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
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

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

Method 1: This method uses the technique of 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 :

Compelxity Analysis:
- Time Complexity: O(n).
  Only one traversal of 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 sum.

Implementation:

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 

Compelxity Analysis:
- Time Complexity: O(n).
  Only one traversal of 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
Using this property, the missing element can be found. Calculate XOR of all the natural number from 1 to n and store it as a. Now calculate XOR of all the elements of the array and store it as b. The missing number will be a ^ b.
^ is XOR operator.
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 array is needed.
- Space Complexity: O(1).
  No extra space is needed.

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

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My Personal Notes

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

C++

C

Java

Python

C#

PHP

C++

Java

Python3

C#

C++

C

Java

Python3

C#

PHP

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