MissingNumber.java

//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.
//
//
//Constraints:
//
//n == nums.length
//1 <= n <= 104
//0 <= nums[i] <= n
//All the numbers of nums are unique.
//
//
//Follow up: Could you implement a solution using only O(1) extra space complexity and O(n) runtime complexity?

public class MissingNumber {
    public int missingNumber(int[] nums) {
        int sum = 0;
        for(int i = 0; i < nums.length; i++){
            sum += nums[i];
        }

        int n = nums.length;
        return n * (n + 1) / 2 - sum;
    }

    public static void main(String[] args) {
        MissingNumber mn = new MissingNumber();
        int[] nums = {9, 6, 4, 2, 3, 5, 7, 0, 1};
//        int[] nums = {0, 1};
//        int[] nums = {3, 0, 1};
        System.out.println(mn.missingNumber(nums));;
    }
}