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.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?
Solution
/**
* @param {number[]} nums
* @return {number}
*/
var missingNumber = function(nums) {
var sum = nums.reduce((s, i) => s + i, 0);
return nums.length * (nums.length + 1) / 2 - sum;
};
Explain:
nope.
Complexity:
- Time complexity : O(n).
- Space complexity : O(1).