<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 i = 0;
        int j = nums.size()-1;

        while (i <= j) {
            if (i + 1 == j && j - 1 == i) {
                if (target > nums[i] && target < nums[j]) {
                    return i+1;
                }
            }

            if (target < nums[i]) return i;
            if (target > nums[j]) return j + 1;

            if (nums[i] == target) return i;
            if (nums[j] == target) return j;

            i++;
            j--;
        }

        return -1;
    }
};
12000