CSES Solutions - Missing Number

Given an array arr[] of size N - 1 containing numbers between 1, 2... N, except one, the task is to find the missing number.

Examples:

Input: N = 5, arr[] = {2, 3, 1, 5}
Output: 4
Explanation: arr[] contains all numbers from 1 to 5 except 4, therefore the answer is 4.

Input: N = 6, arr[] = {4, 5, 1, 2, 3}
Output: 6
Explanation: arr[] contains all numbers from 1 to 6 except 6, therefore the answer is 6.

Approach: To solve the problem, follow the below idea:

The problem can be solved using bitwise XOR operation. We can take the XOR of all the elements of arr[] and all numbers from 1 to N. Except for the missing number, every number from 1 to N will occur two times: once in arr[] and once while iterating from 1 to N. Since taking XOR of any number with itself leads to 0, so the XOR of all the numbers which occur two times will become 0 leaving behind the only number which occurred once while iterating from 1 to N but not in arr[].

Step-by-step algorithm:

• Initialize a variable XOR to store the bitwise XOR of all the elements of arr[] and all the numbers from 1 to N.
• After taking the XOR of all the numbers and elements of arr[], the missing number is equal to XOR.
• Return XOR as the answer.

Below is the implementation of the algorithm:

4

Time Complexity: O(N), where N is the size of array arr[].
Auxiliary Space: O(1)

Approach2 (Using Direct Formula approach)

In this approach we will create Function to find the missing number using the sum of natural numbers formula. First we will Calculate the total sum of the first N natural numbers using formula n * (n + 1) / 2. Now we calculate sum of all elements in given array. Subtract the total sum with sum of all elements in given array and return the missing number.

Example : To demonstrate finding the missing number Using Direct Formula approach

The missing number from the array is: 3

Time Complexity: O(n)
Auxiliary Space: O(1)