Given a number N and an array arr[] that consist of merging N length sequence of distinct integers any number of times maintaining the relative order of elements in the initial sequence. The task is to find the initial sequence of length N maintaining the right order.
Examples:
Input: N = 4, arr[] = {1, 13, 1, 24, 13, 24, 2, 2}
Output: 1 13 24 2
Explanation:
Here the given sequence is obtained by merging 1 13 24 2 maintaining the relative order of elements. Therefore, the answer is 1 13 24 2.Input: N = 3, arr[] = {3, 2, 3, 1, 2, 3, 2, 1, 1}
Output: 3 2 1
Explanation:
Here the given sequence is obtained by merging 3 2 1 maintaining the relative order of elements. Therefore, the answer is 3 2 1.
Approach: The idea is to observe that the element occurring first in the given sequence is the first element of the resultant restored sequence. Take that element in our restored sequence and don’t include duplicate from the given sequence. Perform the same for the rest of the elements until we reach the end. The idea can be implemented by both Map and Set in C++.
Using Map
- Traverse through the given sequence from left to right.
- The element coming for the first time in the sequence is taken into account and marked using a map.
- Elements that are marked while traversing are ignored.
- Step 2 and Step 3 is continued until the end of the given sequence is reached.
Below is the implementation of the above approach:
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Function that returns the restored // permutation vector< int > restore( int arr[], int N)
{ // Vector to store the result
vector< int > result;
// Map to mark the elements
// which are taken in result
map< int , int > mp;
for ( int i = 0; i < N; i++) {
// Check if the element is
// coming first time
if (mp[arr[i]] == 0) {
// Push in result vector
result.push_back(arr[i]);
// Mark it in the map
mp[arr[i]]++;
}
}
// Return the answer
return result;
} // Function to print the result void print_result(vector< int > result)
{ for ( int i = 0; i < result.size(); i++)
cout << result[i] << " " ;
} // Driver Code int main()
{ // Given Array
int arr[] = { 1, 13, 1, 24, 13, 24, 2, 2 };
int N = sizeof (arr) / sizeof (arr[0]);
// Function Call
print_result(restore(arr, N));
return 0;
} |
// Java program for the above approach import java.util.*;
class GFG{
// Function that returns the restored // permutation static Vector<Integer> restore( int arr[], int N)
{ // Vector to store the result
Vector<Integer> result = new Vector<>();
// Map to mark the elements
// which are taken in result
HashMap<Integer,
Integer> mp = new HashMap<Integer,
Integer>();
for ( int i = 0 ; i < N; i++)
{
// Check if the element is
// coming first time
if (mp.containsKey(arr[i]) &&
mp.get(arr[i]) == 0 )
{
// Push in result vector
result.add(arr[i]);
// Mark it in the map
if (mp.containsKey(arr[i]))
{
mp.put(arr[i], mp.get(arr[i]) + 1 );
}
else
{
mp.put(arr[i], 1 );
}
}
else
mp.put(arr[i], 0 );
}
// Return the answer
return result;
} // Function to print the result static void print_result(Vector<Integer> result)
{ for ( int i = 0 ; i < result.size(); i++)
System.out.print(result.get(i) + " " );
} // Driver Code public static void main(String[] args)
{ // Given Array
int arr[] = { 1 , 13 , 1 , 24 , 13 , 24 , 2 , 2 };
int N = arr.length;
// Function Call
print_result(restore(arr, N));
} } // This code is contributed by Rajput-Ji |
# Python3 program for the above approach # Function that returns the restored # permutation def restore(arr, N):
# List to store the result
result = []
# Map to mark the elements
# which are taken in result
mp = {}
for i in range (N):
# Checking if the element is
# coming first time
if not arr[i] in mp:
# Push in result vector
result.append(arr[i])
# Mark it in the map
mp[arr[i]] = 1
# Return the answer
return result
# Function to print the result def print_result(result):
for i in range ( len (result)):
print (result[i], end = " " )
# Driver code def main():
# Given array
arr = [ 1 , 13 , 1 , 24 , 13 , 24 , 2 , 2 ]
N = len (arr)
# Function call
print_result(restore(arr, N))
main() # This code is contributed by Stuti Pathak |
// C# program for the above approach using System;
using System.Collections.Generic;
class GFG{
// Function that returns the restored // permutation static List< int > restore( int []arr, int N)
{ // List to store the result
List< int > result = new List< int >();
// Map to mark the elements
// which are taken in result
Dictionary< int ,
int > mp = new Dictionary< int ,
int >();
for ( int i = 0; i < N; i++)
{
// Check if the element is
// coming first time
if (mp.ContainsKey(arr[i]) &&
mp[arr[i]] == 0)
{
// Push in result vector
result.Add(arr[i]);
// Mark it in the map
if (mp.ContainsKey(arr[i]))
{
mp[arr[i]] = mp[arr[i]] + 1;
}
else
{
mp.Add(arr[i], 1);
}
}
else
mp.Add(arr[i], 0);
}
// Return the answer
return result;
} // Function to print the result static void print_result(List< int > result)
{ for ( int i = 0; i < result.Count; i++)
Console.Write(result[i] + " " );
} // Driver Code public static void Main(String[] args)
{ // Given Array
int []arr = { 1, 13, 1, 24, 13, 24, 2, 2 };
int N = arr.Length;
// Function call
print_result(restore(arr, N));
} } // This code is contributed by 29AjayKumar |
<script> // Javascript program for the above approach // Function that returns the restored // permutation function restore(arr, N)
{ // Vector to store the result
var result = [];
// Map to mark the elements
// which are taken in result
var mp = new Map();
for ( var i = 0; i < N; i++) {
// Check if the element is
// coming first time
if (!mp.has(arr[i])) {
// Push in result vector
result.push(arr[i]);
// Mark it in the map
mp.set(arr[i], mp.get(arr[i])+1);
}
}
// Return the answer
return result;
} // Function to print the result function print_result(result)
{ for ( var i = 0; i < result.length; i++)
{
document.write( result[i] + " " );
}
} // Driver Code // Given Array var arr = [1, 13, 1, 24, 13, 24, 2, 2];
var N = arr.length;
// Function Call print_result(restore(arr, N)); </script> |
1 13 24 2
Time Complexity: O(N)
Auxiliary Space: O(N)
Using Set
- Traverse through the given sequence from left to right.
- A counter is taken and initialized as 1.
- Insert the elements into the set one by one. If at some point that the size of the set and the counter is the same, the element is taken into account and the counter is increased by 1.
- Step 3 and Step 4 is continued until the end of the given sequence is reached.
Below is the implementation of the above approach:
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Function that returns the restored // permutation vector< int > restore( int arr[], int N)
{ // Vector to store the result
vector< int > result;
int count1 = 1;
// Set to insert unique elements
set< int > s;
for ( int i = 0; i < N; i++) {
s.insert(arr[i]);
// Check if the element is
// coming first time
if (s.size() == count1) {
// Push in result vector
result.push_back(arr[i]);
count1++;
}
}
return result;
} // Function to print the result void print_result(vector< int > result)
{ for ( int i = 0; i < result.size(); i++)
cout << result[i] << " " ;
} // Driver Code int main()
{ // Given Array
int arr[] = { 1, 13, 1, 24, 13, 24, 2, 2 };
int N = sizeof (arr) / sizeof (arr[0]);
// Function Call
print_result(restore(arr, N));
return 0;
} |
// Java program for the above approach import java.util.*;
class GFG{
// Function that returns the restored // permutation static Vector<Integer> restore( int arr[], int N)
{ // Vector to store the result
Vector<Integer> result = new Vector<Integer>();
int count1 = 1 ;
// Set to insert unique elements
HashSet<Integer> s = new HashSet<Integer>();
for ( int i = 0 ; i < N; i++)
{
s.add(arr[i]);
// Check if the element is
// coming first time
if (s.size() == count1)
{
// Push in result vector
result.add(arr[i]);
count1++;
}
}
return result;
} // Function to print the result static void print_result(Vector<Integer> result)
{ for ( int i = 0 ; i < result.size(); i++)
System.out.print(result.get(i) + " " );
} // Driver Code public static void main(String[] args)
{ // Given Array
int arr[] = { 1 , 13 , 1 , 24 , 13 , 24 , 2 , 2 };
int N = arr.length;
// Function call
print_result(restore(arr, N));
} } // This code is contributed by Princi Singh |
# Python3 program for the # above approach # Function that returns the # restored permutation def restore(arr, N):
# Vector to store the
# result
result = []
count1 = 1
# Set to insert unique
# elements
s = set ([])
for i in range (N):
s.add(arr[i])
# Check if the element is
# coming first time
if ( len (s) = = count1):
# Push in result vector
result.append(arr[i])
count1 + = 1
return result
# Function to print the # result def print_result(result):
for i in range ( len (result)):
print (result[i],
end = " " )
# Driver Code if __name__ = = "__main__" :
# Given Array
arr = [ 1 , 13 , 1 , 24 ,
13 , 24 , 2 , 2 ]
N = len (arr)
# Function Call
print_result(restore(arr, N))
# This code is contributed by Chitranayal |
// C# program for the above approach using System;
using System.Collections.Generic;
class GFG{
// Function that returns the restored // permutation static List< int > restore( int []arr, int N)
{ // List to store the result
List< int > result = new List< int >();
int count1 = 1;
// Set to insert unique elements
HashSet< int > s = new HashSet< int >();
for ( int i = 0; i < N; i++)
{
s.Add(arr[i]);
// Check if the element is
// coming first time
if (s.Count == count1)
{
// Push in result vector
result.Add(arr[i]);
count1++;
}
}
return result;
} // Function to print the result static void print_result(List< int > result)
{ for ( int i = 0; i < result.Count; i++)
Console.Write(result[i] + " " );
} // Driver Code public static void Main(String[] args)
{ // Given Array
int []arr = { 1, 13, 1, 24, 13, 24, 2, 2 };
int N = arr.Length;
// Function call
print_result(restore(arr, N));
} } // This code is contributed by PrinciRaj1992 |
<script> // Javascript program for the above approach // Function that returns the restored // permutation function restore(arr, N)
{ // Vector to store the result
let result = [];
let count1 = 1;
// Set to insert unique elements
let s = new Set();
for (let i = 0; i < N; i++)
{
s.add(arr[i]);
// Check if the element is
// coming first time
if (s.size == count1)
{
// Push in result vector
result.push(arr[i]);
count1++;
}
}
return result;
} // Function to print the result function print_result(result)
{ for (let i = 0; i < result.length; i++)
document.write(result[i] + " " );
} // Driver Code let arr = [ 1, 13, 1, 24, 13, 24, 2, 2 ]; let N = arr.length; // Function call print_result(restore(arr, N)); // This code is contributed by avanitrachhadiya2155 </script> |
1 13 24 2
Time Complexity: O(N)
Auxiliary Space: O(N)