Split array into two equal length subsets such that all repetitions of a number lies in a single subset

Given an array arr[] consisting of N integers, the task is to check if it is possible to split the integers into two equal length subsets such that all repetitions of any array element belong to the same subset. If found to be true, print “Yes”. Otherwise, print “No”.

Examples:

Input: arr[] = {2, 1, 2, 3}
Output: Yes
Explanation:
One possible way of dividing the array is {1, 3} and {2, 2}

Input: arr[] = {1, 1, 1, 1}
Output: No

Naive Approach: The simplest approach to solve the problem is to try all possible combinations of splitting the array into two equal subsets. For each combination, check whether every repetition belongs to only one of the two sets or not. If found to be true, then print “Yes”. Otherwise, print “No”.

Time Complexity: O(2^N), where N is the size of the given integer.
Auxiliary Space: O(N)

Efficient Approach: The above approach can be optimized by storing the frequency of all elements of the given array in an array freq[]. For elements to be divided into two equal sets, N/2 elements must be present in each set. Therefore, to divide the given array arr[] into 2 equal parts, there must be some subset of integers in freq[] having sum N/2. Follow the steps below to solve the problem:

1. Store the frequency of each element in Map M.
2. Now, create an auxiliary array aux[] and insert it into it, all the frequencies stored from the Map.
3. The given problem reduces to finding a subset in the array aux[] having a given sum N/2.
4. If there exists any such subset in the above step, then print “Yes”. Otherwise, print “No”.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to create the frequency
// array of the given array arr[]
vector<int> findSubsets(vector<int> arr, int N)
{
  // Hashmap to store the
  // frequencies
  map<int,int> M;

  // Store freq for each element
  for (int i = 0; i < N; i++)
  {
      M[arr[i]]++;
  }

  // Get the total frequencies
  vector<int> subsets;
  int I = 0;

  // Store frequencies in
  // subset[] array
  for(auto playerEntry = M.begin(); playerEntry != M.end(); playerEntry++)
  {
      subsets.push_back(playerEntry->second);
      I++;
  }

  // Return frequency array
  return subsets;
}

// Function to check is sum
// N/2 can be formed using
// some subset
bool subsetSum(vector<int> subsets, int N, int target)
{
  // dp[i][j] store the answer to
  // form sum j using 1st i elements
  bool dp[N + 1][target + 1];

  // Initialize dp[][] with true
  for (int i = 0; i < N + 1; i++)
    dp[i][0] = true;

  // Fill the subset table in the
  // bottom up manner
  for (int i = 1; i <= N; i++)
  {
    for (int j = 1; j <= target; j++)
    {
      dp[i][j] = dp[i - 1][j];

      // If current element is
      // less than j
      if (j >= subsets[i - 1])
      {
        // Update current state
        dp[i][j] |= dp[i - 1][j - subsets[i - 1]];
      }
    }
  }

  // Return the result
  return dp[N][target];
}

// Function to check if the given
// array can be split into required sets
void divideInto2Subset(vector<int> arr, int N)
{
  // Store frequencies of arr[]
  vector<int> subsets = findSubsets(arr, N);

  // If size of arr[] is odd then
  // print "Yes"
  if ((N) % 2 == 1)
  {
    cout << "No" << endl;
    return;
  }

  int subsets_size = subsets.size();

  // Check if answer is true or not
  bool isPossible = subsetSum(subsets, subsets_size, N / 2);

  // Print the result
  if (isPossible)
  {
    cout << "Yes" << endl;
  }
  else
  {
    cout << "No" << endl;
  }
}

int main()
{
      // Given array arr[]
      vector<int> arr{2, 1, 2, 3};

      int N = arr.size();

      // Function Call
      divideInto2Subset(arr, N);

    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.util.*;

class GFG {

    // Function to create the frequency
    // array of the given array arr[]
    private static int[] findSubsets(int[] arr)
    {

        // Hashmap to store the frequencies
        HashMap<Integer, Integer> M
            = new HashMap<>();

        // Store freq for each element
        for (int i = 0; i < arr.length; i++) {
            M.put(arr[i],
                  M.getOrDefault(arr[i], 0) + 1);
        }

        // Get the total frequencies
        int[] subsets = new int[M.size()];
        int i = 0;

        // Store frequencies in subset[] array
        for (
            Map.Entry<Integer, Integer> playerEntry :
            M.entrySet()) {
            subsets[i++]
                = playerEntry.getValue();
        }

        // Return frequency array
        return subsets;
    }

    // Function to check is sum N/2 can be
    // formed using some subset
    private static boolean
    subsetSum(int[] subsets,
              int target)
    {

        // dp[i][j] store the answer to
        // form sum j using 1st i elements
        boolean[][] dp
            = new boolean[subsets.length
                          + 1][target + 1];

        // Initialize dp[][] with true
        for (int i = 0; i < dp.length; i++)
            dp[i][0] = true;

        // Fill the subset table in the
        // bottom up manner
        for (int i = 1;
             i <= subsets.length; i++) {

            for (int j = 1;
                 j <= target; j++) {
                dp[i][j] = dp[i - 1][j];

                // If current element is
                // less than j
                if (j >= subsets[i - 1]) {

                    // Update current state
                    dp[i][j]
                        |= dp[i - 1][j
                                     - subsets[i - 1]];
                }
            }
        }

        // Return the result
        return dp[subsets.length][target];
    }

    // Function to check if the given
    // array can be split into required sets
    public static void
    divideInto2Subset(int[] arr)
    {
        // Store frequencies of arr[]
        int[] subsets = findSubsets(arr);

        // If size of arr[] is odd then
        // print "Yes"
        if ((arr.length) % 2 == 1) {
            System.out.println("No");
            return;
        }

        // Check if answer is true or not
        boolean isPossible
            = subsetSum(subsets,
                        arr.length / 2);

        // Print the result
        if (isPossible) {
            System.out.println("Yes");
        }
        else {
            System.out.println("No");
        }
    }

    // Driver Code
    public static void main(String[] args)
    {
        // Given array arr[]
        int[] arr = { 2, 1, 2, 3 };

        // Function Call
        divideInto2Subset(arr);
    }
}

// This code is contributed by divyesh072019

Python3

# Python3 program for the
# above approach
from collections import defaultdict

# Function to create the
# frequency array of the
# given array arr[]
def findSubsets(arr):

    # Hashmap to store
    # the frequencies
    M = defaultdict (int)

    # Store freq for each element
    for i in range (len(arr)):
        M[arr[i]] += 1

    # Get the total frequencies
    subsets = [0] * len(M)
    i = 0

    # Store frequencies in
    # subset[] array
    for j in M:
        subsets[i] = M[j]
        i += 1

    # Return frequency array
    return subsets

# Function to check is
# sum N/2 can be formed
# using some subset
def subsetSum(subsets, target):

    # dp[i][j] store the answer to
    # form sum j using 1st i elements
    dp = [[0 for x in range(target + 1)]
             for y in range(len(subsets) + 1)]

    # Initialize dp[][] with true
    for i in range(len(dp)):
        dp[i][0] = True

    # Fill the subset table in the
    # bottom up manner
    for i in range(1, len(subsets) + 1):
        for j in range(1, target + 1):
            dp[i][j] = dp[i - 1][j]

            # If current element is
            # less than j
            if (j >= subsets[i - 1]):

                # Update current state
                dp[i][j] |= (dp[i - 1][j -
                             subsets[i - 1]])

    # Return the result
    return dp[len(subsets)][target]

# Function to check if the given
# array can be split into required sets
def divideInto2Subset(arr):

    # Store frequencies of arr[]
    subsets = findSubsets(arr)

    # If size of arr[] is odd then
    # print "Yes"
    if (len(arr) % 2 == 1):
        print("No")
        return

    # Check if answer is true or not
    isPossible = subsetSum(subsets,
                           len(arr) // 2)

    # Print the result
    if (isPossible):
        print("Yes")
    else :
        print("No")

# Driver Code
if __name__ == "__main__":

    # Given array arr
    arr = [2, 1, 2, 3]

    # Function Call
    divideInto2Subset(arr)

# This code is contributed by Chitranayal

C#

// C# program for the above
// approach
using System;
using System.Collections.Generic;
class GFG{

// Function to create the frequency
// array of the given array arr[]
static int[] findSubsets(int[] arr)
{
  // Hashmap to store the
  // frequencies
  Dictionary<int,
             int> M =
             new Dictionary<int,
                            int>();

  // Store freq for each element
  for (int i = 0; i < arr.Length; i++)
  {
    if(M.ContainsKey(arr[i]))
    {
      M[arr[i]]++;
    }
    else
    {
      M[arr[i]] = 1;
    }
  }

  // Get the total frequencies
  int[] subsets = new int[M.Count];
  int I = 0;

  // Store frequencies in
  // subset[] array
  foreach(KeyValuePair<int,
                       int>
          playerEntry in M)
  {
    subsets[I] = playerEntry.Value;
    I++;
  }

  // Return frequency array
  return subsets;
}

// Function to check is sum
// N/2 can be formed using
// some subset
static bool subsetSum(int[] subsets,
                      int target)
{
  // dp[i][j] store the answer to
  // form sum j using 1st i elements
  bool[,] dp = new bool[subsets.Length + 1,
                        target + 1];

  // Initialize dp[][] with true
  for (int i = 0;
           i < dp.GetLength(0); i++)
    dp[i, 0] = true;

  // Fill the subset table in the
  // bottom up manner
  for (int i = 1;
           i <= subsets.Length; i++)
  {
    for (int j = 1; j <= target; j++)
    {
      dp[i, j] = dp[i - 1, j];

      // If current element is
      // less than j
      if (j >= subsets[i - 1])
      {
        // Update current state
        dp[i, j] |= dp[i - 1,
                       j - subsets[i - 1]];
      }
    }
  }

  // Return the result
  return dp[subsets.Length,
            target];
}

// Function to check if the given
// array can be split into required sets
static void divideInto2Subset(int[] arr)
{
  // Store frequencies of arr[]
  int[] subsets = findSubsets(arr);

  // If size of arr[] is odd then
  // print "Yes"
  if ((arr.Length) % 2 == 1)
  {
    Console.WriteLine("No");
    return;
  }

  // Check if answer is true or not
  bool isPossible = subsetSum(subsets,
                              arr.Length / 2);

  // Print the result
  if (isPossible)
  {
    Console.WriteLine("Yes");
  }
  else
  {
    Console.WriteLine("No");
  }
}

// Driver code
static void Main()
{
  // Given array arr[]
  int[] arr = {2, 1, 2, 3};

  // Function Call
  divideInto2Subset(arr);
}
}

// This code is contributed by divyeshrabadiya07

Javascript

<script>

// JavaScript program for the above approach

// Function to create the frequency
// array of the given array arr[]
function findSubsets( arr, N)
{
  // Hashmap to store the
  // frequencies
  let M = new Map();

  // Store freq for each element
  for (let i = 0; i < N; i++)
  {
      if(M[arr[i]])
      M[arr[i]]++;
      else
      M[arr[i]] = 1
  }

  // Get the total frequencies
  let subsets = [];
  let I = 0;

  // Store frequencies in
  // subset[] array
  for(var it in M)
  {
      subsets.push(M[it]);
  }

  // Return frequency array
  return subsets;
}

// Function to check is sum
// N/2 can be formed using
// some subset
function subsetSum( subsets, N, target)
{
  // dp[i][j] store the answer to
  // form sum j using 1st i elements
  var dp = [],
    H = N+1; // 4 rows
    W = target+1; // of 6 cells

for ( var y = 0; y < H; y++ ) {
    dp[ y ] = [];
    for ( var x = 0; x < W; x++ ) {
        dp[ y ][ x ] = false;
    }
}

  // Initialize dp[][] with true
  for (let i = 0; i < N + 1; i++)
    dp[i][0] = true;

  // Fill the subset table in the
  // bottom up manner
  for (let i = 1; i <= N; i++)
  {
    for (let j = 1; j <= target; j++)
    {
      dp[i][j] = dp[i - 1][j];

      // If current element is
      // less than j
      if (j >= subsets[i - 1])
      {
        // Update current state
        dp[i][j] |= dp[i - 1][j - subsets[i - 1]];
      }
    }
  }

  // Return the result
  return dp[N][target];
}

// Function to check if the given
// array can be split into required sets
function divideInto2Subset( arr, N)
{
  // Store frequencies of arr[]
  let subsets = findSubsets(arr, N);

  // If size of arr[] is odd then
  // print "Yes"
  if ((N) % 2 == 1)
  {
    document.write( "No<br>");
    return;
  }

  let subsets_size = subsets.length;

  // Check if answer is true or not
 let isPossible = subsetSum(subsets,
 subsets_size, Math.floor(N / 2));

  // Print the result
  if (isPossible)
  {
    document.write( "Yes<br>");
  }
  else
  {
    document.write( "No<br>");
  }
}

      // Given array arr[]
      let arr = [2, 1, 2, 3];

      let N = arr.length;

      // Function Call
      divideInto2Subset(arr, N);


</script>

Output:

Yes

Time Complexity: O(N*M), where N is the size of the array and M is the total count of distinct elements in the given array.
Auxiliary Space: O(N)