Find Two Missing Numbers | Set 1 (An Interesting Linear Time Solution)

Given an array of n unique integers where each element in the array is in the range [1, n]. The array has all distinct elements and the size of the array is (n-2). Hence Two numbers from the range are missing from this array. Find the two missing numbers.

Examples :

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

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

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

Method 1 – O(n) time complexity and O(n) Extra Space

Step 1: Take a boolean array mark that keeps track of all the elements present in the array. 
Step 2: Iterate from 1 to n, check for every element if it is marked as true in the boolean array, if not then simply display that element.

C++

// C++ Program to find two Missing Numbers using O(n)
// extra space
#include <bits/stdc++.h>
using namespace std;

// Function to find two missing numbers in range
// [1, n]. This function assumes that size of array
// is n-2 and all array elements are distinct
void findTwoMissingNumbers(int arr[], int n)
{
    // Create a boolean vector of size n+1 and
    // mark all present elements of arr[] in it.
    vector<bool> mark(n+1, false);
    for (int i = 0; i < n-2; i++)
        mark[arr[i]] = true;

    // Print two unmarked elements
    cout << "Two Missing Numbers are\n";
    for (int i = 1; i <= n; i++)
       if (! mark[i])
           cout << i << " ";

    cout << endl;
}

// Driver program to test above function
int main()
{
    int arr[] = {1, 3, 5, 6};

    // Range of numbers is 2 plus size of array
    int n = 2 + sizeof(arr)/sizeof(arr[0]);

    findTwoMissingNumbers(arr, n);

    return 0;
}

Java

// Java Program to find two Missing Numbers using O(n)
// extra space
import java.util.*;

class GFG
{

// Function to find two missing numbers in range
// [1, n]. This function assumes that size of array
// is n-2 and all array elements are distinct
static void findTwoMissingNumbers(int arr[], int n)
{
    // Create a boolean vector of size n+1 and
    // mark all present elements of arr[] in it.
    boolean []mark = new boolean[n+1];
    for (int i = 0; i < n-2; i++)
        mark[arr[i]] = true;

    // Print two unmarked elements
    System.out.println("Two Missing Numbers are");
    for (int i = 1; i <= n; i++)
    if (! mark[i])
        System.out.print(i + " ");

    System.out.println();
}

// Driver code
public static void main(String[] args)
{
    int arr[] = {1, 3, 5, 6};

    // Range of numbers is 2 plus size of array
    int n = 2 + arr.length;

    findTwoMissingNumbers(arr, n);
}
}

// This code is contributed by 29AjayKumar

Python3

# Python3 program to find two Missing Numbers using O(n)
# extra space

# Function to find two missing numbers in range
# [1, n]. This function assumes that size of array
# is n-2 and all array elements are distinct
def findTwoMissingNumbers(arr, n):
    # Create a boolean vector of size n+1 and
    # mark all present elements of arr[] in it.
    mark = [False for i in range(n+1)]
    for i in range(0,n-2,1):
        mark[arr[i]] = True

    # Print two unmarked elements
    print("Two Missing Numbers are")
    for i in range(1,n+1,1):
        if (mark[i] == False):
            print(i,end = " ")

    print("\n")

# Driver program to test above function
if __name__ == '__main__':
    arr = [1, 3, 5, 6]

    # Range of numbers is 2 plus size of array
    n = 2 + len(arr)


    findTwoMissingNumbers(arr, n);

# This code is contributed by
# Surendra_Gangwar

C#

// C# Program to find two Missing Numbers
// using O(n) extra space
using System;
using System.Collections.Generic;

class GFG
{

// Function to find two missing numbers in range
// [1, n]. This function assumes that size of array
// is n-2 and all array elements are distinct
static void findTwoMissingNumbers(int []arr, int n)
{
    // Create a boolean vector of size n+1 and
    // mark all present elements of arr[] in it.
    Boolean []mark = new Boolean[n + 1];
    for (int i = 0; i < n - 2; i++)
        mark[arr[i]] = true;

    // Print two unmarked elements
    Console.WriteLine("Two Missing Numbers are");
    for (int i = 1; i <= n; i++)
    if (! mark[i])
        Console.Write(i + " ");

    Console.WriteLine();
}

// Driver code
public static void Main(String[] args)
{
    int []arr = {1, 3, 5, 6};

    // Range of numbers is 2 plus size of array
    int n = 2 + arr.Length;

    findTwoMissingNumbers(arr, n);
}
}

// This code contributed by Rajput-Ji

Javascript

<script>

    // Javascript Program to find two
    // Missing Numbers using O(n) extra space

    // Function to find two missing numbers in range
    // [1, n]. This function assumes that size of array
    // is n-2 and all array elements are distinct
    function findTwoMissingNumbers(arr, n)
    {
        // Create a boolean vector of size n+1 and
        // mark all present elements of arr[] in it.
        let mark = new Array(n+1);
        for (let i = 0; i < n-2; i++)
            mark[arr[i]] = true;

        // Print two unmarked elements
        document.write("Two Missing Numbers are" + "</br>");
        for (let i = 1; i <= n; i++)
            if (!mark[i])
                document.write(i + " ");

        document.write("</br>");
    }

    let arr = [1, 3, 5, 6];

    // Range of numbers is 2 plus size of array
    let n = 2 + arr.length;

    findTwoMissingNumbers(arr, n);

</script>

Output

Two Missing Numbers are
2 4 

Method 2 – O(n) time complexity and O(1) Extra Space

The idea is based on this popular solution for finding one missing number. We extend the solution so that two missing elements are printed. 
Let’s find out the sum of 2 missing numbers:

arrSum => Sum of all elements in the array

sum (Sum of 2 missing numbers) = (Sum of integers from 1 to n) - arrSum
                               = ((n)*(n+1))/2 – arrSum 

avg (Average of 2 missing numbers) = sum / 2;

• One of the numbers will be less than or equal to avg while the other one will be strictly greater than avg. Two numbers can never be equal since all the given numbers are distinct.
• We can find the first missing number as a sum of natural numbers from 1 to avg, i.e., avg*(avg+1)/2 minus the sum of array elements smaller than avg
• We can find the second missing number by subtracting the first missing number from the sum of missing numbers

Consider an example for better clarification 

Input : 1 3 5 6, n = 6
Sum of missing integers = n*(n+1)/2 - (1+3+5+6) = 6.
Average of missing integers = 6/2 = 3.
Sum of array elements less than or equal to average = 1 + 3 = 4
Sum of natural numbers from 1 to avg = avg*(avg + 1)/2
                                     = 3*4/2 = 6
First missing number = 6 - 4 = 2
Second missing number = Sum of missing integers-First missing number
Second missing number = 6-2= 4

Below is the implementation of the above idea.

C++

// C++ Program to find 2 Missing Numbers using O(1)
// extra space
#include <iostream>
using namespace std;

// Returns the sum of the array
int getSum(int arr[],int n)
{
    int sum = 0;
    for (int i = 0; i < n; i++)
        sum += arr[i];
    return sum;
}

// Function to find two missing numbers in range
// [1, n]. This function assumes that size of array
// is n-2 and all array elements are distinct
void findTwoMissingNumbers(int arr[],int n)
{
    // Sum of 2 Missing Numbers
    int sum = (n*(n + 1)) /2 - getSum(arr, n-2);

    // Find average of two elements
    int avg = (sum / 2);

    // Find sum of elements smaller than average (avg)
    // and sum of elements greater than average (avg)
    int sumSmallerHalf = 0, sumGreaterHalf = 0;
    for (int i = 0; i < n-2; i++)
    {
        if (arr[i] <= avg)
            sumSmallerHalf += arr[i];
        else
            sumGreaterHalf += arr[i];
    }

    cout << "Two Missing Numbers are\n";

    // The first (smaller) element = (sum of natural
    // numbers upto avg) - (sum of array elements
    // smaller than or equal to avg)
    int totalSmallerHalf = (avg*(avg + 1)) / 2;
    int smallerElement = totalSmallerHalf - sumSmallerHalf;
    cout << smallerElement << " ";

    // The second (larger) element = (sum of both
    // the elements) - smaller element
    cout << sum - smallerElement;
}

// Driver program to test above function
int main()
{
    int arr[] = {1, 3, 5, 6};

    // Range of numbers is 2 plus size of array
    int n = 2 + sizeof(arr)/sizeof(arr[0]);

    findTwoMissingNumbers(arr, n);

    return 0;
}

Java

// Java Program to find 2 Missing
// Numbers using O(1) extra space
import java.io.*;

class GFG
{

// Returns the sum of the array
static int getSum(int arr[], int n)
{
    int sum = 0;
    for (int i = 0; i < n; i++)
        sum += arr[i];
    return sum;
}

// Function to find two missing
// numbers in range [1, n]. This
// function assumes that size of
// array is n-2 and all array
// elements are distinct
static void findTwoMissingNumbers(int arr[],
                                  int n)
{
    // Sum of 2 Missing Numbers
    int sum = (n * (n + 1)) /
               2 - getSum(arr, n - 2);

    // Find average of two elements
    int avg = (sum / 2);

    // Find sum of elements smaller
    // than average (avg) and sum of
    // elements greater than average (avg)
    int sumSmallerHalf = 0,
        sumGreaterHalf = 0;
    for (int i = 0; i < n - 2; i++)
    {
        if (arr[i] <= avg)
            sumSmallerHalf += arr[i];
        else
            sumGreaterHalf += arr[i];
    }

    System.out.println("Two Missing " +
                       "Numbers are");

    // The first (smaller) element =
    // (sum of natural numbers upto
    // avg) - (sum of array elements
    // smaller than or equal to avg)
    int totalSmallerHalf = (avg *
                           (avg + 1)) / 2;
    System.out.println(totalSmallerHalf -
                         sumSmallerHalf);

    // The first (smaller) element =
    // (sum of natural numbers from
    // avg+1 to n) - (sum of array
    // elements greater than avg)
    System.out.println(((n * (n + 1)) / 2 -
                        totalSmallerHalf) -
                           sumGreaterHalf);
}

// Driver Code
public static void main (String[] args)
{
int arr[] = {1, 3, 5, 6};

// Range of numbers is 2
// plus size of array
int n = 2 + arr.length;

findTwoMissingNumbers(arr, n);
}
}

// This code is contributed by aj_36

Python3

# Python Program to find 2 Missing
# Numbers using O(1) extra space

# Returns the sum of the array
def getSum(arr,n):

    sum = 0;
    for i in range(0, n):
        sum += arr[i]
    return sum

# Function to find two missing
# numbers in range [1, n]. This
# function assumes that size of
# array is n-2 and all array
# elements are distinct
def findTwoMissingNumbers(arr, n):

    # Sum of 2 Missing Numbers
    sum = ((n * (n + 1)) / 2 -
           getSum(arr, n - 2));

    #Find average of two elements
    avg = (sum / 2);

    # Find sum of elements smaller
    # than average (avg) and sum
    # of elements greater than
    # average (avg)
    sumSmallerHalf = 0
    sumGreaterHalf = 0;
    for i in range(0, n - 2):

        if (arr[i] <= avg):
            sumSmallerHalf += arr[i]
        else:
            sumGreaterHalf += arr[i]

    print("Two Missing Numbers are")

    # The first (smaller) element = (sum
    # of natural numbers upto avg) - (sum
    # of array elements smaller than or
    # equal to avg)
    totalSmallerHalf = (avg * (avg + 1)) / 2
    print(str(totalSmallerHalf -
              sumSmallerHalf) + " ")

    # The first (smaller) element = (sum
    # of natural numbers from avg+1 to n) -
    # (sum of array elements greater than avg)
    print(str(((n * (n + 1)) / 2 -
               totalSmallerHalf) -
               sumGreaterHalf))

# Driver Code
arr = [1, 3, 5, 6]

# Range of numbers is 2
# plus size of array
n = 2 + len(arr)

findTwoMissingNumbers(arr, n)

# This code is contributed
# by Yatin Gupta

C#

// C# Program to find 2 Missing
// Numbers using O(1) extra space
using System;

class GFG
{

// Returns the sum of the array
static int getSum(int []arr, int n)
{
    int sum = 0;
    for (int i = 0; i < n; i++)
        sum += arr[i];
    return sum;
}

// Function to find two missing
// numbers in range [1, n]. This
// function assumes that size of
// array is n-2 and all array
// elements are distinct
static void findTwoMissingNumbers(int []arr,
                                  int n)
{
    // Sum of 2 Missing Numbers
    int sum = (n * (n + 1)) / 2 -
              getSum(arr, n - 2);

    // Find average of two elements
    int avg = (sum / 2);

    // Find sum of elements smaller
    // than average (avg) and sum of
    // elements greater than average (avg)
    int sumSmallerHalf = 0,
        sumGreaterHalf = 0;
    for (int i = 0; i < n - 2; i++)
    {
        if (arr[i] <= avg)
            sumSmallerHalf += arr[i];
        else
            sumGreaterHalf += arr[i];
    }

    Console.WriteLine("Two Missing " +
                      "Numbers are ");

    // The first (smaller) element =
    // (sum of natural numbers upto
    // avg) - (sum of array elements
    // smaller than or equal to avg)
    int totalSmallerHalf = (avg *
                           (avg + 1)) / 2;
    Console.WriteLine(totalSmallerHalf -
                        sumSmallerHalf);

    // The first (smaller) element =
    // (sum of natural numbers from
    // avg+1 to n) - (sum of array
    // elements greater than avg)
    Console.WriteLine(((n * (n + 1)) / 2 -
                        totalSmallerHalf) -
                        sumGreaterHalf);
}

// Driver Code
static public void Main ()
{
    int []arr = {1, 3, 5, 6};

    // Range of numbers is 2
    // plus size of array
    int n = 2 + arr.Length;

    findTwoMissingNumbers(arr, n);
}
}

// This code is contributed by ajit

PHP

<?php
// PHP Program to find 2 Missing
// Numbers using O(1) extra space

// Returns the sum of the array
function getSum($arr, $n)
{
    $sum = 0;
    for ($i = 0; $i < $n; $i++)
        $sum += $arr[$i];
    return $sum;
}

// Function to find two missing
// numbers in range [1, n]. This
// function assumes that size of
// array is n-2 and all array
// elements are distinct
function findTwoMissingNumbers($arr, $n)
{
    // Sum of 2 Missing Numbers
    $sum = ($n * ($n + 1)) /2 -
            getSum($arr, $n - 2);

    // Find average of two elements
    $avg = ($sum / 2);

    // Find sum of elements smaller
    // than average (avg) and sum of
    // elements greater than average (avg)
    $sumSmallerHalf = 0;
    $sumGreaterHalf = 0;
    for ($i = 0; $i < $n - 2; $i++)
    {
        if ($arr[$i] <= $avg)
            $sumSmallerHalf += $arr[$i];
        else
            $sumGreaterHalf += $arr[$i];
    }

    echo "Two Missing Numbers are\n";

    // The first (smaller) element =
    // (sum of natural numbers upto avg) -
    // (sum of array elements smaller
    // than or equal to avg)
    $totalSmallerHalf = ($avg * ($avg + 1)) / 2;
    echo ($totalSmallerHalf -
          $sumSmallerHalf) , " ";

    // The first (smaller) element =
    // (sum of natural numbers from avg +
    // 1 to n) - (sum of array elements
    // greater than avg)
    echo ((($n * ($n + 1)) / 2 - $totalSmallerHalf) -
                                 $sumGreaterHalf);
}

// Driver Code
$arr= array (1, 3, 5, 6);

// Range of numbers is
// 2 plus size of array
$n = 2 + sizeof($arr);

findTwoMissingNumbers($arr, $n);

// This code is contributed by aj_36
?>

Javascript

<script>

    // Javascript Program to find 2 Missing
    // Numbers using O(1) extra space

    // Returns the sum of the array
    function getSum(arr, n)
    {
        let sum = 0;
        for (let i = 0; i < n; i++)
            sum += arr[i];
        return sum;
    }

    // Function to find two missing
    // numbers in range [1, n]. This
    // function assumes that size of
    // array is n-2 and all array
    // elements are distinct
    function findTwoMissingNumbers(arr, n)
    {
        // Sum of 2 Missing Numbers
        let sum = (n * (n + 1)) / 2 -
        getSum(arr, n - 2);

        // Find average of two elements
        let avg = (sum / 2);

        // Find sum of elements smaller
        // than average (avg) and sum of
        // elements greater than average (avg)
        let sumSmallerHalf = 0,
        sumGreaterHalf = 0;
        for (let i = 0; i < n - 2; i++)
        {
            if (arr[i] <= avg)
                sumSmallerHalf += arr[i];
            else
                sumGreaterHalf += arr[i];
        }

        document.write(
        "Two Missing " + "Numbers are " + "</br>"
        );

        // The first (smaller) element =
        // (sum of natural numbers upto
        // avg) - (sum of array elements
        // smaller than or equal to avg)
        let totalSmallerHalf = (avg * (avg + 1)) / 2;
        document.write(
        (totalSmallerHalf - sumSmallerHalf) + " "
        );

        // The first (smaller) element =
        // (sum of natural numbers from
        // avg+1 to n) - (sum of array
        // elements greater than avg)
        document.write(
        ((n * (n + 1)) / 2 - totalSmallerHalf) -
        sumGreaterHalf + "</br>"
        );
    }

    let arr = [1, 3, 5, 6];

    // Range of numbers is 2
    // plus size of array
    let n = 2 + arr.length;

    findTwoMissingNumbers(arr, n);

</script>

Output

Two Missing Numbers are
2 4

Note: There can be overflow issues in the above solution. 

In below set 2, another solution that is O(n) time, O(1) space, and doesn’t cause overflow issues is discussed.
Find Two Missing Numbers | Set 2 (XOR based solution)