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268.missing-number.py
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69 lines (64 loc) · 1.97 KB
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# Tag: Array, Hash Table, Math, Binary Search, Bit Manipulation, Sorting, Index Sort
# Time: O(N)
# Space: O(1)
# Ref: -
# Note: -
# 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?
#
class Solution:
def missingNumber(self, nums: List[int]) -> int:
n = len(nums)
res = 0
for i in range(n):
res = res ^ nums[i] ^ (i + 1)
return res
class Solution:
def missingNumber(self, nums: List[int]) -> int:
n = len(nums)
res = (n + 1) * n // 2
for i in range(n):
res -= nums[i]
return res
class Solution:
def missingNumber(self, nums: List[int]) -> int:
nums = [-1] + nums
n = len(nums)
i = 1
while i < n:
j = nums[i]
if nums[i] >= 0 and nums[i] < n and nums[i] != nums[j]:
nums[i], nums[j] = nums[j], nums[i]
else:
i += 1
for i in range(n):
if nums[i] != i:
return i