<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 {
public:
    int searchInsert(vector<int>& nums, int target) {
        int left = 0;
        int right = nums.size() - 1;
        while (left <= right) {
            int new_index = (right - left) / 2 + left;
            if (nums[new_index] == target) {
                return new_index;
            } else if (nums[new_index] > target) {
                right = new_index - 1;
            } else { // must be < target
                left = new_index + 1;
            }
        }
        return left;
    }
};
12400