268. Missing Number

EasyLeetCode

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.

How to solve the problem

• Mathematical Approach (Sum)
• Bit Manipulation (XOR)

Code

class Solution:
    def missingNumber(self, nums: list[int]) -> int:
        n = len(nums)
        expected_sum = n * (n + 1) // 2
        actual_sum = sum(nums)
        return expected_sum - actual_sum

class Solution:
    def missingNumber(self, nums: list[int]) -> int:
        missing = len(nums)
        for i, num in enumerate(nums):
            missing ^= i ^ num
        return missing

Complexity

• Approach 1: Mathematical Approach (Sum)

• Time complexity: O(n)

  • We need to calculate the sum of the array, which takes O(n) time

• Space complexity: O(1)

  • We only use a constant amount of extra space

• Approach 2: Bit Manipulation (XOR)

• Time complexity: O(n)

  • We traverse the array once, which takes O(n) time

• Space complexity: O(1)

  • We only use a constant amount of extra space