Find missing elements from an Array

Given a list of integers from the range [1, N] with some of the elements missing. The task is to find the missing elements. Note that there can be duplicates in the list.

Examples:

Input: arr[] = {1, 3, 3, 3, 5}
Output: 2 4

Input: arr[] = {1, 2, 3, 4, 4, 7, 7} 
Output: 5 6

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: In the given range [1, N] there should be an element corresponding to each index. If an element is missing then its index will never be visited.

Traverse the array:
For each element:
    if array[element] > 0:
          Mark the element as visited
Again, traverse the array:
     if element isNot Visited:
           add it as missing element

Below is the implementation of the above approach:

C++

// C++ implementation of the approach 
#include <bits/stdc++.h> 
using namespace std; 
  
// Function to find the missing elements 
vector<int> missing_elements(vector<int> vec) 
{ 
  
    // Vector to store the list 
    // of missing elements 
    vector<int> mis; 
  
    // For every given element 
    for (int i = 0; i < vec.size(); i++) { 
  
        // Find its index 
        int temp = abs(vec[i]) - 1; 
  
        // Update the element at the found index 
        vec[temp] = vec[temp] > 0 ? -vec[temp] : vec[temp]; 
    } 
    for (int i = 0; i < vec.size(); i++) 
  
        // Current element was not present 
        // in the original vector 
        if (vec[i] > 0) 
            mis.push_back(i + 1); 
  
    return mis; 
} 
  
// Driver code 
int main() 
{ 
    vector<int> vec = { 3, 3, 3, 5, 1 }; 
  
    // Vector to store the returned 
    // list of missing elements 
    vector<int> miss_ele = missing_elements(vec); 
  
    // Print the list of elements 
    for (int i = 0; i < miss_ele.size(); i++) 
        cout << miss_ele[i] << " "; 
  
    return 0; 
} 

Java

// Java implementation of the above approach  
import java.util.*; 
  
class GFG 
{ 
      
    // Function to find the missing elements  
    static Vector missing_elements(Vector vec)  
    {  
      
        // Vector to store the list  
        // of missing elements  
        Vector mis = new Vector(); 
      
        // For every given element  
        for (int i = 0; i < vec.size(); i++)  
        {  
      
            // Find its index  
            int temp = Math.abs((int)vec.get(i)) - 1;  
      
            // Update the element at the found index  
            if ((int)vec.get(temp) > 0) 
                vec.set(temp,-(int)vec.get(temp)); 
            else
                vec.set(temp,vec.get(temp)); 
        }  
        for (int i = 0; i < vec.size(); i++)  
        { 
            // Current element was not present  
            // in the original vector  
            if ((int)vec.get(i) > 0)  
                mis.add(i + 1);  
        } 
        return mis;  
    }  
      
    // Driver code  
    public static void main(String args[])  
    {  
        Vector vec = new Vector();  
        vec.add(3); 
        vec.add(3); 
        vec.add(3); 
        vec.add(5); 
        vec.add(1); 
          
        // Vector to store the returned  
        // list of missing elements  
        Vector miss_ele = missing_elements(vec);  
      
        // Print the list of elements  
        for (int i = 0; i < miss_ele.size(); i++)  
            System.out.print(miss_ele.get(i) + " ");  
    }  
} 
  
// This code is contributed by AnkitRai01 

Python3

# Python3 implementation of the approach 
  
# Function to find the missing elements 
def missing_elements(vec): 
  
    # Vector to store the list 
    # of missing elements 
    mis = [] 
  
    # For every given element 
    for i in range(len(vec)): 
  
        # Find its index 
        temp = abs(vec[i]) - 1
  
        # Update the element at the found index 
        if vec[temp] > 0: 
            vec[temp] = -vec[temp] 
  
    for i in range(len(vec)): 
  
        # Current element was not present 
        # in the original vector 
        if (vec[i] > 0): 
            mis.append(i + 1) 
  
    return mis 
  
# Driver code 
vec = [3, 3, 3, 5, 1] 
  
# Vector to store the returned 
# list of missing elements 
miss_ele = missing_elements(vec) 
  
# Print the list of elements 
for i in range(len(miss_ele)): 
    print(miss_ele[i], end = " ") 
  
# This code is contributed by Mohit Kumar 

C#

// C# implementation of the approach 
using System; 
using System.Collections.Generic;              
  
class GFG 
{ 
      
    // Function to find the missing elements  
    static List<int> missing_elements(List<int> vec)  
    {  
      
        // List<int> to store the list  
        // of missing elements  
        List<int> mis = new List<int>(); 
      
        // For every given element  
        for (int i = 0; i < vec.Count; i++)  
        {  
      
            // Find its index  
            int temp = Math.Abs((int)vec[i]) - 1;  
      
            // Update the element at the found index  
            if ((int)vec[temp] > 0) 
                vec[temp] = -(int)vec[temp]; 
            else
                vec[temp] = vec[temp]; 
        }  
        for (int i = 0; i < vec.Count; i++)  
        { 
            // Current element was not present  
            // in the original vector  
            if ((int)vec[i] > 0)  
                mis.Add(i + 1);  
        } 
        return mis;  
    }  
      
    // Driver code  
    public static void Main(String []args)  
    {  
        List<int> vec = new List<int>();  
        vec.Add(3); 
        vec.Add(3); 
        vec.Add(3); 
        vec.Add(5); 
        vec.Add(1); 
          
        // List to store the returned  
        // list of missing elements  
        List<int> miss_ele = missing_elements(vec);  
      
        // Print the list of elements  
        for (int i = 0; i < miss_ele.Count; i++)  
            Console.Write(miss_ele[i] + " ");  
    }  
} 
  
// This code is contributed by 29AjayKumar 

Output:

2 4

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My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

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