<p>Given a sorted array of distinct integers and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order.</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 = [1,3,5,6], target = 5
<strong>Output:</strong> 2
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,3,5,6], target = 2
<strong>Output:</strong> 1
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,3,5,6], target = 7
<strong>Output:</strong> 4
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>4</sup></code></li>
	<li><code>-10<sup>4</sup> &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
	<li><code>nums</code> contains <strong>distinct</strong> values sorted in <strong>ascending</strong> order.</li>
	<li><code>-10<sup>4</sup> &lt;= target &lt;= 10<sup>4</sup></code></li>
</ul>

class Solution:
    def searchInsert(self, nums: List[int], target: int) -> int:
        if target < nums[0]:
            return 0
        for i in range(len(nums)-1):
            if nums[i] == target:
                return i
            if nums[i] < target < nums[i + 1]:
                return i + 1
        if nums[-1] == target:
            return len(nums) -1
        return len(nums)

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