<p>Given an array of integers <code>nums</code> sorted in non-decreasing order, find the starting and ending position of a given <code>target</code> value.</p>

<p>If <code>target</code> is not found in the array, return <code>[-1, -1]</code>.</p>

<p>You must&nbsp;write an algorithm with&nbsp;<code>O(log n)</code> runtime complexity.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> nums = [5,7,7,8,8,10], target = 8
<strong>Output:</strong> [3,4]
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> nums = [5,7,7,8,8,10], target = 6
<strong>Output:</strong> [-1,-1]
</pre><p><strong class="example">Example 3:</strong></p>
<pre><strong>Input:</strong> nums = [], target = 0
<strong>Output:</strong> [-1,-1]
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>0 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>-10<sup>9</sup>&nbsp;&lt;= nums[i]&nbsp;&lt;= 10<sup>9</sup></code></li>
	<li><code>nums</code> is a non-decreasing array.</li>
	<li><code>-10<sup>9</sup>&nbsp;&lt;= target&nbsp;&lt;= 10<sup>9</sup></code></li>
</ul>

class Solution:
    def searchRange(self, nums: List[int], target: int) -> List[int]:
        def binary_search(target):
            l = 0
            r = len(nums) - 1

            while l < r:
                mid = l + (r - l) // 2

                if nums[mid] == target:
                    return mid

                if nums[mid] < target:
                    l = mid + 1
                else:
                    r = mid - 1

            return -1 if nums[l] != target else l

        if not nums:
            return [-1, -1]

        index = binary_search(target)
        if index == -1:
            return [-1, -1]

        start = index - 1
        while start >= 0 and nums[start] == target:
            start -= 1       

        end = index + 1        
        while end < len(nums) and nums[end] == target:
            end += 1

        return [start + 1, end - 1]

17600