268. Missing Number (Easy) (https://leetcode.com/problems/missing-number/)
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.
Constraints: n == nums.length 1 <= n <= 10^4 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?
function missingNumber(nums: number[]): number {
const n = nums.length;
const expectedSum = (n * (n + 1)) / 2;
const actualSum = nums.reduce((acc, num) => acc + num, 0);
return expectedSum - actualSum;
}
// Local check:
console.log(missingNumber([3, 0, 1]));
console.log(missingNumber([0, 1]));
console.log(missingNumber([9, 6, 4, 2, 3, 5, 7, 0, 1]));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.