Partition an array of non-negative integers into two subsets such that average of both the subsets is equal

Given an array of size N. The task is to partition the given array into two subsets such that the average of all the elements in both subsets is equal. If no such partition exists print -1. Otherwise, print the partitions. If multiple solutions exist, print the solution where the length of the first subset is minimum. If there is still a tie then print the partitions where the first subset is lexicographically smallest.

Examples: 

Input : vec[] = {1, 7, 15, 29, 11, 9}
Output : [9, 15] [1, 7, 11, 29]
Explanation : Average of the both the subsets is 12
 
Input : vec[] = {1, 2, 3, 4, 5, 6}
Output : [1, 6] [2, 3, 4, 5]. 
Explanation : Another possible solution is [3, 4] [1, 2, 5, 6], 
but print the  solution whose first subset is lexicographically 
smallest.

Observation : 
If we directly compute the average of a certain subset and compare it with another subset’s average, due to precision issues with compilers, unexpected results will occur. For example, 5/3 = 1.66666.. and 166/100 = 1.66. Some compilers might treat them as same, whereas some others won’t.
Let the sum of two subsets under consideration be sub1 and sub2, and let their sizes be s1 and s2. If their averages are equal, sub1/s1 = sub2/s2 . Which means sub1*s2 = sub2*s1. 
Also total sum of the above two subsets = sub1+sub2, and s2= total size – s1. 
On simplifying the above, we get 

(sub1/s1) = (sub1+sub2)/ (s1+s2) = (total sum) / (total size).

Now this problem reduces to the fact that if we can select a particular size 
of subset whose sum is equal to the current subset’s sum, we are done.



Approach : 
Let us define the function partition(ind, curr_sum, curr_size), which returns true if it is possible to construct subset using elements with index equals to ind and having size equals to curr_size and sum equals to curr_sum.
This recursive relation can be defined as:

partition(ind, curr_sum, curr_size) = partition(ind+1, curr_sum, curr_size) || partition(ind+1, curr_sum – val[ind], curr_size-1).

Two parts on the right side of the above equations represent whether we including the element at index ind or not. 
This is a deviation from the classic subset sum problem, in which subproblems are being evaluated again and again. Therefore we memorize the subproblems and turn it into a Dynamic Programming solution. 

C++14

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ program to Partition an array of
// non-negative integers into two subsets
// such that average of both the subsets are equal
#include <bits/stdc++.h>
using namespace std;
 
vector<vector<vector<bool> > > dp;
vector<int> res;
vector<int> original;
int total_size;
 
// Function that returns true if it is possible to
// use elements with index = ind to construct a set of s
// ize = curr_size whose sum is curr_sum.
bool possible(int index, int curr_sum, int curr_size)
{
    // base cases
    if (curr_size == 0)
        return (curr_sum == 0);
    if (index >= total_size)
        return false;
 
    // Which means curr_sum cant be found for curr_size
    if (dp[index][curr_sum][curr_size] == false)
        return false;
 
    if (curr_sum >= original[index]) {
        res.push_back(original[index]);
        // Checks if taking this element at
        // index i leads to a solution
        if (possible(index + 1, curr_sum -
        original[index],
                     curr_size - 1))
            return true;
 
        res.pop_back();
    }
    // Checks if not taking this element at
    // index i leads to a solution
    if (possible(index + 1, curr_sum, curr_size))
        return true;
 
    // If no solution has been found
    return dp[index][curr_sum][curr_size] = false;
}
 
// Function to find two Partitions having equal average
vector<vector<int> > partition(vector<int>& Vec)
{
    // Sort the vector
    sort(Vec.begin(), Vec.end());
    original.clear();
    original = Vec;
    dp.clear();
    res.clear();
 
    int total_sum = 0;
    total_size = Vec.size();
 
    for (int i = 0; i < total_size; ++i)
        total_sum += Vec[i];
    // building the memoization table
    dp.resize(original.size(), vector<vector<bool> >
(total_sum + 1, vector<bool>(total_size, true)));
 
    for (int i = 1; i < total_size; i++) {
        // Sum_of_Set1 has to be an integer
        if ((total_sum * i) % total_size != 0)
            continue;
        int Sum_of_Set1 = (total_sum * i) / total_size;
 
        // We build our solution vector if its possible
        // to find subsets that match our criteria
        // using a recursive function
        if (possible(0, Sum_of_Set1, i)) {
 
            // Find out the elements in Vec, not in
            // res and return the result.
            int ptr1 = 0, ptr2 = 0;
            vector<int> res1 = res;
            vector<int> res2;
            while (ptr1 < Vec.size() || ptr2 < res.size())
            {
                if (ptr2 < res.size() &&
                        res[ptr2] == Vec[ptr1])
                {
                    ptr1++;
                    ptr2++;
                    continue;
                }
                res2.push_back(Vec[ptr1]);
                ptr1++;
            }
 
            vector<vector<int> > ans;
            ans.push_back(res1);
            ans.push_back(res2);
            return ans;
        }
    }
    // If we havent found any such subset.
    vector<vector<int> > ans;
    return ans;
}
 
// Function to print partitions
void Print_Partition(vector<vector<int> > sol)
{
    // Print two partitions
    for (int i = 0; i < sol.size(); i++) {
        cout << "[";
        for (int j = 0; j < sol[i].size(); j++) {
            cout << sol[i][j];
            if (j != sol[i].size() - 1)
                cout << " ";
        }
        cout << "] ";
    }
}
 
// Driver code
int main()
{
    vector<int> Vec = { 1, 7, 15, 29, 11, 9 };
 
    vector<vector<int> > sol = partition(Vec);
 
    // If partition possible
    if (sol.size())
        Print_Partition(sol);
    else
        cout << -1;
 
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java program to Partition an array of
// non-negative integers into two subsets
// such that average of both the subsets are equal
import java.io.*;
import java.util.*;
 
class GFG
{
 
    static boolean[][][] dp;
    static Vector<Integer> res = new Vector<>();
    static int[] original;
    static int total_size;
 
    // Function that returns true if it is possible to
    // use elements with index = ind to construct a set of s
    // ize = curr_size whose sum is curr_sum.
    static boolean possible(int index, int curr_sum,
                                        int curr_size)
    {
 
        // base cases
        if (curr_size == 0)
            return (curr_sum == 0);
        if (index >= total_size)
            return false;
 
        // Which means curr_sum cant be found for curr_size
        if (dp[index][curr_sum][curr_size] == false)
            return false;
 
        if (curr_sum >= original[index])
        {
            res.add(original[index]);
 
            // Checks if taking this element at
            // index i leads to a solution
            if (possible(index + 1, curr_sum - original[index],
                                                curr_size - 1))
                return true;
 
            res.remove(res.size() - 1);
        }
 
        // Checks if not taking this element at
        // index i leads to a solution
        if (possible(index + 1, curr_sum, curr_size))
            return true;
 
        // If no solution has been found
        return dp[index][curr_sum][curr_size] = false;
    }
 
    // Function to find two Partitions having equal average
    static Vector<Vector<Integer>> partition(int[] Vec)
    {
 
        // Sort the vector
        Arrays.sort(Vec);
        original = Vec;
        res.clear();
 
        int total_sum = 0;
        total_size = Vec.length;
 
        for (int i = 0; i < total_size; ++i)
            total_sum += Vec[i];
 
        // building the memoization table
        dp = new boolean[original.length][total_sum + 1][total_size];
 
        for (int i = 0; i < original.length; i++)
            for (int j = 0; j < total_sum + 1; j++)
                for (int k = 0; k < total_size; k++)
                    dp[i][j][k] = true;
 
        for (int i = 1; i < total_size; i++)
        {
 
            // Sum_of_Set1 has to be an integer
            if ((total_sum * i) % total_size != 0)
                continue;
            int Sum_of_Set1 = (total_sum * i) / total_size;
 
            // We build our solution vector if its possible
            // to find subsets that match our criteria
            // using a recursive function
            if (possible(0, Sum_of_Set1, i))
            {
 
                // Find out the elements in Vec, not in
                // res and return the result.
                int ptr1 = 0, ptr2 = 0;
                Vector<Integer> res1 = res;
                Vector<Integer> res2 = new Vector<>();
                while (ptr1 < Vec.length || ptr2 < res.size())
                {
                    if (ptr2 < res.size() &&
                        res.elementAt(ptr2) == Vec[ptr1])
                    {
                        ptr1++;
                        ptr2++;
                        continue;
                    }
                    res2.add(Vec[ptr1]);
                    ptr1++;
                }
 
                Vector<Vector<Integer>> ans = new Vector<>();
                ans.add(res1);
                ans.add(res2);
                return ans;
            }
        }
 
        // If we havent found any such subset.
        Vector<Vector<Integer>> ans = new Vector<>();
        return ans;
    }
 
    // Function to print partitions
    static void Print_Partition(Vector<Vector<Integer>> sol)
    {
 
        // Print two partitions
        for (int i = 0; i < sol.size(); i++)
        {
            System.out.print("[");
            for (int j = 0; j < sol.elementAt(i).size(); j++)
            {
                System.out.print(sol.elementAt(i).elementAt(j));
                if (j != sol.elementAt(i).size() - 1)
                    System.out.print(" ");
            }
            System.out.print("]");
        }
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int[] Vec = { 1, 7, 15, 29, 11, 9 };
        Vector<Vector<Integer>> sol = partition(Vec);
 
        // If partition possible
        if (sol.size() > 0)
            Print_Partition(sol);
        else
            System.out.println("-1");
    }
}
 
// This code is contributed by
// sanjeev2552

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 program to partition an array of
# non-negative integers into two subsets
# such that average of both the subsets are equal
dp = []
res = []
original = []
total_size = int(0)
 
# Function that returns true if it is possible
# to use elements with index = ind to construct
# a set of s ize = curr_size whose sum is curr_sum.
def possible(index, curr_sum, curr_size):
     
    index = int(index)
    curr_sum = int(curr_sum)
    curr_size = int(curr_size)
     
    global dp, res
     
    # Base cases
    if curr_size == 0:
        return (curr_sum == 0)
    if index >= total_size:
        return False
     
    # Which means curr_sum cant be
    # found for curr_size
    if dp[index][curr_sum][curr_size] == False:
        return False
     
    if curr_sum >= original[index]:
        res.append(original[index])
         
        # Checks if taking this element
        # at index i leads to a solution
        if possible(index + 1,
                    curr_sum - original[index],
                    curr_size - 1):
            return True
             
        res.pop()
     
    # Checks if not taking this element at
    # index i leads to a solution
    if possible(index + 1, curr_sum, curr_size):
        return True
     
    # If no solution has been found
    dp[index][curr_sum][curr_size] = False
    return False
 
# Function to find two partitions
# having equal average
def partition(Vec):
     
    global dp, original, res, total_size
     
    # Sort the vector
    Vec.sort()
     
    if len(original) > 0:
        original.clear()
     
    original = Vec
     
    if len(dp) > 0:
        dp.clear()
    if len(res) > 0:
        res.clear()
 
    total_sum = 0
    total_size = len(Vec)
 
    for i in range(total_size):
        total_sum += Vec[i]
     
    # Building the memoization table
    dp = [[[True for _ in range(total_size)]
                 for _ in range(total_sum + 1)]
                 for _ in range(len(original))]
     
    for i in range(1, total_size):
         
        # Sum_of_Set1 has to be an integer
        if (total_sum * i) % total_size != 0:
            continue
             
        Sum_of_Set1 = (total_sum * i) / total_size
 
        # We build our solution vector if its possible
        # to find subsets that match our criteria
        # using a recursive function
        if possible(0, Sum_of_Set1, i):
 
            # Find out the elements in Vec,
            # not in res and return the result.
            ptr1 = 0
            ptr2 = 0
            res1 = res
            res2 = []
             
            while ptr1 < len(Vec) or ptr2 < len(res):
                if (ptr2 < len(res) and
                    res[ptr2] == Vec[ptr1]):
                    ptr1 += 1
                    ptr2 += 1
                    continue
                     
                res2.append(Vec[ptr1])
                ptr1 += 1
 
            ans = []
            ans.append(res1)
            ans.append(res2)
             
            return ans
     
    # If we havent found any such subset.
    ans = []
    return ans
 
# Driver code
Vec = [ 1, 7, 15, 29, 11, 9 ]
 
sol = partition(Vec)
 
if len(sol) > 0:
    print(sol)
else:
    print("-1")
     
# This code is contributed by saishashank1

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# program to Partition an array of 
// non-negative integers into two subsets 
// such that average of both the subsets are equal 
using System;
using System.Collections; 
 
class GFG{
   
static bool[,,] dp;
static ArrayList res = new ArrayList();
static int[] original;
static int total_size;
 
// Function that returns true if it is possible to
// use elements with index = ind to construct a set of s
// ize = curr_size whose sum is curr_sum.
static bool possible(int index, int curr_sum, 
                                int curr_size) 
{
     
    // base cases
    if (curr_size == 0)
        return (curr_sum == 0);
    if (index >= total_size)
        return false;
 
    // Which means curr_sum cant be
    // found for curr_size
    if (dp[index, curr_sum, curr_size] == false)
        return false;
 
    if (curr_sum >= original[index])
    {
        res.Add(original[index]);
 
        // Checks if taking this element at
        // index i leads to a solution
        if (possible(index + 1, curr_sum -
                     original[index], curr_size - 1))
            return true;
 
        res.Remove(res[res.Count - 1]);
    }
 
    // Checks if not taking this element at
    // index i leads to a solution
    if (possible(index + 1, curr_sum, curr_size))
        return true;
     
    dp[index, curr_sum, curr_size] = false;
     
    // If no solution has been found
    return dp[index, curr_sum, curr_size];
}
 
// Function to find two Partitions
// having equal average
static ArrayList partition(int[] Vec)
{
     
    // Sort the vector
    Array.Sort(Vec);
    original = Vec;
    res.Clear();
 
    int total_sum = 0;
    total_size = Vec.Length;
 
    for(int i = 0; i < total_size; ++i)
        total_sum += Vec[i];
 
    // Building the memoization table
    dp = new bool[original.Length,
                  total_sum + 1,
                  total_size];
 
    for(int i = 0; i < original.Length; i++)
        for(int j = 0; j < total_sum + 1; j++)
            for(int k = 0; k < total_size; k++)
                dp[i, j, k] = true;
 
    for(int i = 1; i < total_size; i++) 
    {
         
        // Sum_of_Set1 has to be an integer
        if ((total_sum * i) % total_size != 0)
            continue;
             
        int Sum_of_Set1 = (total_sum * i) / total_size;
 
        // We build our solution vector if its possible
        // to find subsets that match our criteria
        // using a recursive function
        if (possible(0, Sum_of_Set1, i)) 
        {
             
            // Find out the elements in Vec, not in
            // res and return the result.
            int ptr1 = 0, ptr2 = 0;
             
            ArrayList res1 = new ArrayList(res);
            ArrayList res2 = new ArrayList();
             
            while (ptr1 < Vec.Length || ptr2 < res.Count)
            {
                if (ptr2 < res.Count && 
                   (int)res[ptr2] == Vec[ptr1])
                {
                    ptr1++;
                    ptr2++;
                    continue;
                }
                res2.Add(Vec[ptr1]);
                ptr1++;
            }
 
            ArrayList ans = new ArrayList();
            ans.Add(res1);
            ans.Add(res2);
            return ans;
        }
    }
 
    // If we havent found any such subset.
    ArrayList ans2 = new ArrayList();
    return ans2;
}
 
// Function to print partitions
static void Print_Partition(ArrayList sol) 
{
     
    // Print two partitions
    for(int i = 0; i < sol.Count; i++) 
    {
        Console.Write("[");
        for(int j = 0; j < ((ArrayList)sol[i]).Count; j++) 
        {
            Console.Write((int)((ArrayList)sol[i])[j]);
            if (j != ((ArrayList)sol[i]).Count - 1)
                Console.Write(" ");
        }
        Console.Write("] ");
    }
}
 
// Driver Code
public static void Main(string[] args)
{
    int[] Vec = { 1, 7, 15, 29, 11, 9 };
    ArrayList sol = partition(Vec);
 
    // If partition possible
    if (sol.Count > 0)
        Print_Partition(sol);
    else
        Console.Write("-1");
}
}
 
// This code is contributed by rutvik_56

chevron_right


Output: 

[9 15] [1 7 11 29]

 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.




My Personal Notes arrow_drop_up

Recommended Posts:


Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.