Question:

Given an array nums containing n distinct numbers in the range [0, n], return the only number in the range that is missing from the array.

Example 1:
Input: nums = [3,0,1]
Output: 2
Explanation: n = 3 since there are 3 numbers, so all numbers are in the range [0,3]. 2 is the missing number in the range since it does not appear in nums.

Example 2:
Input: nums = [0,1]
Output: 2
Explanation: n = 2 since there are 2 numbers, so all numbers are in the range [0,2]. 2 is the missing number in the range since it does not appear in nums.

Example 3:
Input: nums = [9,6,4,2,3,5,7,0,1]
Output: 8
Explanation: n = 9 since there are 9 numbers, so all numbers are in the range [0,9]. 8 is the missing number in the range since it does not appear in nums.

My Thoughts

Its easy, just sum all items, and subtract from expected; this will be O(n); the process to calculate expected can be done either using individual for loop, or Arithematic Progression ‘EXPECTED_SUM= (n*(n+1))/2’

Initial Solution:

    class Solution {
        public int missingNumber(int[] nums) {
            int sum =0;
            int expected = 0;
            for(int i=0;i<nums.length;i++){
                expected += i+1;
                sum += nums[i];
            }
            return expected - sum;
        }
    }

A bit Optimized Solution with Arithmatic Progression (A.P.)

class Solution {
    public int missingNumber(int[] nums) {
        int sum =0;
        int expected = (nums.length * (nums.length+1))/2;
        for(int i=0;i<nums.length;i++){
            sum += nums[i];
        }
        return expected - sum;
    }
}