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++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function that returns the restored// permutationvector<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 resultvoid print_result(vector<int> result){ for (int i = 0; i < result.size(); i++) cout << result[i] << " ";}// Driver Codeint 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
// Java program for the above approachimport java.util.*;class GFG{// Function that returns the restored// permutationstatic 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 resultstatic void print_result(Vector<Integer> result){ for (int i = 0; i < result.size(); i++) System.out.print(result.get(i) + " ");}// Driver Codepublic 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
# Python3 program for the above approach# Function that returns the restored# permutationdef 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 resultdef print_result(result): for i in range(len(result)): print(result[i], end = " ") # Driver codedef 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#
// C# program for the above approachusing System;using System.Collections.Generic;class GFG{// Function that returns the restored// permutationstatic 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 resultstatic void print_result(List<int> result){ for(int i = 0; i < result.Count; i++) Console.Write(result[i] + " ");}// Driver Codepublic 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 |
Javascript
<script> // Javascript program for the above approach// Function that returns the restored// permutationfunction 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 resultfunction print_result(result){ for (var i = 0; i < result.length; i++) { document.write( result[i] + " "); } }// Driver Code// Given Arrayvar arr = [1, 13, 1, 24, 13, 24, 2, 2];var N = arr.length;// Function Callprint_result(restore(arr, N));</script> |
Output:
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++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function that returns the restored// permutationvector<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 resultvoid print_result(vector<int> result){ for (int i = 0; i < result.size(); i++) cout << result[i] << " ";}// Driver Codeint 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
// Java program for the above approachimport java.util.*;class GFG{// Function that returns the restored// permutationstatic 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 resultstatic void print_result(Vector<Integer> result){ for(int i = 0; i < result.size(); i++) System.out.print(result.get(i) + " ");}// Driver Codepublic 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
# Python3 program for the # above approach# Function that returns the# restored permutationdef 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 # resultdef print_result(result): for i in range(len(result)): print(result[i], end = " ")# Driver Codeif __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#
// C# program for the above approachusing System;using System.Collections.Generic;class GFG{// Function that returns the restored// permutationstatic 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 resultstatic void print_result(List<int> result){ for(int i = 0; i < result.Count; i++) Console.Write(result[i] + " ");}// Driver Codepublic 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 |
Javascript
<script>// Javascript program for the above approach// Function that returns the restored// permutationfunction 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 resultfunction print_result(result){ for(let i = 0; i < result.length; i++) document.write(result[i] + " ");}// Driver Codelet arr = [ 1, 13, 1, 24, 13, 24, 2, 2 ];let N = arr.length;// Function callprint_result(restore(arr, N));// This code is contributed by avanitrachhadiya2155</script> |
Output:
1 13 24 2
Time Complexity: O(N)
Auxiliary Space: O(N)
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