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[LeetCode 268] Missing Number
Problem
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.length1 <= n <= 10^40 <= nums[i] <= n- All the numbers of
numsare unique.
Follow up
- Could you implement a solution using only O(1) extra space complexity and O(n) runtime complexity?
Thoughts
- expectedSum - currentSum
TypeScript
function missingNumber(nums: number[]): number {
const LEN = nums.length + 1
let target: number = LEN*(LEN - 1)/2;
return target - nums.reduce((acc, num) => acc + num)
};