Given two unsorted arrays, check whether both arrays have the same set of elements or not.
Examples:
Input : A = {2, 5, 6, 8, 10, 2, 2} B = {2, 5, 5, 6, 8, 5, 6} Output : No Input : A = {2, 5, 6, 8, 2, 10, 2} B = {2, 5, 6, 8, 2, 10, 2} Output : Yes Input : A = {2, 5, 8, 6, 10, 2, 2} B = {2, 5, 6, 8, 2, 10, 2} Output : Yes
Method 1 (Simple):
A simple solution to this problem is to check if each element of A is present in B. But this approach will lead to a wrong answer in case of multiple instances of an element is present in B. To overcome this issue, we mark visited instances of B[] using an auxiliary array visited[].
// C++ implementation of the above approach #include <bits/stdc++.h> using namespace std;
// Function to check if both arrays are same bool areSameSet(vector< int > A, vector< int > B)
{ int n = A.size();
if (B.size() != n)
return false ;
// visited array is used to handle duplicates
vector< bool > visited(n, false );
// each element of A is matched
// against each element of B
for ( int i = 0; i < n; i++) {
int j = 0;
for (j = 0; j < n; j++)
{
if (A[i] == B[j] && visited[j] == false )
{
visited[j] = true ;
break ;
}
}
// If we could not find A[i] in B[]
if (j == n)
return false ;
}
return true ;
} // Driver code int main()
{ vector< int > A, B;
A.push_back(2);
A.push_back(5);
A.push_back(10);
A.push_back(6);
A.push_back(8);
A.push_back(2);
A.push_back(2);
B.push_back(2);
B.push_back(5);
B.push_back(6);
B.push_back(8);
B.push_back(10);
B.push_back(2);
B.push_back(2);
areSameSet(A, B)? cout << "Yes" : cout << "No" ;
} |
// Java implementation of the above approach import java.util.*;
class GFG
{ // Function to check if both arrays are same
static boolean areSameSet(Vector<Integer> A, Vector<Integer> B)
{
int n = A.size();
if (B.size() != n)
{
return false ;
}
// visited array is used to handle duplicates
Vector<Boolean> visited = new Vector<Boolean>();
for ( int i = 0 ; i < n; i++)
{
visited.add(i, Boolean.FALSE);
}
// each element of A is matched
// against each element of B
for ( int i = 0 ; i < n; i++)
{
int j = 0 ;
for (j = 0 ; j < n; j++)
{
if (A.get(i) == B.get(j) && visited.get(j) == false )
{
visited.add(j, Boolean.TRUE);
break ;
}
}
// If we could not find A[i] in B[]
if (j == n)
{
return false ;
}
}
return true ;
}
// Driver code
public static void main(String[] args)
{
Vector<Integer> A = new Vector<>();
Vector<Integer> B = new Vector<>();
A.add( 2 );
A.add( 5 );
A.add( 10 );
A.add( 6 );
A.add( 8 );
A.add( 2 );
A.add( 2 );
B.add( 2 );
B.add( 5 );
B.add( 6 );
B.add( 8 );
B.add( 10 );
B.add( 2 );
B.add( 2 );
if (areSameSet(A, B))
{
System.out.println( "Yes" );
}
else
{
System.out.println( "No" );
}
}
} /* This code contributed by PrinciRaj1992 */ |
# Python3 implementation of the above approach # Function to check if both arrays are same def areSameSet(A, B):
n = len (A)
if ( len (B) ! = n):
return False
# visited array is used to handle duplicates
visited = [ False for i in range (n)]
# each element of A is matched
# against each element of B
for i in range (n):
j = 0
for j in range (n):
if (A[i] = = B[j] and
visited[j] = = False ):
visited[j] = True
break
# If we could not find A[i] in B[]
if (j = = n):
return False
return True
# Driver code A = []
B = []
A.append( 2 )
A.append( 5 )
A.append( 10 )
A.append( 6 )
A.append( 8 )
A.append( 2 )
A.append( 2 )
B.append( 2 )
B.append( 5 )
B.append( 6 )
B.append( 8 )
B.append( 10 )
B.append( 2 )
B.append( 2 )
if (areSameSet(A, B)):
print ( "Yes" )
else :
print ( "No" )
# This code is contributed # by mohit kumar |
// C# implementation of the above approach using System;
using System.Collections.Generic;
class GFG
{ // Function to check if both arrays are same
static Boolean areSameSet(List< int > A, List< int > B)
{
int n = A.Count;
if (B.Count != n)
{
return false ;
}
// visited array is used to handle duplicates
List<Boolean> visited = new List<Boolean>();
for ( int i = 0; i < n; i++)
{
visited.Insert(i, false );
}
// each element of A is matched
// against each element of B
for ( int i = 0; i < n; i++)
{
int j = 0;
for (j = 0; j < n; j++)
{
if (A[i] == B[j] && visited[j] == false )
{
visited.Insert(j, true );
break ;
}
}
// If we could not find A[i] in B[]
if (j == n)
{
return false ;
}
}
return true ;
}
// Driver code
public static void Main(String[] args)
{
List< int > A = new List< int >();
List< int > B = new List< int >();
A.Add(2);
A.Add(5);
A.Add(10);
A.Add(6);
A.Add(8);
A.Add(2);
A.Add(2);
B.Add(2);
B.Add(5);
B.Add(6);
B.Add(8);
B.Add(10);
B.Add(2);
B.Add(2);
if (areSameSet(A, B))
{
Console.WriteLine( "Yes" );
}
else
{
Console.WriteLine( "No" );
}
}
} // This code has been contributed by 29AjayKumar |
<script> // Javascript implementation of the above approach // Function to check if both arrays are same
function areSameSet(A,B)
{
let n = A.length;
if (B.length != n)
{
return false ;
}
// visited array is used to handle duplicates
let visited = [];
for (let i = 0; i < n; i++)
{
visited.push( false );
}
// each element of A is matched
// against each element of B
for (let i = 0; i < n; i++)
{
let j = 0;
for (j = 0; j < n; j++)
{
if (A[i] == B[j] && visited[j] == false )
{
visited[j]= true ;
break ;
}
}
// If we could not find A[i] in B[]
if (j == n)
{
return false ;
}
}
return true ;
}
// Driver code
let A=[];
let B=[];
A.push(2);
A.push(5);
A.push(10);
A.push(6);
A.push(8);
A.push(2);
A.push(2);
B.push(2);
B.push(5);
B.push(6);
B.push(8);
B.push(10);
B.push(2);
B.push(2);
if (areSameSet(A, B))
{
document.write( "Yes" );
}
else
{
document.write( "No" );
}
// This code is contributed by patel2127 </script> |
Output:
Yes
Time complexity: O(n^2).
Method 2 (Sorting):
Sort both the arrays and compare corresponding elements of each array.
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std;
// Function to check if both arrays are same bool areSameSet(vector< int > A, vector< int > B)
{ int n = A.size();
if (B.size() != n)
return false ;
sort(A.begin(), A.end());
sort(B.begin(), B.end());
// Compare corresponding elements
for ( int i = 0; i < n; i++)
if (A[i] != B[i])
return false ;
return true ;
} int main()
{ vector< int > A, B;
A.push_back(2);
A.push_back(5);
A.push_back(10);
A.push_back(6);
A.push_back(8);
A.push_back(2);
A.push_back(2);
B.push_back(2);
B.push_back(5);
B.push_back(6);
B.push_back(8);
B.push_back(10);
B.push_back(2);
B.push_back(2);
areSameSet(A, B)? cout << "Yes" : cout << "No" ;
} |
// Java implementation of the approach import java.util.*;
class GFG
{ // Function to check if both arrays are same
static boolean areSameSet(Vector<Integer> A,
Vector<Integer> B)
{
int n = A.size();
if (B.size() != n)
{
return false ;
}
Collections.sort(A);
Collections.sort(B);
// Compare corresponding elements
for ( int i = 0 ; i < n; i++)
{
if (A.get(i) != B.get(i))
{
return false ;
}
}
return true ;
}
// Driver code
public static void main(String[] args)
{
Vector<Integer> A = new Vector<>();
Vector<Integer> B = new Vector<>();
A.add( 2 );
A.add( 5 );
A.add( 10 );
A.add( 6 );
A.add( 8 );
A.add( 2 );
A.add( 2 );
B.add( 2 );
B.add( 5 );
B.add( 6 );
B.add( 8 );
B.add( 10 );
B.add( 2 );
B.add( 2 );
if (areSameSet(A, B))
{
System.out.println( "Yes" );
}
else
{
System.out.println( "No" );
}
}
} // This code is contributed by 29AjayKumar |
# Python3 implementation of the approach # Function to check if # both arrays are same def areSameSet(A, B):
n = len (A)
if ( len (B) ! = n):
return False
A.sort()
B.sort()
# Compare corresponding
# elements
for i in range (n):
if (A[i] ! = B[i]):
return False return True
# Driver code if __name__ = = "__main__" :
A = []
B = []
A.append( 2 )
A.append( 5 )
A.append( 10 )
A.append( 6 )
A.append( 8 )
A.append( 2 )
A.append( 2 )
B.append( 2 )
B.append( 5 )
B.append( 6 )
B.append( 8 )
B.append( 10 )
B.append( 2 )
B.append( 2 )
if areSameSet(A, B):
print ( "Yes" )
else :
print ( "No" )
# This code is contributed by Chitranayal |
// C# implementation of the approach using System;
using System.Collections.Generic;
class GFG
{ // Function to check if both arrays are same
static Boolean areSameSet(List< int > A,
List< int > B)
{
int n = A.Count;
if (B.Count!= n)
{
return false ;
}
A.Sort();
B.Sort();
// Compare corresponding elements
for ( int i = 0; i < n; i++)
{
if (A[i] != B[i])
{
return false ;
}
}
return true ;
}
// Driver code
public static void Main(String[] args)
{
List< int > A = new List< int >();
List< int > B = new List< int >();
A.Add(2);
A.Add(5);
A.Add(10);
A.Add(6);
A.Add(8);
A.Add(2);
A.Add(2);
B.Add(2);
B.Add(5);
B.Add(6);
B.Add(8);
B.Add(10);
B.Add(2);
B.Add(2);
if (areSameSet(A, B))
{
Console.WriteLine( "Yes" );
}
else
{
Console.WriteLine( "No" );
}
}
} /* This code contributed by PrinciRaj1992 */ |
<script> // Javascript implementation of the approach // Function to check if both arrays are same function areSameSet(A, B)
{ let n = A.length;
if (B.length != n)
{
return false ;
}
A.sort( function (a, b){ return a - b;});
B.sort( function (a, b){ return a - b;});
// Compare corresponding elements
for (let i = 0; i < n; i++)
{
if (A[i] != B[i])
{
return false ;
}
}
return true ;
} // Driver code let A = []; let B = []; A.push(2); A.push(5); A.push(10); A.push(6); A.push(8); A.push(2); A.push(2); B.push(2); B.push(5); B.push(6); B.push(8); B.push(10); B.push(2); B.push(2); if (areSameSet(A, B))
{ document.write( "Yes" );
} else { document.write( "No" );
} // This code is contributed by unknown2108 </script> |
Output:
Yes
Time Complexity: O(n*log(n)).
Method 3 (Hashing):
We can decrease the time complexity of the above problem by using a Hash table. First, we iterate through A and mark the number of instances of each element of A in a Hash Table. Then we iterate through B and decrease the corresponding value in the hash table. If in the end if all the entries of the hash table are zero, the answer will be “Yes” else “No”.
// C++ program to implement Naive approach // to remove duplicates #include <bits/stdc++.h> using namespace std;
bool areSameSet(vector< int > A, vector< int > B){
int n = A.size();
// If the size of vector A and vector B is not equal return False
if (B.size() != n)
return false ;
// Create a hash table to
// number of instances
unordered_map< int , int > m;
// for each element of A
// increase it's instance by 1.
for ( int i = 0; i < n; i++)
m[A[i]]++;
// for each element of B
// decrease it's instance by 1.
for ( int i = 0; i < n; i++)
m[B[i]]--;
// Iterate through map and check if
// any entry is non-zero
for ( auto i : m){
if (i.second != 0){
return false ;
}
}
return true ;
} // driver code to test above function int main(){
// initializing vector A
vector< int > A = {2, 5, 10, 6, 8, 2, 2};
// initializing vector B
vector< int > B = {2, 5, 6, 8, 10, 2, 2};
// Function call
areSameSet(A, B)? cout << "Yes" : cout << "No" ;
} |
// Java program to implement Naive approach // to remove duplicates import java.util.HashMap;
class GFG
{ static boolean areSameSet( int [] A, int [] B)
{ int n = A.length;
if (B.length != n)
return false ;
// Create a hash table to
// number of instances
HashMap<Integer,
Integer> m = new HashMap<>();
// for each element of A
// increase it's instance by 1.
for ( int i = 0 ; i < n; i++)
m.put(A[i], m.get(A[i]) == null ? 1 :
m.get(A[i]) + 1 );
// for each element of B
// decrease it's instance by 1.
for ( int i = 0 ; i < n; i++)
m.put(B[i], m.get(B[i]) - 1 );
// Iterate through map and check if
// any entry is non-zero
for (HashMap.Entry<Integer,
Integer> entry : m.entrySet())
if (entry.getValue() != 0 )
return false ;
return true ;
} // Driver Code public static void main(String[] args)
{ int [] A = { 2 , 5 , 10 , 6 , 8 , 2 , 2 };
int [] B = { 2 , 5 , 6 , 8 , 10 , 2 , 2 };
if (areSameSet(A, B))
System.out.println( "Yes" );
else
System.out.println( "No" );
} } // This code is contributed by // sanjeev2552 |
# Python3 program to implement Naive # approach to remove duplicates def areSameSet(A, B):
n = len (A)
if ( len (B) ! = n):
return False
# Create a hash table to
# number of instances
m = {}
# For each element of A
# increase it's instance by 1.
for i in range (n):
if A[i] not in m:
m[A[i]] = 1
else :
m[A[i]] + = 1
# For each element of B
# decrease it's instance by 1.
for i in range (n):
if B[i] in m:
m[B[i]] - = 1
# Iterate through map and check if
# any entry is non-zero
for i in m:
if (m[i] ! = 0 ):
return False
return True
# Driver Code A = []
B = []
A.append( 2 )
A.append( 5 )
A.append( 10 )
A.append( 6 )
A.append( 8 )
A.append( 2 )
A.append( 2 )
B.append( 2 )
B.append( 5 )
B.append( 6 )
B.append( 8 )
B.append( 10 )
B.append( 2 )
B.append( 2 )
if (areSameSet(A, B)):
print ( "Yes" )
else :
print ( "No" )
# This code is contributed by avanitrachhadiya2155 |
// C# program to implement Naive approach // to remove duplicates using System;
using System.Collections.Generic;
class GFG
{ static bool areSameSet( int [] A, int [] B)
{ int n = A.Length;
if (B.Length != n)
return false ;
// Create a hash table to
// number of instances
Dictionary< int , int > m = new Dictionary< int , int >();
// for each element of A
// increase it's instance by 1.
for ( int i = 0; i < n; i++)
if (m.ContainsKey(A[i]))
m[A[i]] = m[A[i]] + 1;
else
m.Add(A[i], 1);
// for each element of B
// decrease it's instance by 1.
for ( int i = 0; i < n; i++)
if (m.ContainsKey(B[i]))
m[B[i]] = m[B[i]] - 1;
// Iterate through map and check if
// any entry is non-zero
foreach (KeyValuePair< int , int > entry in m)
if (entry.Value != 0)
return false ;
return true ;
} // Driver Code public static void Main(String[] args)
{ int [] A = { 2, 5, 10, 6, 8, 2, 2 };
int [] B = { 2, 5, 6, 8, 10, 2, 2 };
if (areSameSet(A, B))
Console.WriteLine( "Yes" );
else
Console.WriteLine( "No" );
} } // This code is contributed by PrinciRaj1992 |
<script> // JavaScript program to implement Naive approach // to remove duplicates function areSameSet(A,B)
{
let n = A.length;
if (B.length != n)
return false ;
// Create a hash table to
// number of instances
let m = new Map();
// for each element of A
// increase it's instance by 1.
for (let i = 0; i < n; i++)
m.set(A[i], m.get(A[i]) == null ? 1 :
m.get(A[i]) + 1);
// for each element of B
// decrease it's instance by 1.
for (let i = 0; i < n; i++)
m.set(B[i], m.get(B[i]) - 1);
// Iterate through map and check if
// any entry is non-zero
for (let [key, value] of m.entries())
if (value != 0)
return false ;
return true ;
}
// Driver Code
let A=[2, 5, 10, 6, 8, 2, 2 ];
let B=[2, 5, 6, 8, 10, 2, 2];
if (areSameSet(A, B))
document.write( "Yes" );
else
document.write( "No" );
// This code is contributed by rag2127 </script> |
Output:
Yes
Time Complexity: O(n), where n is the number of elements in the given vector.
Auxiliary Space: O(n)