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(2N), 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:
- Store the frequency of each element in Map M.
- Now, create an auxiliary array aux[] and insert it into it, all the frequencies stored from the Map.
- The given problem reduces to finding a subset in the array aux[] having a given sum N/2.
- 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> |
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)
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