Explore diverse LeetCode solutions in Python, C++, JavaScript, SQL, and TypeScript. Ideal for interview prep, learning, and code practice in multiple programming languages. Github Repo Link
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 <= 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?
Python
# time complexity: O(n) # space complexity: O(1) from typing import List class Solution: def missingNumber(self, nums: List[int]) -> int: numSet = set(nums) for i in range( len(numSet) + 1): if i not in numSet: return i nums = [9, 6, 4, 2, 3, 5, 7, 0, 1] print(Solution().missingNumber(nums))