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
// 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 averagevector<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 partitionsvoid 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 codeint 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;} |
Java
// 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 |
Python3
# 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 codeVec = [ 1, 7, 15, 29, 11, 9 ]sol = partition(Vec)if len(sol) > 0: print(sol)else: print("-1") # This code is contributed by saishashank1 |
C#
// 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 |
Javascript
// JS program to Partition an array of// non-negative integers into two subsets// such that average of both the subsets are equallet dp = [];let res = [];let original = [];let 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.function possible(index, curr_sum, 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(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 dp[index][curr_sum][curr_size];}// Function to find two Partitions having equal averagefunction partition(Vec){ // Sort the vector Vec.sort(); original = []; original = Vec; dp = []; res = []; let total_sum = 0; total_size = Vec.length; for (var i = 0; i < total_size; ++i) total_sum += Vec[i]; // building the memoization table dp = new Array(original.length); for (var i = 0; i < original.length; i++) { dp[i] = new Array(total_sum + 1); for (var j = 0; j <= total_sum; j++) { dp[i][j] = new Array(total_size).fill(true); } } for (var i = 1; i < total_size; i++) { // Sum_of_Set1 has to be an integer if ((total_sum * i) % total_size != 0) continue; var 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. var ptr1 = 0, ptr2 = 0; var res1 = res; var res2 = []; while (ptr1 < Vec.length || ptr2 < res.length) { if (ptr2 < res.length && res[ptr2] == Vec[ptr1]) { ptr1++; ptr2++; continue; } res2.push(Vec[ptr1]); ptr1++; } let ans = []; ans.push(res1); ans.push(res2); return ans; } } // If we havent found any such subset. return -1;}// Driver codelet Vec = [1, 7, 15, 29, 11, 9 ];let sol = partition(Vec);console.log(sol) |
Output:
[9 15] [1 7 11 29]
Time Complexity: O(n3)
Auxiliary Space: O(n3)
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