Codility - Lesson 3 Time complexity - 1. PermMissingElem

Source Link:
https://codility.com/programmers/lessons/3-time_complexity/perm_missing_elem/

Question:

A zero-indexed array A consisting of N different integers is given. The array contains integers in the range [1..(N + 1)], which means that exactly one element is missing.
Your goal is to find that missing element.
Write a function:

class Solution { public int solution(int[] A); }

that, given a zero-indexed array A, returns the value of the missing element.
For example, given array A such that:
A[0] = 2 A[1] = 3 A[2] = 1 A[3] = 5 the function should return 4, as it is the missing element.
Assume that:

• N is an integer within the range [0..100,000];
• the elements of A are all distinct;
• each element of array A is an integer within the range [1..(N + 1)].

Complexity:

• expected worst-case time complexity is O(N);
• expected worst-case space complexity is O(1), beyond input storage (not counting the storage required for input arguments).

Elements of input arrays can be modified.

My Solution:
Notes:
1.  Using the concept of "Sum = (ceiling + floor) * height /2"
     So---> Sum = (1 + N+1) * N /2
     the missing element can be found by minus other elements
2.  Need to use "long" to avoid potential bugs (large numbers)
3. Because there is one element "missing" , need to plus extra "1" (be careful about this) 
long height = A.length + 1; 
4. Minus other elements 
for(int i=0; i<A.length; i++){
missing_number = missing_number - A[i]; // minus other elements
}
5. return the missing element (long->int)
return (int)missing_number;