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 the = 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.

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// CPP 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;
}

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Output:

[9 15] [1 7 11 29]


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