Find the Missing Number

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 list. One of the integers is missing in the list. Write an efficient code to find the missing integer.

Example :
I/P    [1, 2, 4, ,6, 3, 7, 8]
O/P    5



METHOD 1(Use sum formula)
Algorithm:



1. Get the sum of numbers 
       total = n*(n+1)/2
2  Subtract all the numbers from sum and
   you will get the missing number.

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 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 : 

3

Time Complexity : O(n)

There can be overflow if n is large. In order to avoid Integer Overflow, we can pick one number from known numbers and subtract one number from given numbers. This way we wont have Integer Overflow ever. Thanks to Sahil Rally for suggesting this improvement.



METHOD 2(Use XOR)



  1) XOR all the array elements, let the result of XOR be X1.
  2) XOR all numbers from 1 to n, let XOR be X2.
  3) XOR of X1 and X2 gives the missing number.

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 : 

3

Time Complexity : O(n)

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Improved By : jit_t, YatinGupta