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| 1 | +#!/usr/bin/python3 |
| 2 | + |
| 3 | +#------------------------------------------------------------------------------ |
| 4 | + |
| 5 | +# This first solution is technically O(log(n)+k) where k is the number of occurances of target |
| 6 | +class Solution: |
| 7 | + def searchRange(self, nums, target): |
| 8 | + """ |
| 9 | + :type nums: List[int] |
| 10 | + :type target: int |
| 11 | + :rtype: List[int] |
| 12 | + """ |
| 13 | + left = self.binarySearch(nums, target, 0, len(nums)-1) |
| 14 | + if left == -1: return [-1, -1] |
| 15 | + right = left |
| 16 | + |
| 17 | + while nums[left] == target and left > 0: |
| 18 | + left -= 1 |
| 19 | + if nums[left] != target: |
| 20 | + left += 1 |
| 21 | + while nums[right] == target and right < len(nums)-1: |
| 22 | + right += 1 |
| 23 | + if nums[right] != target: |
| 24 | + right -= 1 |
| 25 | + |
| 26 | + return [left, right] |
| 27 | + |
| 28 | + def binarySearch(self, nums, target, left, right): |
| 29 | + """ |
| 30 | + Binary Search for log(n) time |
| 31 | + """ |
| 32 | + if right >= left: |
| 33 | + mid = (left + right ) // 2 |
| 34 | + |
| 35 | + if nums[mid] == target: |
| 36 | + return mid |
| 37 | + elif nums[mid] > target: |
| 38 | + return self.binarySearch(nums, target, left, mid-1) |
| 39 | + else: |
| 40 | + return self.binarySearch(nums, target, mid+1, right) |
| 41 | + else: |
| 42 | + return -1 |
| 43 | + |
| 44 | +#------------------------------------------------------------------------------ |
| 45 | + |
| 46 | +# This second solution is O(2log(n)) |
| 47 | +class Solution: |
| 48 | + def searchRange(self, nums, target): |
| 49 | + def binarySearchLeft(A, x): |
| 50 | + left, right = 0, len(A) - 1 |
| 51 | + while left <= right: |
| 52 | + mid = (left + right) // 2 |
| 53 | + if x > A[mid]: left = mid + 1 |
| 54 | + else: right = mid - 1 |
| 55 | + return left |
| 56 | + |
| 57 | + def binarySearchRight(A, x): |
| 58 | + left, right = 0, len(A) - 1 |
| 59 | + while left <= right: |
| 60 | + mid = (left + right) // 2 |
| 61 | + if x >= A[mid]: left = mid + 1 |
| 62 | + else: right = mid - 1 |
| 63 | + return right |
| 64 | + |
| 65 | + left, right = binarySearchLeft(nums, target), binarySearchRight(nums, target) |
| 66 | + return (left, right) if left <= right else [-1, -1] |
| 67 | + |
| 68 | +#------------------------------------------------------------------------------ |
| 69 | +#Testing |
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