/*

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.

*/

public class LeetCode268 {

static public int missingNumber(int[] nums) {

int n = nums.length;

int i=0;

int sumOfNums=0;

while(i<n){

sumOfNums+=nums[i++];

}

return (n*(n+1))/2 - sumOfNums;

}

public static void main(String[] args) {

System.out.println(missingNumber(new int[]{9,6,4,2,3,5,7,0,1}));

}

}