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++ 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 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 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# 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 |
<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*Target) where target is N/2
Auxiliary Space: O(N*Target)
Efficient Approach : Space optimization
In previous approach the current computation Dp[i][j] is only depend upon the current row and previous row of DP. So we can optimize the space by using a 1D array to store these computations.
Implementation Steps:
- Create a DP array of size target+1 and initialize it with False.
- Set base case by initializing dp[0] = true.
- Now iterate over subproblems with the help of nested loops and get the current value from the previous computations.
- At last return final answer stored in dp[target].
Implementation:
// C++ program for 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;
// Store frequencies in
// subset[] array
for ( auto playerEntry = M.begin(); playerEntry != M.end(); playerEntry++)
{
subsets.push_back(playerEntry->second);
}
// 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[] store the answer to
// form sum j using 1st i elements
bool dp[target + 1];
// Initialize dp[] with true
memset (dp, false , sizeof (dp));
dp[0] = true ;
// Fill the subset table in the
// bottom up manner
for ( int i = 1; i <= N; i++)
{
for ( int j = target; j >= 1; j--)
{
// If current element is
// less than j
if (j >= subsets[i - 1])
{
// Update current state
dp[j] |= dp[j - subsets[i - 1]];
}
}
}
// Return the result
return dp[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 "No"
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;
} // this code is contributed by bhardwajji |
import java.util.*;
public class Main { // Function to create the frequency
// array of the given array arr[]
static List<Integer> findSubsets(List<Integer> arr,
int N)
{
// Hashmap to store the
// frequencies
Map<Integer, Integer> M = new HashMap<>();
// Store freq for each element
for ( int i = 0 ; i < N; i++) {
int element = arr.get(i);
if (M.containsKey(element)) {
M.put(element, M.get(element) + 1 );
}
else {
M.put(element, 1 );
}
}
// Get the total frequencies
List<Integer> subsets = new ArrayList<>();
// Store frequencies in
// subset[] array
for (Map.Entry<Integer, Integer> entry :
M.entrySet()) {
subsets.add(entry.getValue());
}
// Return frequency array
return subsets;
}
// Function to check is sum
// N/2 can be formed using
// some subset
static boolean subsetSum(List<Integer> subsets, int N,
int target)
{
// dp[] store the answer to
// form sum j using 1st i elements
boolean [] dp = new boolean [target + 1 ];
// Initialize dp[] with true
Arrays.fill(dp, false );
dp[ 0 ] = true ;
// Fill the subset table in the
// bottom up manner
for ( int i = 1 ; i <= N; i++) {
for ( int j = target; j >= 1 ; j--) {
// If current element is
// less than j
if (j >= subsets.get(i - 1 )) {
// Update current state
dp[j] |= dp[j - subsets.get(i - 1 )];
}
}
}
// Return the result
return dp[target];
}
// Function to check if the given
// array can be split into required sets
static void divideInto2Subset(List<Integer> arr, int N)
{
// Store frequencies of arr[]
List<Integer> subsets = findSubsets(arr, N);
// If size of arr[] is odd then
// print "No"
if ((N) % 2 == 1 ) {
System.out.println( "No" );
return ;
}
int subsets_size = subsets.size();
// Check if answer is true or not
boolean isPossible
= subsetSum(subsets, subsets_size, N / 2 );
// Print the result
if (isPossible) {
System.out.println( "Yes" );
}
else {
System.out.println( "No" );
}
}
public static void main(String[] args)
{
// Given array arr[]
List<Integer> arr
= new ArrayList<>(Arrays.asList( 2 , 1 , 2 , 3 ));
int N = arr.size();
// Function Call
divideInto2Subset(arr, N);
}
} |
# Python 3 program for above approach # Function to create the frequency # array of the given array arr[] def findSubsets(arr, N):
# Hashmap to store the
# frequencies
M = {}
# Store freq for each element
for i in range (N):
if arr[i] not in M:
M[arr[i]] = 1
else :
M[arr[i]] + = 1
# Get the total frequencies
subsets = []
# Store frequencies in
# subset[] array
for playerEntry in M:
subsets.append(M[playerEntry])
# Return frequency array
return subsets
# Function to check is sum # N/2 can be formed using # some subset def subsetSum(subsets, N, target):
# dp[] store the answer to
# form sum j using 1st i elements
dp = [ False ] * (target + 1 )
# Initialize dp[] with true
dp[ 0 ] = True
# Fill the subset table in the
# bottom up manner
for i in range ( 1 , N + 1 ):
for j in range (target, 0 , - 1 ):
# If current element is
# less than j
if j > = subsets[i - 1 ]:
# Update current state
dp[j] | = dp[j - subsets[i - 1 ]]
# Return the result
return dp[target]
# Function to check if the given # array can be split into required sets def divideInto2Subset(arr, N):
# Store frequencies of arr[]
subsets = findSubsets(arr, N)
# If size of arr[] is odd then
# print "No"
if (N) % 2 = = 1 :
print ( "No" )
return
subsets_size = len (subsets)
# Check if answer is true or not
isPossible = subsetSum(subsets, subsets_size, N / / 2 )
# Print the result
if isPossible:
print ( "Yes" )
else :
print ( "No" )
# Given array arr[] arr = [ 2 , 1 , 2 , 3 ]
N = len (arr)
# Function Call divideInto2Subset(arr, N) |
// C# program for above approach // Function to create the frequency // array of the given array arr[] using System;
using System.Collections.Generic;
public class Program {
public static List< int > findSubsets(List< int > arr, int N) {
// Dictionary to store the
// frequencies
Dictionary< int , int > M = new Dictionary< int , int >();
// Store freq for each element
for ( int i = 0; i < N; i++) {
if (!M.ContainsKey(arr[i])) {
M[arr[i]] = 1;
}
else {
M[arr[i]] += 1;
}
}
// Get the total frequencies
List< int > subsets = new List< int >();
// Store frequencies in
// subset[] array
foreach (KeyValuePair< int , int > playerEntry in M) {
subsets.Add(playerEntry.Value);
}
// Return frequency array
return subsets;
}
// Function to check is sum
// N/2 can be formed using
// some subset
public static bool subsetSum(List< int > subsets, int N, int target) {
bool [] dp = new bool [target+1];
// dp[] store the answer to
// form sum j using 1st i elements
// Initialize dp[] with true
dp[0] = true ;
// Fill the subset table in the
// bottom up manner
for ( int i = 1; i <= N; i++) {
for ( int j = target; j > 0; j--) {
// If current element is
// less than j
if (j >= subsets[i - 1]) {
// Update current state
dp[j] |= dp[j - subsets[i - 1]];
}
}
}
// Return the result
return dp[target];
}
// Function to check if the given
// array can be split into required sets
public static void divideInto2Subset(List< int > arr, int N) {
// Store frequencies of List
List< int > subsets = findSubsets(arr, N);
// If size of arr[] is odd then
// print "No"
if (N % 2 == 1) {
Console.WriteLine( "No" );
return ;
}
int subsets_size = subsets.Count;
// Check if answer is true or not
bool isPossible = subsetSum(subsets, subsets_size, N / 2);
// Print the result
if (isPossible) {
Console.WriteLine( "Yes" );
} else {
Console.WriteLine( "No" );
}
}
public static void Main( string [] args) {
// Given List
List< int > arr = new List< int >() {2, 1, 2, 3};
int N = arr.Count;
// Function Call
divideInto2Subset(arr, N);
}
} // This code is cintributed by shiv1o43g |
// Javascript code addition // 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.has(arr[i])) {
M.set(arr[i], M.get(arr[i]) + 1);
} else {
M.set(arr[i], 1);
}
}
// Get the total frequencies
let subsets = [];
// Store frequencies in subset[] array
for (let [key, value] of M) {
subsets.push(value);
}
// Return frequency array
return subsets;
} // Function to check is sum N/2 can be formed using some subset function subsetSum(subsets, N, target) {
// dp[] store the answer to form sum j using 1st i elements
let dp = new Array(target + 1).fill( false );
dp[0] = true ;
// Fill the subset table in the bottom up manner
for (let i = 1; i <= N; i++) {
for (let j = target; j >= 1; j--) {
// If current element is less than j
if (j >= subsets[i - 1]) {
// Update current state
dp[j] |= dp[j - subsets[i - 1]];
}
}
}
// Return the result
return dp[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 "No"
if (N % 2 === 1) {
console.log( "No" );
return ;
}
let subsets_size = subsets.length;
// Check if answer is true or not
let isPossible = subsetSum(subsets, subsets_size, N / 2);
// Print the result
if (isPossible) {
console.log( "Yes" );
} else {
console.log( "No" );
}
} // Given array arr[] let arr = [2, 1, 2, 3]; let N = arr.length; // Function Call divideInto2Subset(arr, N); // The code is contributed by Nidhi goel. |
Output
Yes
Time Complexity: O(N*Target) where target is N/2
Auxiliary Space: O(Target)