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

• Difficulty Level : Medium
• Last Updated : 05 Jul, 2022

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 ``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

 ``

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.
Letâ€™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 â€“ 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 ``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

 ``

## Javascript

 ``

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)

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