Find all missing numbers from a given sorted array

Given a sorted array arr[] of N integers, The task is to find the multiple missing elements in the array between the ranges [arr[0], arr[N-1]].

Examples:

Input: arr[] = {6, 7, 10, 11, 13}
Output: 8 9 12
Explanation:
The elements of the array are present in the range of the maximum and minimum array element [6, 13]. Therefore, the total values will be {6, 7, 8, 9, 10, 11, 12, 13}. 
The elements from the above range which are missing from the array are {8, 9, 12}. 

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

Naive Approach: The naive idea is to iterate over the difference between the consecutive pair of elements and the print all the numbers in this range if the difference is non-zero. Below are the steps:

1. Initialize the variable diff which is equal to arr[0] – 0.
2. Now traverse the array and see if the difference between arr[i] – i and diff is zero or not.
3. If the difference is not equal to zero in the above steps, then the missing element is found.
4. To find the multiple missing elements run a loop inside it and see if the diff is less than arr[i] – i then print the missing element i.e., i + diff.
5. Now increment the diff as the difference is increased now.
6. Repeat from step 2 until all the missing numbers are not found.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find the missing elements
void printMissingElements(int arr[], int N)
{

    // Initialize diff
    int diff = arr[0] - 0;

    for (int i = 0; i < N; i++) {

        // Check if diff and arr[i]-i
        // both are equal or not
        if (arr[i] - i != diff) {

            // Loop for consecutive
            // missing elements
            while (diff < arr[i] - i) {
                cout << i + diff << " ";
                diff++;
            }
        }
    }
}

// Driver Code
int main()
{
    // Given array arr[]
    int arr[] = { 6, 7, 10, 11, 13 };

    int N = sizeof(arr) / sizeof(int);

    // Function Call
    printMissingElements(arr, N);
    return 0;
}

Java

// Java program for the above approach
import java.util.*;

class GFG{

// Function to find the missing elements
static void printMissingElements(int arr[],
                                 int N)
{

    // Initialize diff
    int diff = arr[0] - 0;

    for(int i = 0; i < N; i++)
    {

        // Check if diff and arr[i]-i
        // both are equal or not
        if (arr[i] - i != diff)
        {

            // Loop for consecutive
            // missing elements
            while (diff < arr[i] - i)
            {
                System.out.print((i + diff) + " ");
                diff++;
            }
        }
    }
}

// Driver Code
public static void main (String[] args)
{

    // Given array arr[]
    int arr[] = { 6, 7, 10, 11, 13 };

    int N = arr.length;

    // Function call
    printMissingElements(arr, N);
}
}

// This code is contributed by offbeat

Python3

# Python3 program for the above approach

# Function to find the missing elements
def printMissingElements(arr, N):

    # Initialize diff
    diff = arr[0]

    for i in range(N):

        # Check if diff and arr[i]-i
        # both are equal or not
        if(arr[i] - i != diff):

            # Loop for consecutive
            # missing elements
            while(diff < arr[i] - i):
                print(i + diff, end = " ")
                diff += 1

# Driver Code

# Given array arr[]
arr = [ 6, 7, 10, 11, 13 ]

N = len(arr)

# Function call
printMissingElements(arr, N)

# This code is contributed by Shivam Singh

C#

// C# program for the above approach
using System;

class GFG{

// Function to find the missing elements
static void printMissingElements(int[] arr,
                                 int N)
{

    // Initialize diff
    int diff = arr[0] - 0;

    for(int i = 0; i < N; i++)
    {

        // Check if diff and arr[i]-i
        // both are equal or not
        if (arr[i] - i != diff)
        {

            // Loop for consecutive
            // missing elements
            while (diff < arr[i] - i)
            {
                Console.Write(i + diff + " ");
                diff++;
            }
        }
    }
}

// Driver code
static void Main()
{

    // Given array arr[]
    int[] arr = { 6, 7, 10, 11, 13 };

    int N = arr.Length;

    // Function call
    printMissingElements(arr, N);
}
}

// This code is contributed by divyeshrabadiya07

Javascript

<script>

// Javascript program to gather characters
// of a string in minimum cost

// Function to find the missing elements
function prletMissingElements(arr, N)
{

    // Initialize diff
    let diff = arr[0] - 0;

    for(let i = 0; i < N; i++)
    {

        // Check if diff and arr[i]-i
        // both are equal or not
        if (arr[i] - i != diff)
        {

            // Loop for consecutive
            // missing elements
            while (diff < arr[i] - i)
            {
                document.write((i + diff) + " ");
                diff++;
            }
        }
    }
}

// Driver Code

     // Given array arr[]
    let arr = [ 6, 7, 10, 11, 13 ];

    let N = arr.length;

    // Function call
    prletMissingElements(arr, N);

</script>

Output

8 9 12 

Time Complexity: O(N²)
Auxiliary Space: O(1)

Efficient Approach: The idea is to use Hashing to optimize the above approach. Create a boolean array(say b[]) of size equal to the maximum element in the array and mark only those positions in the array b[] which are present in the given array. Print all the index in the array b[] that are not marked. 
Below are the steps:  

1. Initialize a boolean array b[] with zero of size equals to the maximum element of the array.
2. Iterate over the given array and mark for each element in the given array mark that index as true in the array b[].
3. Now traverse the given array b[] from index arr[0] and print those index whose value is false as they are the element that is missing in the given array.

Below is the implementation of the above approach:

C++

// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to find the missing elements
void printMissingElements(int arr[], int N)
{
    // Initialize an array with zero
    // of size equals to the maximum
    // element in the array
    int b[arr[N - 1] + 1] = { 0 };

    // Make b[i]=1 if i is present
    // in the array
    for (int i = 0; i < N; i++) {

        // If the element is present
        // make b[arr[i]]=1
        b[arr[i]] = 1;
    }

    // Print the indices where b[i]=0
    for (int i = arr[0]; i <= arr[N - 1]; i++) {

        if (b[i] == 0) {
            cout << i << " ";
        }
    }
}

// Driver Code
int main()
{
    // Given array arr[]
    int arr[] = { 6, 7, 10, 11, 13 };

    int N = sizeof(arr) / sizeof(int);

    // Function Call
    printMissingElements(arr, N);

    return 0;
}

Java

// Java program for the above approach
import java.util.*;

class GFG{

// Function to find the missing elements
static void printMissingElements(int arr[],
                                int N)
{

    // Initialize an array with zero
    // of size equals to the maximum
    // element in the array
    int[] b = new int[arr[N - 1] + 1];

    // Make b[i]=1 if i is present
    // in the array
    for(int i = 0; i < N; i++)
    {

        // If the element is present
        // make b[arr[i]]=1
        b[arr[i]] = 1;
    }

    // Print the indices where b[i]=0
    for(int i = arr[0]; i <= arr[N - 1]; i++)
    {
        if (b[i] == 0)
        {
            System.out.print(i + " ");
        }
    }
}

// Driver Code
public static void main (String[] args)
{

    // Given array arr[]
    int arr[] = { 6, 7, 10, 11, 13 };

    int N = arr.length;

    // Function call
    printMissingElements(arr, N);
}
}

// This code is contributed by offbeat

Python3

# Python3 program to implement
# the above approach

# Function to find the missing elements
def printMissingElements(arr, N):

    # Initialize an array with zero
    # of size equals to the maximum
    # element in the array
    b = [0] * (arr[N - 1] + 1)

    # Make b[i]=1 if i is present
    # in the array
    for i in range(N):

        # If the element is present
        # make b[arr[i]]=1
        b[arr[i]] = 1

    # Print the indices where b[i]=0
    for i in range(arr[0], arr[N - 1] + 1):
        if(b[i] == 0):
            print(i, end = " ")

# Driver Code

# Given array arr[]
arr = [ 6, 7, 10, 11, 13 ]

N = len(arr)

# Function call
printMissingElements(arr, N)

# This code is contributed by Shivam Singh

C#

// C# program for
// the above approach
using System;
class GFG{

// Function to find the missing elements
static void printMissingElements(int []arr,
                                 int N)
{
  // Initialize an array with zero
  // of size equals to the maximum
  // element in the array
  int[] b = new int[arr[N - 1] + 1];

  // Make b[i]=1 if i is present
  // in the array
  for(int i = 0; i < N; i++)
  {
    // If the element is present
    // make b[arr[i]]=1
    b[arr[i]] = 1;
  }

  // Print the indices where b[i]=0
  for(int i = arr[0]; i <= arr[N - 1];
          i++)
  {
    if (b[i] == 0)
    {
      Console.Write(i + " ");
    }
  }
}

// Driver Code
public static void Main(String[] args)
{
  // Given array []arr
  int []arr = {6, 7, 10, 11, 13};

  int N = arr.Length;

  // Function call
  printMissingElements(arr, N);
}
}

// This code is contributed by Princi Singh

Javascript

<script>

// JavaScript program for the above approach

// Function to find the missing elements
function printMissingElements(arr, N)
{
    // Initialize an array with zero
    // of size equals to the maximum
    // element in the array
    let b  = new Uint8Array(arr[N - 1] + 1);

    // Make b[i]=1 if i is present
    // in the array
    for (let i = 0; i < N; i++) {

        // If the element is present
        // make b[arr[i]]=1
        b[arr[i]] = 1;
    }

    // Print the indices where b[i]=0
    for (let i = arr[0]; i <= arr[N - 1]; i++) {

        if (b[i] == 0) {
            document.write( i + " ");
        }
    }
}

// Driver Code

    // Given array arr[]
    let arr = [ 6, 7, 10, 11, 13 ];

    let N = arr.length;

    // Function Call
    printMissingElements(arr, N);


//This code is contributed by Mayank Tyagi
</script>

Output

8 9 12 

Time Complexity: O(M), where M is the maximum element of the array. 
Auxiliary Space: O(M)

Most Efficient  And Simple Approach: 

In below approach simple we create a variable (cnt) this variable keeps the track of element present in array

1. We need to traverse the arr[0] to arr[N] to find missing number between it.

2. In for loop if arr[cnt] match to current element then we do not print that element and skip that element because it is present in array 

once we found element then we increment the cnt++ for pointing next element in array 

3. In else part we just print the element which does not match or present in array

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find the missing elements
void printMissingElements(int arr[], int N)
{
    int cnt = 0;
    for (int i = arr[0]; i <= arr[N - 1]; i++) {
        // Check if number is equal to the first element in
        // given array if array element match skip it increment for next element
        if (arr[cnt] == i) {
            // Increment the count to check next element
            cnt++;
        }
        else {
            // Print missing number
            cout << i << " ";
        }
    }
}
// Driver Code
int main()
{
    // Given array arr[]
    int arr[] = { 6, 7, 10, 11, 13 };
    int N = sizeof(arr) / sizeof(arr[0]);
    // Function Call
    printMissingElements(arr, N);

    return 0;
}
//This code is contributed by Kuldeep Kushwaha

Java

//Java program for the above approach
import java.io.*;
class GFG {

  // Function to find the missing elements
  public static void printMissingElements(int arr[], int N)
  {
    int cnt = 0;
    for (int i = arr[0]; i <= arr[N - 1]; i++)
    {

      // Check if number is equal to the first element in
      // given array if array element match skip it increment for next element
      if (arr[cnt] == i)
      {

        // Increment the count to check next element
        cnt++;
      }
      else
      {

        // Print missing number
        System.out.print(i + " ");
      }
    }
  }

  // Driver Code
  public static void main (String[] args)
  {

    // Given array arr[]
    int arr[] = { 6, 7, 10, 11, 13 };
    int N = arr.length;

    // Function Call
    printMissingElements(arr, N);
  }
}

// This code is contributed by Shubham Singh

Python3

# Python program for the above approach

# Function to find the missing elements
def printMissingElements(arr, N):
    cnt = 0
    for i in range(arr[0], arr[N - 1]+1):

        # Check if number is equal to the first element in
        # given array if array element match skip it increment for next element
        if (arr[cnt] == i):

            # Increment the count to check next element
            cnt += 1

        else:

            # Print missing number
            print(i , end = " ")

# Driver Code
# Given array arr[]
arr = [ 6, 7, 10, 11, 13 ]
N = len(arr)

# Function Call
printMissingElements(arr, N)

# This code is contributed by Shubham Singh

C#

//C# program for the above approach
using System;
using System.Linq;

public class GFG{

  // Function to find the missing elements
  public static void printMissingElements(int[] arr, int N)
  {
    int cnt = 0;
    for (int i = arr[0]; i <= arr[N - 1]; i++)
    {

      // Check if number is equal to the first element in
      // given array if array element match skip it increment for next element
      if (arr[cnt] == i)
      {

        // Increment the count to check next element
        cnt++;
      }
      else
      {

        // Print missing number
        Console.Write(i + " ");
      }
    }
  }

  // Driver Code
  static public void Main ()
  {

    // Given array arr[]
    int[] arr = { 6, 7, 10, 11, 13 };
    int N = arr.Length;

    // Function Call
    printMissingElements(arr, N);
  }
}

// This code is contributed by Shubham Singh

Javascript

<script>
// Javascript program for the above approach

// Function to find the missing elements
function printMissingElements(arr, N)
{
    var cnt = 0;
    for (var i = arr[0]; i <= arr[N - 1]; i++) {

        // Check if number is equal to the first element in
        // given array if array element match skip it increment for next element
        if (arr[cnt] == i)
        {

            // Increment the count to check next element
            cnt++;
        }
        else
        {

            // Print missing number
            document.write(i + " ");
        }
    }
}

// Driver Code

// Given array arr[]
var arr = [ 6, 7, 10, 11, 13 ];
var N = arr.length;

// Function Call
printMissingElements(arr, N);

// This code is contributed by Shubham Singh
</script>

Output

8 9 12 

Time Complexity: O(N), where N is the maximum element of the array. 

Auxiliary Space: O(1) Because of this method the overflow of hash or extra space will be saved.