Kth Largest Element in a Stream
<p>You are part of a university admissions office and need to keep track of the <code>kth</code> highest test score from applicants in real-time. This helps to determine cut-off marks for interviews and admissions dynamically as new applicants submit their scores.</p>

<p>You are tasked to implement a class which, for a given integer&nbsp;<code>k</code>, maintains a stream of test scores and continuously returns the&nbsp;<code>k</code>th highest test score&nbsp;<strong>after</strong>&nbsp;a new score has been submitted. More specifically, we are looking for the <code>k</code>th highest score in the sorted list of all scores.</p>

<p>Implement the&nbsp;<code>KthLargest</code> class:</p>

<ul>
	<li><code>KthLargest(int k, int[] nums)</code> Initializes the object with the integer <code>k</code> and the stream of test scores&nbsp;<code>nums</code>.</li>
	<li><code>int add(int val)</code> Adds a new test score&nbsp;<code>val</code> to the stream and returns the element representing the <code>k<sup>th</sup></code> largest element in the pool of test scores so far.</li>
</ul>

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

<div class="example-block">
<p><strong>Input:</strong><br />
<span class="example-io">[&quot;KthLargest&quot;, &quot;add&quot;, &quot;add&quot;, &quot;add&quot;, &quot;add&quot;, &quot;add&quot;]<br />
[[3, [4, 5, 8, 2]], [3], [5], [10], [9], [4]]</span></p>

<p><strong>Output:</strong> <span class="example-io">[null, 4, 5, 5, 8, 8]</span></p>

<p><strong>Explanation:</strong></p>

<p>KthLargest kthLargest = new KthLargest(3, [4, 5, 8, 2]);<br />
kthLargest.add(3); // return 4<br />
kthLargest.add(5); // return 5<br />
kthLargest.add(10); // return 5<br />
kthLargest.add(9); // return 8<br />
kthLargest.add(4); // return 8</p>
</div>

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

<div class="example-block">
<p><strong>Input:</strong><br />
<span class="example-io">[&quot;KthLargest&quot;, &quot;add&quot;, &quot;add&quot;, &quot;add&quot;, &quot;add&quot;]<br />
[[4, [7, 7, 7, 7, 8, 3]], [2], [10], [9], [9]]</span></p>

<p><strong>Output:</strong> <span class="example-io">[null, 7, 7, 7, 8]</span></p>

<p><strong>Explanation:</strong></p>
KthLargest kthLargest = new KthLargest(4, [7, 7, 7, 7, 8, 3]);<br />
kthLargest.add(2); // return 7<br />
kthLargest.add(10); // return 7<br />
kthLargest.add(9); // return 7<br />
kthLargest.add(9); // return 8</div>

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

<ul>
	<li><code>0 &lt;= nums.length &lt;= 10<sup>4</sup></code></li>
	<li><code>1 &lt;= k &lt;= nums.length + 1</code></li>
	<li><code>-10<sup>4</sup> &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
	<li><code>-10<sup>4</sup> &lt;= val &lt;= 10<sup>4</sup></code></li>
	<li>At most <code>10<sup>4</sup></code> calls will be made to <code>add</code>.</li>
</ul>
Tree; Design; Binary Search Tree; Binary Tree; Data Stream; Heap (Priority Queue)
class KthLargest:

    def __init__(self, k: int, nums: List[int]):
        self.k = k
        self.min_q = []
        for x in nums:
            self.add(x)

    def add(self, val: int) -> int:
        heappush(self.min_q, val)
        if len(self.min_q) > self.k:
            heappop(self.min_q)
        return self.min_q[0]


# Your KthLargest object will be instantiated and called as such:
# obj = KthLargest(k, nums)
# param_1 = obj.add(val)
