Given an array arr[] consisting of N integers, the task is to print the smaller of the two subsets obtained by splitting the array into two subsets such that the sum of the smaller subset is maximized.
Examples:
Input: arr[] = {5, 3, 2, 4, 1, 2}
Output: 4 5
Explanation:
Split the array into two subsets as {4, 5} and {1, 2, 2, 3}.
The subset {4, 5} is of minimum length, i.e. 2, having maximum sum = 4 + 5 = 9.Input: arr[] = {20, 15, 20, 50, 20}
Output: 15 50
Approach: The given problem can be solved by using Hashing and Sorting.
Follow the steps below to solve the problem:
- Initialize a HashMap, say M, to store the frequency of each character of the array arr[].
- Traverse the array arr[] and increment the count of every character in the HashMap M.
- Initialize 2 variables, say S, and flag, to store the sum of the first subset and to store if an answer exists or not respectively.
- Sort the array arr[] in ascending order.
- Initialize an ArrayList, say ans, to store the elements of the resultant subset.
- Traverse the array arr[] in reverse order and perform the following steps:
- Store the frequency of the current character in a variable, say F.
- If (F + ans.size()) is less than (N – (F + ans.size())) then append the element arr[i] in the ArrayList ans F number of times.
- Decrement the value of i by F.
- If the value of S is greater than the sum of the array elements, then mark the flag as true and then break.
- After completing the above steps, if the value of flag is true, then print the ArrayList ans as the resultant subset. Otherwise, print -1.the
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to split array elements// into two subsets having sum of// the smaller subset maximizedstatic void findSubset(vector<int> arr){ // Stores the size of the array int N = arr.size(); // Stores the frequency // of array elements map<int,int> mp; // Stores the total // sum of the array int totSum = 0; // Stores the sum of // the resultant set int s = 0; // Stores if it is possible // to split the array that // satisfies the conditions int flag = 0; // Stores the elements // of the first subseta vector<int> ans; // Traverse the array arr[] for (int i = 0; i < arr.size(); i++) { // Increment total sum totSum += arr[i]; // Increment count of arr[i] mp[arr[i]]=mp[arr[i]]+1; } // Sort the array arr[] sort(arr.begin(),arr.end()); // Stores the index of the // last element of the array int i = N - 1; // Traverse the array arr[] while (i >= 0) { // Stores the frequency // of arr[i] int frq = mp[arr[i]]; // If frq + ans.size() is // at most remaining size if ((frq + ans.size()) < (N - (frq + ans.size()))) { for (int k = 0; k < frq; k++) { // Append arr[i] to ans ans.push_back(arr[i]); // Decrement totSum by arr[i] totSum -= arr[i]; // Increment s by arr[i] s += arr[i]; i--; } } // Otherwise, decrement i // by frq else { i -= frq; } // If s is greater // than totSum if (s > totSum) { // Mark flag 1 flag = 1; break; } } // If flag is equal to 1 if (flag == 1) { // Print the arrList ans for (i = ans.size() - 1; i >= 0; i--) { cout<<ans[i]<<" "; } } // Otherwise, print "-1" else { cout<<-1; }}// Driver Codeint main(){ vector<int> arr = { 5, 3, 2, 4, 1, 2 }; findSubset(arr);}// This code is contributed by mohit kumar 29. |
Java
// Java program for above approachimport java.io.*;import java.lang.*;import java.util.*;class GFG { // Function to split array elements // into two subsets having sum of // the smaller subset maximized static void findSubset(int[] arr) { // Stores the size of the array int N = arr.length; // Stores the frequency // of array elements Map<Integer, Integer> map = new HashMap<>(); // Stores the total // sum of the array int totSum = 0; // Stores the sum of // the resultant set int s = 0; // Stores if it is possible // to split the array that // satisfies the conditions int flag = 0; // Stores the elements // of the first subset ArrayList<Integer> ans = new ArrayList<>(); // Traverse the array arr[] for (int i = 0; i < arr.length; i++) { // Increment total sum totSum += arr[i]; // Increment count of arr[i] map.put(arr[i], map.getOrDefault( arr[i], 0) + 1); } // Sort the array arr[] Arrays.sort(arr); // Stores the index of the // last element of the array int i = N - 1; // Traverse the array arr[] while (i >= 0) { // Stores the frequency // of arr[i] int frq = map.get(arr[i]); // If frq + ans.size() is // at most remaining size if ((frq + ans.size()) < (N - (frq + ans.size()))) { for (int k = 0; k < frq; k++) { // Append arr[i] to ans ans.add(arr[i]); // Decrement totSum by arr[i] totSum -= arr[i]; // Increment s by arr[i] s += arr[i]; i--; } } // Otherwise, decrement i // by frq else { i -= frq; } // If s is greater // than totSum if (s > totSum) { // Mark flag 1 flag = 1; break; } } // If flag is equal to 1 if (flag == 1) { // Print the arrList ans for (i = ans.size() - 1; i >= 0; i--) { System.out.print( ans.get(i) + " "); } } // Otherwise, print "-1" else { System.out.print(-1); } } // Driver Code public static void main(String[] args) { int[] arr = { 5, 3, 2, 4, 1, 2 }; findSubset(arr); }} |
Python3
# Python 3 program for the above approachfrom collections import defaultdict# Function to split array elements# into two subsets having sum of# the smaller subset maximizeddef findSubset(arr): # Stores the size of the array N = len(arr) # Stores the frequency # of array elements mp = defaultdict(int) # Stores the total # sum of the array totSum = 0 # Stores the sum of # the resultant set s = 0 # Stores if it is possible # to split the array that # satisfies the conditions flag = 0 # Stores the elements # of the first subseta ans = [] # Traverse the array arr[] for i in range(len(arr)): # Increment total sum totSum += arr[i] # Increment count of arr[i] mp[arr[i]] = mp[arr[i]]+1 # Sort the array arr[] arr.sort() # Stores the index of the # last element of the array i = N - 1 # Traverse the array arr[] while (i >= 0): # Stores the frequency # of arr[i] frq = mp[arr[i]] # If frq + ans.size() is # at most remaining size if ((frq + len(ans)) < (N - (frq + len(ans)))): for k in range(frq): # Append arr[i] to ans ans.append(arr[i]) # Decrement totSum by arr[i] totSum -= arr[i] # Increment s by arr[i] s += arr[i] i -= 1 # Otherwise, decrement i # by frq else: i -= frq # If s is greater # than totSum if (s > totSum): # Mark flag 1 flag = 1 break # If flag is equal to 1 if (flag == 1): # Print the arrList ans for i in range(len(ans) - 1, -1, -1): print(ans[i], end = " ") # Otherwise, print "-1" else: print(-1)# Driver Codeif __name__ == "__main__": arr = [5, 3, 2, 4, 1, 2] findSubset(arr) # This code is contributed by ukasp. |
C#
// C# program for the above approachusing System;using System.Collections.Generic;class GFG{ // Function to split array elements// into two subsets having sum of// the smaller subset maximizedstatic void findSubset(List<int> arr){ // Stores the size of the array int N = arr.Count; int i; // Stores the frequency // of array elements Dictionary<int,int> mp = new Dictionary<int,int>(); // Stores the total // sum of the array int totSum = 0; // Stores the sum of // the resultant set int s = 0; // Stores if it is possible // to split the array that // satisfies the conditions int flag = 0; // Stores the elements // of the first subseta List<int> ans = new List<int>(); // Traverse the array arr[] for (i = 0; i < arr.Count; i++) { // Increment total sum totSum += arr[i]; // Increment count of arr[i] if(mp.ContainsKey(arr[i])) mp[arr[i]]=mp[arr[i]]+1; else mp.Add(arr[i],1); } // Sort the array arr[] arr.Sort(); // Stores the index of the // last element of the array i = N - 1; // Traverse the array arr[] while (i >= 0) { // Stores the frequency // of arr[i] int frq = mp[arr[i]]; // If frq + ans.size() is // at most remaining size if ((frq + ans.Count) < (N - (frq + ans.Count))) { for (int k = 0; k < frq; k++) { // Append arr[i] to ans ans.Add(arr[i]); // Decrement totSum by arr[i] totSum -= arr[i]; // Increment s by arr[i] s += arr[i]; i--; } } // Otherwise, decrement i // by frq else { i -= frq; } // If s is greater // than totSum if (s > totSum) { // Mark flag 1 flag = 1; break; } } // If flag is equal to 1 if (flag == 1) { // Print the arrList ans for (i = ans.Count - 1; i >= 0; i--) { Console.Write(ans[i]+" "); } } // Otherwise, print "-1" else { Console.Write(-1); }}// Driver Codepublic static void Main(){ List<int> arr = new List<int>(){ 5, 3, 2, 4, 1, 2 }; findSubset(arr);}}// This code is contributed by ipg2016107. |
Javascript
<script>// Javascript program for the above approach// Function to split array elements// into two subsets having sum of// the smaller subset maximizedfunction findSubset(arr){ // Stores the size of the array var N = arr.length; // Stores the frequency // of array elements var mp = new Map(); // Stores the total // sum of the array var totSum = 0; // Stores the sum of // the resultant set var s = 0; // Stores if it is possible // to split the array that // satisfies the conditions var flag = 0; // Stores the elements // of the first subseta var ans = []; // Traverse the array arr[] for (var i = 0; i < arr.length; i++) { // Increment total sum totSum += arr[i]; // Increment count of arr[i] if(mp.has(arr[i])) mp.set(arr[i], mp.get(arr[i])+1) else mp.set(arr[i], 1); } // Sort the array arr[] arr.sort((a,b)=> a-b) // Stores the index of the // last element of the array var i = N - 1; // Traverse the array arr[] while (i >= 0) { // Stores the frequency // of arr[i] var frq = mp.get(arr[i]); // If frq + ans.size() is // at most remaining size if ((frq + ans.length) < (N - (frq + ans.length))) { for (var k = 0; k < frq; k++) { // Append arr[i] to ans ans.push(arr[i]); // Decrement totSum by arr[i] totSum -= arr[i]; // Increment s by arr[i] s += arr[i]; i--; } } // Otherwise, decrement i // by frq else { i -= frq; } // If s is greater // than totSum if (s > totSum) { // Mark flag 1 flag = 1; break; } } // If flag is equal to 1 if (flag == 1) { // Print the arrList ans for (i = ans.length - 1; i >= 0; i--) { document.write( ans[i] + " "); } } // Otherwise, print "-1" else { document.write(-1); }}// Driver Codevar arr = [5, 3, 2, 4, 1, 2 ];findSubset(arr);// This code is contributed by rutvik_56.</script> |
Output:
4 5
Time Complexity: O(N*log N)
Auxiliary Space: O(N)
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