Find the Missing Number - GeeksforGeeks

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

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

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

METHOD 1(Use sum formula)
Algorithm:

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

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 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 :

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 won’t have Integer Overflow ever. Thanks to Sahil Rally for suggesting this improvement.

Here is the implementation of the same

C++

#include <iostream> 
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 

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 

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 <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 :

Time Complexity : O(n)

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Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

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