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Tested Python solution for LeetCode 41 with 20 pytest cases. Generate a practice environment with lcpy.
lcpy gen -n 41 # by problem number
lcpy gen -s first_missing_positive # by problem name
nums. Return the smallest positive integer that is not present in nums.
You must implement an algorithm that runs in O(n) time and uses O(1) auxiliary space.
Input: nums = [1,2,0]
Output: 3
Explanation: The numbers in the range [1,2] are all in the array.
Input: nums = [3,4,-1,1]
Output: 2
Explanation: 1 is in the array but 2 is missing.
Input: nums = [7,8,9,11,12]
Output: 1
Explanation: The smallest positive integer 1 is missing.
1 <= nums.length <= 10^5-2^31 <= nums[i] <= 2^31 - 1class Solution:
# Time: O(n)
# Space: O(1)
def first_missing_positive(self, nums: list[int]) -> int:
n = len(nums)
# Place each number in its correct position
i = 0
while i < n:
correct_idx = nums[i] - 1
if 1 <= nums[i] <= n and nums[i] != nums[correct_idx]:
nums[i], nums[correct_idx] = nums[correct_idx], nums[i]
else:
i += 1
# Find the first missing positive
for i in range(n):
if nums[i] != i + 1:
return i + 1
return n + 1
| Time | Space |
|---|---|
| O(n) | O(1) |