34. Find First and Last Position of Element in Sorted Array
Medium
Given an array of integers 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.
Example 1:
Input: nums = [5,7,7,8,8,10], target = 8 Output: [3,4]
Example 2:
Input: nums = [5,7,7,8,8,10], target = 6 Output: [-1,-1]
Example 3:
Input: nums = [], target = 0 Output: [-1,-1]
Constraints:
0 <= nums.length <= 105-109 <= nums[i] <= 109numsis a non-decreasing array.-109 <= target <= 109
class Solution:
def searchRange(self, nums: List[int], target: int) -> List[int]:
# start, end = -1,-1
# for i in range(len(nums)):
# if nums[i] == target:
# if start == -1:
# start = i
# end = i
# return start, end
start = self.find_left(nums, target)
end = -1
if start != -1:
end = self.find_right(nums, start, target)
return [start, end]
def find_right(self, nums, l, target):
r = len(nums)-1
while l <= r:
mid = l + (r-l)//2
if nums[mid] == target:
if mid == r or nums[mid+1] > target:
return mid
else:
l = mid+1
elif nums[mid] > target:
r = mid - 1
else:
l = mid + 1
def find_left(self, nums, target):
l, r = 0, len(nums)-1
while l <=r:
mid = l + (r-l)//2
if nums[mid] == target:
if mid == l or nums[mid-1] < target:
return mid
else:
r = mid - 1
elif nums[mid] > target:
r = mid - 1
else:
l = mid + 1
return -1