leetcode_gym/_problems/searchForARange.md

layout, title, last_solved, category, leetcode_url, status

layout	title	last_solved	category	leetcode_url	status
page	Find First and Last Position of Element in Sorted Array		binary search	https://leetcode.com/problems/find-first-and-last-position-of-element-in-sorted-array/	Attempted

Problem

Given an array of integers nums sorted in ascending order, find the starting and ending position of a given target value.

Your algorithm's runtime complexity must be in the order of O(log n).

If the target is not found in the array, return [-1, -1].

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]

Solution

nick white

Run binary search twice:

• find the starting position
• find the ending position

Take note of boundary conditions and termination conditions when implementing binary search.

{% highlight python %}

def searchRange(self, nums: List[int], target: int) -> List[int]:

def findStartIndex(nums, target):
    
    index = -1
    
    left, right = 0, len(nums)-1

    while left<=right:

        mid = left + (right-left) // 2
        if nums[mid]>=target:   # >= becuz we want to find teh first occurance of the duplicates
            right = mid-1
        else:
            left = mid+1

        if nums[mid]==target:
            index = mid
    
    return index



def findEndIndex(nums, target):
    
    index = -1
    
    left, right = 0, len(nums)-1

    while left<=right:

        mid = left + (right-left) // 2
        if nums[mid]<=target:
            left = mid+1
        else:
            right = mid-1

        if nums[mid]==target:
            index = mid
    
    return index
    
start = findStartIndex(nums, target)
end = findEndIndex(nums, target)

return [start,end]

{% endhighlight %}