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83 lines
1.5 KiB
Python
83 lines
1.5 KiB
Python
"""
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268. Missing Number
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Difficulty: Easy
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https://leetcode.com/problems/missing-number/
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──────────────────────────────────────────────────
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Given an array nums containing n distinct numbers in the range [0,
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n], return the only number in the range that is missing from the
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array.
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Example 1:
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Input: nums = [3,0,1]
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Output: 2
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Explanation:
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n = 3 since there are 3 numbers, so all numbers are in the range
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[0,3]. 2 is the missing number in the range since it does not appear
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in nums.
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Example 2:
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Input: nums = [0,1]
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Output: 2
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Explanation:
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n = 2 since there are 2 numbers, so all numbers are in the range
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[0,2]. 2 is the missing number in the range since it does not appear
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in nums.
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Example 3:
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Input: nums = [9,6,4,2,3,5,7,0,1]
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Output: 8
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Explanation:
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n = 9 since there are 9 numbers, so all numbers are in the range
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[0,9]. 8 is the missing number in the range since it does not appear
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in nums.
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Constraints:
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• n == nums.length
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• 1 <= n <= 10^4
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• 0 <= nums[i] <= n
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• All the numbers of nums are unique.
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Follow up: Could you implement a solution using only O(1) extra space
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complexity and O(n) runtime complexity?
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"""
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class Solution:
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def missingNumber(self, nums: List[int]) -> int:
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n = len(nums)
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expected_sum = (n * (n + 1)) // 2
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actual_sum = sum(nums)
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return expected_sum - actual_sum
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