<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(std::vector<int>& nums, int target) 
    {
        int result = 0;

        for (int i = 0; i < nums.size(); i++)
        {
            

            if (nums[i] == target)
            {
                
                result= i;
                break;
            }
            
            else if (nums[nums.size()-1] < target)
            {
                nums.push_back(target);
                result =  nums.size()-1;
                break;
            }
            else if (nums[0] > target)
            {
                nums.insert(nums.begin(), target);
                result = 0;
                break;
            }

            else if ((nums[i] < target) && (nums[i + 1] > target))
            {
                nums.insert(nums.begin() + i + 1, target);
                result = i + 1;
                break;
            }


        }

        return result;
    }
};
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