 2, 3, 4 and 5. The minimum four elements are 1, 2, 3 and 4. The difference between two sums is (2 + 3 + 4 + 5) - (1 + 2 + 3 + 4) = 4 Input : arr[] = 5 8 11 40 15 m = 2Output : 42The difference is (40 + 15) - (5 + 8). [Naive Solution] - Using Sorting - O(n Log n) Time and O(1) SpaceThe idea is to first sort the array, then find sum of first m elements and sum of last m elements. Finally return difference between two sums. C++ #include <bits/stdc++.h> using namespace std; // utility function int findDiff(vector<int>& arr, int m) { int max = 0, min = 0; // sort array sort(arr.begin(), arr.end()); // Traverse m elements from both ends int n = arr.size(); for (int i = 0; i < m; i++) { min += arr[i]; max += arr[n - 1 - i]; } return (max - min); } // Driver code int main() { vector<int> arr = { 1, 2, 3, 4, 5 }; int m = 4; cout << findDiff(arr, m); return 0; } Java import java.util.Arrays; public class Main { // utility function public static int findDiff(int[] arr, int m) { int max = 0, min = 0; // sort array Arrays.sort(arr); // Traverse m elements from both ends int n = arr.length; for (int i = 0;i < m; i++) { min += arr[i]; max += arr[n - 1 - i]; } return (max - min); } // Driver code public static void main(String[] args) { int[] arr = {1, 2, 3, 4, 5}; int m = 4; System.out.println(findDiff(arr, m)); } } Python def find_diff(arr, m): max_sum = 0 min_sum = 0 # sort array arr.sort() # Traverse m elements from both ends n = len(arr) for i in range(m): min_sum += arr[i] max_sum += arr[n - 1 - i] return max_sum - min_sum # Driver code arr = [1, 2, 3, 4, 5] m = 4 print(find_diff(arr, m)) C# using System; using System.Collections.Generic; using System.Linq; class Program { // utility function static int FindDiff(List<int> arr, int m) { int max = 0, min = 0; // sort array arr.Sort(); // Traverse m elements from both ends int n = arr.Count; for (int i = 0; i < m; i++) { min += arr[i]; max += arr[n - 1 - i]; } return max - min; } // Driver code static void Main() { List<int> arr = new List<int> { 1, 2, 3, 4, 5 }; int m = 4; Console.WriteLine(FindDiff(arr, m)); } } JavaScript function findDiff(arr, m) { let max = 0, min = 0; // sort array arr.sort((a, b) => a - b); // Traverse m elements from both ends let n = arr.length; for (let i = 0; i < m; i++) { min += arr[i]; max += arr[n - 1 - i]; } return max - min; } // Driver code let arr = [1, 2, 3, 4, 5]; let m = 4; console.log(findDiff(arr, m)); Output4[Naive Solution] - Using Heap - O(n Log m) Time and O(m) SpaceThe idea is based on the solution discussed in k largest elements in an arrayCreate a Min Heap to Store m largest. We first push first m elements. For remaining n-m elements, we compare every element with the heap top. If Heap top is smaller, we insert the new element in the min heapIn the same way, create a Max Heap to Store m smallest.Find the sums of elements in both the heaps and return the difference of two sums. C++ #include <iostream> #include <queue> #include <vector> using namespace std; int findDiff(vector<int>& arr, int m) { // Min heap to store the 'm' largest elements priority_queue<int, vector<int>, greater<int>> mnH; // Max heap to store the 'm' smallest elements priority_queue<int> mxH; // Maintain 'm' smallest elements in mxH for (int x : arr) { mxH.push(x); if (mxH.size() > m) mxH.pop(); } // Maintain 'm' largest elements in mnH for (int x : arr) { mnH.push(x); if (mnH.size() > m) mnH.pop(); } // Compute sum of 'm' smallest and 'm' largest int minSum = 0, maxSum = 0; while (!mxH.empty()) minSum += mxH.top(), mxH.pop(); while (!mnH.empty()) maxSum += mnH.top(), mnH.pop(); return maxSum - minSum; } int main() { vector<int> arr = {1, 2, 3, 4, 5}; int m = 4; cout << findDiff(arr, m) << endl; return 0; } Java import java.util.*; class GfG { static int findDiff(int[] arr, int m) { // Min heap to store the 'm' largest elements PriorityQueue<Integer> mnH = new PriorityQueue<>(); // Max heap to store the 'm' smallest elements PriorityQueue<Integer> mxH = new PriorityQueue<>(Collections.reverseOrder()); // Maintain 'm' smallest elements in mxH for (int x : arr) { mxH.add(x); if (mxH.size() > m) mxH.poll(); } // Maintain 'm' largest elements in mnH for (int x : arr) { mnH.add(x); if (mnH.size() > m) mnH.poll(); } // Compute sum of 'm' smallest and 'm' largest int minSum = 0, maxSum = 0; while (!mxH.isEmpty()) minSum += mxH.poll(); while (!mnH.isEmpty()) maxSum += mnH.poll(); return maxSum - minSum; } public static void main(String[] args) { int[] arr = {1, 2, 3, 4, 5}; int m = 4; System.out.println(findDiff(arr, m)); } } Python # Function to find the difference between the sum of m largest and m smallest elements import heapq def find_diff(arr, m): # Min heap to store the 'm' largest elements mnH = [] # Max heap to store the 'm' smallest elements mxH = [] # Maintain 'm' smallest elements in mxH for x in arr: heapq.heappush(mxH, -x) if len(mxH) > m: heapq.heappop(mxH) # Maintain 'm' largest elements in mnH for x in arr: heapq.heappush(mnH, x) if len(mnH) > m: heapq.heappop(mnH) # Compute sum of 'm' smallest and 'm' largest min_sum = -sum(mxH) max_sum = sum(mnH) return max_sum - min_sum arr = [1, 2, 3, 4, 5] m = 4 print(find_diff(arr, m)) Output4 Time Complexity: O(n * log m), this solution can work in O(m + (n-m) Log m) as build heap take linear time. Comment More infoAdvertise with us Next Article Maximum difference between two subsets of m elements T TheGameOf256 Follow Improve Article Tags : Misc Searching Heap DSA Arrays +1 More Practice Tags : ArraysHeapMiscSearching Similar Reads Heap Data Structure A Heap is a complete binary tree data structure that satisfies the heap property: for every node, the value of its children is greater than or equal to its own value. 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