<p>Given two strings <code>first</code> and <code>second</code>, consider occurrences in some text of the form <code>&quot;first second third&quot;</code>, where <code>second</code> comes immediately after <code>first</code>, and <code>third</code> comes immediately after <code>second</code>.</p>

<p>Return <em>an array of all the words</em> <code>third</code> <em>for each occurrence of</em> <code>&quot;first second third&quot;</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> text = "alice is a good girl she is a good student", first = "a", second = "good"
<strong>Output:</strong> ["girl","student"]
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> text = "we will we will rock you", first = "we", second = "will"
<strong>Output:</strong> ["we","rock"]
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= text.length &lt;= 1000</code></li>
	<li><code>text</code> consists of lowercase English letters and spaces.</li>
	<li>All the words in <code>text</code> are separated by <strong>a single space</strong>.</li>
	<li><code>1 &lt;= first.length, second.length &lt;= 10</code></li>
	<li><code>first</code> and <code>second</code> consist of lowercase English letters.</li>
	<li><code>text</code> will not have any leading or trailing spaces.</li>
</ul>

class Solution:
    def findOcurrences(self, text: str, first: str, second: str) -> List[str]:
        lst = text.split(' ')
        l = len(lst)
        ans = []
        
        for i in range(l):
            if i - 2 >= 0 and lst[i - 1] == second and lst[i - 2] == first:
                ans.append(lst[i])
                
                
        return ans
        
16500