<p>Given an array <code>arr</code> of positive integers sorted in a <strong>strictly increasing order</strong>, and an integer <code>k</code>.</p>

<p>Return <em>the</em> <code>k<sup>th</sup></code> <em><strong>positive</strong> integer that is <strong>missing</strong> from this array.</em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> arr = [2,3,4,7,11], k = 5
<strong>Output:</strong> 9
<strong>Explanation: </strong>The missing positive integers are [1,5,6,8,9,10,12,13,...]. The 5<sup>th</sup>&nbsp;missing positive integer is 9.
</pre>

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

<pre>
<strong>Input:</strong> arr = [1,2,3,4], k = 2
<strong>Output:</strong> 6
<strong>Explanation: </strong>The missing positive integers are [5,6,7,...]. The 2<sup>nd</sup> missing positive integer is 6.
</pre>

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

<ul>
	<li><code>1 &lt;= arr.length &lt;= 1000</code></li>
	<li><code>1 &lt;= arr[i] &lt;= 1000</code></li>
	<li><code>1 &lt;= k &lt;= 1000</code></li>
	<li><code>arr[i] &lt; arr[j]</code> for <code>1 &lt;= i &lt; j &lt;= arr.length</code></li>
</ul>

<p>&nbsp;</p>
<p><strong>Follow up:</strong></p>

<p>Could you solve this problem in less than O(n) complexity?</p>

class Solution:
    def findKthPositive(self, arr: List[int], k: int) -> int:
        missing=[]
        i=1
        while i>0:
            if i not in arr:
                missing.append(i)
                i=i+1
                if len(missing)==k:
                    return missing[-1]
            else:
                i=i+1
        return





        
16500