Find the Missing Number

Difficulty Level : Easy
Last Updated : 21 Nov, 2020

You are given a list of n-1 integers and these integers are in the range of 1 to n. There are no duplicates in the list. One of the integers is missing in the list. Write an efficient code to find the missing integer.
Example:

Input: arr[] = {1, 2, 4, 6, 3, 7, 8}
Output: 5
Explanation: The missing number from 1 to 8 is 5

Input: arr[] = {1, 2, 3, 5}
Output: 4
Explanation: The missing number from 1 to 5 is 4

Method 1: This method uses the technique of the summation formula.

Approach: The length of the array is n-1. So the sum of all n elements, i.e sum of numbers from 1 to n can be calculated using the formula n*(n+1)/2. Now find the sum of all the elements in the array and subtract it from the sum of first n natural numbers, it will be the value of the missing element.
Algorithm:
1. Calculate the sum of first n natural numbers as sumtotal= n*(n+1)/2
2. Create a variable sum to store the sum of array elements.
3. Traverse the array from start to end.
4. Update the value of sum as sum = sum + array[i]
5. Print the missing number as sumtotal – sum
Implementation:

C++

#include <bits/stdc++.h>
using namespace std;
 
// Function to get the missing number
int getMissingNo(int a[], int n)
{
 
    int total = (n + 1) * (n + 2) / 2;
    for (int i = 0; i < n; i++)
        total -= a[i];
    return total;
}
 
// Driver Code
int main()
{
    int arr[] = { 1, 2, 4, 5, 6 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int miss = getMissingNo(arr, n);
    cout << miss;
}

C

#include <stdio.h>
 
/* getMissingNo takes array and size of array as arguments*/
int getMissingNo(int a[], int n)
{
    int i, total;
    total = (n + 1) * (n + 2) / 2;
    for (i = 0; i < n; i++)
        total -= a[i];
    return total;
}
 
/*program to test above function */
int main()
{
    int a[] = { 1, 2, 4, 5, 6 };
    int miss = getMissingNo(a, 5);
    printf("%d", miss);
    getchar();
}

Java

// Java program to find missing Number
 
class Main {
    // Function to ind missing number
    static int getMissingNo(int a[], int n)
    {
        int i, total;
        total = (n + 1) * (n + 2) / 2;
        for (i = 0; i < n; i++)
            total -= a[i];
        return total;
    }
 
    /* program to test above function */
    public static void main(String args[])
    {
        int a[] = { 1, 2, 4, 5, 6 };
        int miss = getMissingNo(a, 5);
        System.out.println(miss);
    }
}

Python

# getMissingNo takes list as argument
def getMissingNo(A):
    n = len(A)
    total = (n + 1)*(n + 2)/2
    sum_of_A = sum(A)
    return total - sum_of_A
 
# Driver program to test the above function
A = [1, 2, 4, 5, 6]
miss = getMissingNo(A)
print(miss)
# This code is contributed by Pratik Chhajer

C#

// C# program to find missing Number
using System;
 
class GFG {
    // Function to ind missing number
    static int getMissingNo(int[] a, int n)
    {
        int total = (n + 1) * (n + 2) / 2;
 
        for (int i = 0; i < n; i++)
            total -= a[i];
 
        return total;
    }
 
    /* program to test above function */
    public static void Main()
    {
        int[] a = { 1, 2, 4, 5, 6 };
        int miss = getMissingNo(a, 5);
        Console.Write(miss);
    }
}
 
// This code is contributed by Sam007_

PHP

<?php
// PHP program to find
// the Missing Number
 
// getMissingNo takes array and
// size of array as arguments
function getMissingNo ($a, $n)
{
    $total = ($n + 1) * ($n + 2) / 2; 
    for ( $i = 0; $i < $n; $i++)
        $total -= $a[$i];
    return $total;
}
 
// Driver Code
$a = array(1, 2, 4, 5, 6);
$miss = getMissingNo($a, 5);
echo($miss);
 
// This code is contributed by Ajit.
?>

Output:

Complexity Analysis:
- Time Complexity: O(n).
  Only one traversal of the array is needed.
- Space Complexity: O(1).
  No extra space is needed

Modification for Overflow

Approach: The approach remains the same but there can be overflow if n is large. In order to avoid integer overflow, pick one number from known numbers and subtract one number from given numbers. This way there won’t have Integer Overflow ever.
Algorithm:
1. Create a variable sum = 1 to which will store the missing number and a counter c = 2.
2. Traverse the array from start to end.
3. Update the value of sum as sum = sum – array[i] + c and update c as c++.
4. Print the missing number as a sum.

C++

#include <bits/stdc++.h>
using namespace std;
 
// a represents the array
// n : Number of elements in array a
int getMissingNo(int a[], int n) 
{ 
    int i, total=1; 
     
    for ( i = 2; i<= (n+1); i++)
    {
        total+=i;
        total -= a[i-2];
    }
    return total; 
} 
 
//Driver Program
int main() {
    int arr[] = {1, 2, 3, 5};
    cout<<getMissingNo(arr,sizeof(arr)/sizeof(arr[0]));
    return 0;
}
 
//This code is contributed by Ankur Goel

Java

// Java implementation 
class GFG
{
 
    // a represents the array
    // n : Number of elements in array a
    static int getMissingNo(int a[], int n) 
    {
        int total = 1;
        for (int i = 2; i <= (n + 1); i++)
        {
            total += i;
            total -= a[i - 2];
        }
        return total;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int[] arr = { 1, 2, 3, 5 };
        System.out.println(getMissingNo(arr, arr.length));
    }
}
 
// This post is contributed 
// by Vivek Kumar Singh

Python3

# a represents the array
# n : Number of elements in array a
def getMissingNo(a, n): 
    i, total = 0, 1
 
    for i in range(2, n + 2):
        total += i
        total -= a[i - 2]
    return total
 
# Driver Code
arr = [1, 2, 3, 5]
print(getMissingNo(arr, len(arr)))
 
# This code is contributed by Mohit kumar

C#

using System;
 
class GFG
{
     
// a represents the array
// n : Number of elements in array a
static int getMissingNo(int[] a, int n) 
{ 
    int i, total = 1; 
     
    for ( i = 2; i <= (n + 1); i++)
    {
        total += i;
        total -= a[i - 2];
    }
    return total; 
} 
 
// Driver Code
public static void Main() 
{
    int[] arr = {1, 2, 3, 5};
    Console.Write(getMissingNo(arr, (arr.Length)));
 
    // Console.Write(getMissingNo(arr, 4));
}
}
// This code is contributed by SoumikMondal

Output:

Complexity Analysis:
- Time Complexity: O(n).
  Only one traversal of the array is needed.
- Space Complexity:O(1).
  No extra space is needed

Thanks to Sahil Rally for suggesting this improvement.
Method 2: This method uses the technique of XOR to solve the problem.

Approach:
XOR has certain properties
- Assume a1 ^ a2 ^ a3 ^ …^ an = a and a1 ^ a2 ^ a3 ^ …^ an-1 = b
- Then a ^ b = an
Algorithm:
1. Create two variables a = 0 and b = 0
2. Run a loop from 1 to n with i as counter.
3. For every index update a as a = a ^ i
4. Now traverse the array from start to end.
5. For every index update b as b = b ^ array[i]
6. Print the missing number as a ^ b.
Implementation:

C++

#include <bits/stdc++.h>
using namespace std;
 
// Function to get the missing number
int getMissingNo(int a[], int n)
{
    // For xor of all the elements in array
    int x1 = a[0];
 
    // For xor of all the elements from 1 to n+1
    int x2 = 1;
 
    for (int i = 1; i < n; i++)
        x1 = x1 ^ a[i];
 
    for (int i = 2; i <= n + 1; i++)
        x2 = x2 ^ i;
 
    return (x1 ^ x2);
}
 
// Driver Code
int main()
{
    int arr[] = { 1, 2, 4, 5, 6 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int miss = getMissingNo(arr, n);
    cout << miss;
}

C

#include <stdio.h>
 
/* getMissingNo takes array and size of array as arguments*/
int getMissingNo(int a[], int n)
{
    int i;
    int x1 = a[0]; /* For xor of all the elements in array */
    int x2 = 1; /* For xor of all the elements from 1 to n+1 */
 
    for (i = 1; i < n; i++)
        x1 = x1 ^ a[i];
 
    for (i = 2; i <= n + 1; i++)
        x2 = x2 ^ i;
 
    return (x1 ^ x2);
}
 
/*program to test above function */
int main()
{
    int a[] = { 1, 2, 4, 5, 6 };
    int miss = getMissingNo(a, 5);
    printf("%d", miss);
    getchar();
}

Java

// Java program to find missing Number
// using xor
class Main {
    // Function to find missing number
    static int getMissingNo(int a[], int n)
    {
        int x1 = a[0];
        int x2 = 1;
 
        /* For xor of all the elements 
           in array */
        for (int i = 1; i < n; i++)
            x1 = x1 ^ a[i];
 
        /* For xor of all the elements 
           from 1 to n+1 */
        for (int i = 2; i <= n + 1; i++)
            x2 = x2 ^ i;
 
        return (x1 ^ x2);
    }
 
    /* program to test above function */
    public static void main(String args[])
    {
        int a[] = { 1, 2, 4, 5, 6 };
        int miss = getMissingNo(a, 5);
        System.out.println(miss);
    }
}

Python3

# Python3 program to find
# the mising Number
# getMissingNo takes list as argument 
 
def getMissingNo(a, n):
    x1 = a[0]
    x2 = 1
     
    for i in range(1, n):
        x1 = x1 ^ a[i]
         
    for i in range(2, n + 2):
        x2 = x2 ^ i
     
    return x1 ^ x2
 
 
# Driver program to test above function
if __name__=='__main__':
 
    a = [1, 2, 4, 5, 6]
    n = len(a)
    miss = getMissingNo(a, n) 
    print(miss)
     
# This code is contributed by Yatin Gupta 

C#

// C# program to find missing Number
// using xor
using System;
 
class GFG {
    // Function to find missing number
    static int getMissingNo(int[] a, int n)
    {
        int x1 = a[0];
        int x2 = 1;
 
        /* For xor of all the elements 
        in array */
        for (int i = 1; i < n; i++)
            x1 = x1 ^ a[i];
 
        /* For xor of all the elements 
        from 1 to n+1 */
        for (int i = 2; i <= n + 1; i++)
            x2 = x2 ^ i;
 
        return (x1 ^ x2);
    }
 
    /* driver program to test above function */
    public static void Main()
    {
        int[] a = { 1, 2, 4, 5, 6 };
        int miss = getMissingNo(a, 5);
        Console.Write(miss);
    }
}
 
// This code is contributed by Sam007_

PHP

<?php
// PHP program to find
// the Misiing Number
// getMissingNo takes array and 
// size of array as arguments
function getMissingNo($a, $n)
{
    // For xor of all the
    // elements in array 
    $x1 = $a[0]; 
     
    // For xor of all the 
    // elements from 1 to n + 1
    $x2 = 1; 
     
    for ($i = 1; $i < $n; $i++)
        $x1 = $x1 ^ $a[$i];
             
    for ($i = 2; $i <= $n + 1; $i++)
        $x2 = $x2 ^ $i;     
     
    return ($x1 ^ $x2);
}
 
// Driver Code
$a = array(1, 2, 4, 5, 6);
$miss = getMissingNo($a, 5);
echo($miss);
 
// This code is contributed by Ajit.
?>

Output:

Complexity Analysis:
- Time Complexity: O(n).
  Only one traversal of the array is needed.
- Space Complexity: O(1).
  No extra space is needed.

Method 3: This method will work only in python.
Approach:
Take the sum of all elements in the array and subtract that from the sum of n+1 elements. For Eg:
If my elements are li=[1,2,4,5] then take the sum of all elements in li and subtract that from the sum of len(li)+1 elements. The following code shows the demonstration.

C++

// C++ program to find
// the mising Number
#include<bits/stdc++.h>
using namespace std;
 
// getMissingNo takes list 
// as argument 
int getMissingNo(int a[] , 
                 int n)
{
  int n_elements_sum = n * (n + 1) / 2 ;
  int sum = 0;
 
  for(int i = 0; i < n - 1; i++)
    sum += a[i];
  return n_elements_sum-sum;
}
 
// Driver code
int main()
{
  int a[] = {1, 2, 4, 5, 6};
  int n = sizeof(a) / 
          sizeof(a[0]) + 1;
  int miss = getMissingNo(a, n);
  cout << (miss);
  return 0;
}
 
// This code is contributed by Chitranayal

Java

// Java program to find
// the missing Number
class GFG{
 
// getMissingNo function for 
// finding missing number
static int getMissingNo(int a[], int n)
{
    int n_elements_sum = n * (n + 1) / 2;
    int sum = 0;
 
    for(int i = 0; i < n - 1; i++)
        sum += a[i];
         
    return n_elements_sum - sum;
}
 
// Driver code
public static void main(String[] args)
{
    int a[] = { 1, 2, 4, 5, 6 };
    int n = a.length + 1;
    int miss = getMissingNo(a, n);
     
    System.out.print(miss);
}
}
 
// This code is contributed by mark_85

Python3

# Python3 program to find
# the mising Number
# getMissingNo takes list as argument 
def getMissingNo(a, n):
    n_elements_sum=n*(n+1)//2
    return n_elements_sum-sum(a)
 
 
# Driver program to test above function
if __name__=='__main__':
 
    a = [1, 2, 4, 5, 6]
    n = len(a)+1
    miss = getMissingNo(a, n) 
    print(miss)

C#

// C# program to implement 
// the above approach
using System; 
class GFG{ 
   
// Function to find missing 
// number 
static int getMissingNo(int[] a, 
                        int n) 
{ 
  int n_elements_sum = (n * (n + 1) / 2);
  int sum = 0; 
 
  for (int i = 0; i < n - 1; i++) 
    sum = sum + a[i]; 
 
  return (n_elements_sum - sum); 
} 
 
// Driver code
public static void Main() 
{ 
  int[] a = {1, 2, 4, 5, 6}; 
  int miss = getMissingNo(a, 5); 
  Console.Write(miss); 
} 
} 
 
// This code is contributed by Virusbuddah

Output:

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

https://youtu.be/lBk6fVhuCyw?list=PLqM7alHXFySEQDk2MDfbwEdjd2svVJH9p

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes

Related Articles

C++

C

Java

Python

C#

PHP

C++

Java

Python3

C#

C++

C

Java

Python3

C#

PHP

C++

Java

Python3

C#