Find first and last position of element in sorted array
class Solution:
def searchRange(self, nums: List[int], target: int) -> List[int]:
def searchLeft(arr, target):
l = 0
r = len(arr) - 1
while l <= r:
mid = (l + r) // 2
if arr[mid] < target:
l = mid + 1
elif arr[mid] > target:
r = mid - 1
elif arr[mid] == target:
r = mid - 1
if l < 0 or l > len(arr) - 1:
return -1
return l if arr[l] == target else -1
def searchRight(arr, target):
l = 0
r = len(arr) - 1
while l <= r:
mid = (l + r) // 2
if arr[mid] < target:
l = mid + 1
elif arr[mid] > target:
r = mid - 1
elif arr[mid] == target:
l = mid + 1
if r < 0 or r > len(arr) - 1:
return -1
return r if arr[r] == target else -1
return [searchLeft(nums, target), searchRight(nums, target)]
Find First and Last Position of Element in Sorted Array
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