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Find missing elements from an Array

  • Difficulty Level : Medium
  • Last Updated : 15 Mar, 2021

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

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

Javascript




<script>
 
    // Javascript implementation of the approach
     
    // Function to find the missing elements
    function missing_elements(vec)
    {
      
        // Vector to store the list
        // of missing elements
        let mis = [];
      
        // For every given element
        for (let i = 0; i < vec.length; i++) {
      
            // Find its index
            let temp = Math.abs(vec[i]) - 1;
      
            // Update the element at the found index
            vec[temp] = vec[temp] > 0 ? -vec[temp] : vec[temp];
        }
        for (let i = 0; i < vec.length; i++)
      
            // Current element was not present
            // in the original vector
            if (vec[i] > 0)
                mis.push(i + 1);
      
        return mis;
    }
     
    let vec = [ 3, 3, 3, 5, 1 ];
  
    // Vector to store the returned
    // list of missing elements
    let miss_ele = missing_elements(vec);
  
    // Print the list of elements
    for (let i = 0; i < miss_ele.length; i++)
        document.write(miss_ele[i] + " ");
 
</script>
Output: 
2 4

 

Time Complexity: O(N), where N is the length of the vector vec.

Auxiliary Space: O(N)

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