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Tested Python solution for LeetCode 34 with 20 pytest cases. Generate a practice environment with lcpy.
lcpy gen -n 34 # by problem number
lcpy gen -s find_first_and_last_position_of_element_in_sorted_array # by problem name
nums sorted in non-decreasing order, find the starting and ending position of a given target value.
If target is not found in the array, return [-1, -1].
You must write an algorithm with O(log n) runtime complexity.
Input: nums = [5,7,7,8,8,10], target = 8
Output: [3,4]
Input: nums = [5,7,7,8,8,10], target = 6
Output: [-1,-1]
Input: nums = [], target = 0
Output: [-1,-1]
0 <= nums.length <= 10^5-10^9 <= nums[i] <= 10^9nums is a non-decreasing array.-10^9 <= target <= 10^9class Solution:
# Time: O(log n)
# Space: O(1)
def search_range(self, nums: list[int], target: int) -> list[int]:
def find_left(nums: list[int], target: int) -> int:
left, right = 0, len(nums) - 1
while left <= right:
mid = (left + right) // 2
if nums[mid] < target:
left = mid + 1
else:
right = mid - 1
return left if left < len(nums) and nums[left] == target else -1
def find_right(nums: list[int], target: int) -> int:
left, right = 0, len(nums) - 1
while left <= right:
mid = (left + right) // 2
if nums[mid] <= target:
left = mid + 1
else:
right = mid - 1
return right if right >= 0 and nums[right] == target else -1
if not nums:
return [-1, -1]
left_pos = find_left(nums, target)
if left_pos == -1:
return [-1, -1]
right_pos = find_right(nums, target)
return [left_pos, right_pos]
| Time | Space |
|---|---|
| O(log n) | O(1) |