diff --git a/DIRECTORY.md b/DIRECTORY.md
index 60ece2b8..6fc63c37 100644
--- a/DIRECTORY.md
+++ b/DIRECTORY.md
@@ -140,6 +140,10 @@
   * Heap
     * Schedule Tasks
       * [Test Schedule Tasks On Minimum Machines](https://github.com/BrianLusina/PythonSnips/blob/master/algorithms/heap/schedule_tasks/test_schedule_tasks_on_minimum_machines.py)
+    * Topkclosesttoorigin
+      * [Test Top K Closest To Origin](https://github.com/BrianLusina/PythonSnips/blob/master/algorithms/heap/topkclosesttoorigin/test_top_k_closest_to_origin.py)
+    * Topklargest
+      * [Test Top K Largest Elements](https://github.com/BrianLusina/PythonSnips/blob/master/algorithms/heap/topklargest/test_top_k_largest_elements.py)
   * Huffman
     * [Decoding](https://github.com/BrianLusina/PythonSnips/blob/master/algorithms/huffman/decoding.py)
     * [Encoding](https://github.com/BrianLusina/PythonSnips/blob/master/algorithms/huffman/encoding.py)
diff --git a/algorithms/heap/topkclosesttoorigin/README.md b/algorithms/heap/topkclosesttoorigin/README.md
new file mode 100644
index 00000000..4fd707d0
--- /dev/null
+++ b/algorithms/heap/topkclosesttoorigin/README.md
@@ -0,0 +1,104 @@
+# K Closest Points to Origin
+
+Given a list of points in the form [[x1, y1], [x2, y2], ... [xn, yn]] and an integer k, find the k closest points to the
+origin (0, 0) on the 2D plane.
+
+The distance between two points (x, y) and (a, b) is calculated using the formula:
+
+√(x1 - a2)2 + (y1 - b2)2
+
+Return the k closest points in any order.
+
+## Examples
+
+```text
+Input:
+
+points = [[3,4],[2,2],[1,1],[0,0],[5,5]]
+k = 3
+
+Output:
+[[2,2],[1,1],[0,0]]
+
+Also Valid:
+
+[[2,2],[0,0],[1,1]]
+[[1,1],[0,0],[2,2]]
+[[1,1],[2,2],[0,0]]
+...
+[[0,0],[1,1],[2,2]]
+```
+
+![Example 1](./images/examples/top_k_closest_to_origin_example_1.png)
+
+```text
+Input: points = [[3,3],[5,-1],[-2,4]], k = 2
+Output: [[3,3],[-2,4]]
+Explanation: The answer [[-2,4],[3,3]] would also be accepted.
+```
+
+```text
+Input: points = [[1,3],[-2,2]], k = 1
+Output: [[-2,2]]
+
+Explanation:
+The distance between (1, 3) and the origin is sqrt(10).
+The distance between (-2, 2) and the origin is sqrt(8).
+Since sqrt(8) < sqrt(10), (-2, 2) is closer to the origin.
+We only want the closest k = 1 points from the origin, so the answer is just [[-2,2]].
+```
+
+## Solution
+
+- [Approach 1](#approach-1-sorting)
+- [Approach 2](#approach-2-max-heap)
+
+### Approach 1: Sorting
+
+The simplest approach is to sort calculate the distance of each point from the origin and sort the points based on their
+distance. This approach has a time complexity of O(n log n) where n is the number of points in the array, and a space
+complexity of O(n) (to store the sorted array of distances).
+
+### Approach 2: Max Heap
+
+This problem can be solved using a similar approach to the one used to solve [Kth Largest Element in an Array](../topklargest/README.md). The key
+difference is that we need to store the k closest points to the origin, rather than the k largest elements. Since we are
+looking for the k smallest elements, we need a max-heap, rather than a min-heap.
+
+By default, python's heapq module implements a min-heap, but we can make it behave like a max-heap by negating the values
+of everything we push onto it.
+
+We add the first k points to the heap by pushing a tuple containing the negative of the distance from the origin, and the
+index of the point. After that is finished, our heap contains the k closest points to the origin that we've seen so far,
+with the point furthest from the origin at the root of the heap.
+
+For each point after the first k, we calculate the distance from the origin and compare it with the root of the heap. If
+the current point is closer to the origin than the root of the heap, we pop the root and push the current point into the
+heap. This way, the heap will always contain the k closest points to the origin we've seen so far.
+
+At the end of the iteration, the heap will contain the k closest points to the origin. We can iterate over each point in
+the heap and return the point associated with each tuple.
+
+![Solution 0](./images/solutions/top_k_closest_to_origin_solution_0.png)
+![Solution 1](./images/solutions/top_k_closest_to_origin_solution_1.png)
+![Solution 2](./images/solutions/top_k_closest_to_origin_solution_2.png)
+![Solution 3](./images/solutions/top_k_closest_to_origin_solution_3.png)
+![Solution 4](./images/solutions/top_k_closest_to_origin_solution_4.png)
+![Solution 5](./images/solutions/top_k_closest_to_origin_solution_5.png)
+![Solution 6](./images/solutions/top_k_closest_to_origin_solution_6.png)
+![Solution 7](./images/solutions/top_k_closest_to_origin_solution_7.png)
+![Solution 8](./images/solutions/top_k_closest_to_origin_solution_8.png)
+![Solution 9](./images/solutions/top_k_closest_to_origin_solution_9.png)
+![Solution 10](./images/solutions/top_k_closest_to_origin_solution_10.png)
+![Solution 11](./images/solutions/top_k_closest_to_origin_solution_11.png)
+
+#### Complexity Analysis
+
+##### Time Complexity: O(n log k)
+
+Where n is the number of points in the array and k is the input parameter. We iterate over
+all points, and in the worst case, we both push and pop each point from the heap, which takes O(log k) time per point.
+
+##### Space Complexity: O(k)
+
+Where k is the input parameter. The space is used by the heap to store the k closest points to the origin.
diff --git a/algorithms/heap/topkclosesttoorigin/__init__.py b/algorithms/heap/topkclosesttoorigin/__init__.py
new file mode 100644
index 00000000..ec62166d
--- /dev/null
+++ b/algorithms/heap/topkclosesttoorigin/__init__.py
@@ -0,0 +1,36 @@
+from typing import List, Tuple
+import heapq
+
+
+def k_closest_to_origin(points: List[List[int]], k: int) -> List[List[int]]:
+    # Max heap will store the top k points closest to the origin in the form (-distance, idx). The distance is negated
+    # because in Python, the heapq module uses min-heap by default. Negating values gives us a maximum heap.
+    # We store the idx of the point for later retrieval from the points array passed in
+    max_heap: List[Tuple[int, int]] = []
+
+    for idx, point in enumerate(points):
+        x, y = point
+        # calculate the distance for this point from the origin
+        distance = x * x + y * y
+
+        # If the contents of the heap are less than the desired top k, then we add the current point's distance and idx
+        if len(max_heap) < k:
+            heapq.heappush(max_heap, (-distance, idx))
+        # We check if the calculated distance of this point is less than the top element in the heap. If this point
+        # is closer to the origin that what is at the root of the heap, we pop the root of the heap and add this
+        # new distance and index.
+        # Note the negation here again to get the actual distance
+        elif distance < -max_heap[0][0]:
+            heapq.heappushpop(max_heap, (-distance, idx))
+
+    # Return the top k points closest to origin. We use 1 to get the index of the point from the original points list
+    # as that is what is stored in the max heap
+    return [points[point[1]] for point in max_heap]
+
+
+def k_closest_to_origin_sorting(points: List[List[int]], k: int) -> List[List[int]]:
+    # Sort the points by the distance from the origin. This incurs a cost of O(n log(n)) and space cost of O(n) due to
+    # timsort
+    sorted_points = sorted(points, key=lambda p: p[0] ** 2 + p[1] ** 2)
+    # Retrieve the top k points closest to the origin
+    return sorted_points[:k]
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diff --git a/algorithms/heap/topkclosesttoorigin/test_top_k_closest_to_origin.py b/algorithms/heap/topkclosesttoorigin/test_top_k_closest_to_origin.py
new file mode 100644
index 00000000..de979b30
--- /dev/null
+++ b/algorithms/heap/topkclosesttoorigin/test_top_k_closest_to_origin.py
@@ -0,0 +1,39 @@
+import unittest
+from typing import List
+from parameterized import parameterized
+from algorithms.heap.topkclosesttoorigin import (
+    k_closest_to_origin,
+    k_closest_to_origin_sorting,
+)
+
+TOP_K_CLOSEST_TO_ORIGIN = [
+    ([[3, 4], [2, 2], [1, 1], [0, 0], [5, 5]], 3, [[2, 2], [1, 1], [0, 0]]),
+    ([[1, 3], [-2, 2]], 1, [[-2, 2]]),
+    ([[3, 3], [5, -1], [-2, 4]], 2, [[3, 3], [-2, 4]]),
+]
+
+
+class TopKClosestToOriginTestCase(unittest.TestCase):
+    @parameterized.expand(TOP_K_CLOSEST_TO_ORIGIN)
+    def test_top_k_closest_to_origin(
+        self, points: List[List[int]], k: int, expected: List[List[int]]
+    ):
+        actual = k_closest_to_origin(points, k)
+
+        sorted_expected = sorted(expected, key=lambda x: (x[0], x[1]))
+        sorted_actual = sorted(actual, key=lambda x: (x[0], x[1]))
+        self.assertEqual(sorted_expected, sorted_actual)
+
+    @parameterized.expand(TOP_K_CLOSEST_TO_ORIGIN)
+    def test_top_k_closest_to_origin_sorting(
+        self, points: List[List[int]], k: int, expected: List[List[int]]
+    ):
+        actual = k_closest_to_origin_sorting(points, k)
+
+        sorted_expected = sorted(expected, key=lambda x: x[0])
+        sorted_actual = sorted(actual, key=lambda x: x[0])
+        self.assertEqual(sorted_expected, sorted_actual)
+
+
+if __name__ == "__main__":
+    unittest.main()
diff --git a/pymath/topkfrequentelements/README.md b/algorithms/heap/topkfrequentelements/README.md
similarity index 100%
rename from pymath/topkfrequentelements/README.md
rename to algorithms/heap/topkfrequentelements/README.md
diff --git a/pymath/topkfrequentelements/__init__.py b/algorithms/heap/topkfrequentelements/__init__.py
similarity index 100%
rename from pymath/topkfrequentelements/__init__.py
rename to algorithms/heap/topkfrequentelements/__init__.py
diff --git a/algorithms/heap/topklargest/README.md b/algorithms/heap/topklargest/README.md
new file mode 100644
index 00000000..aa97298e
--- /dev/null
+++ b/algorithms/heap/topklargest/README.md
@@ -0,0 +1,106 @@
+# Top-K Largest Elements in an Array
+
+Given an integer array nums, return the 3 largest elements in the array in any order.
+
+## Example
+
+```text
+Input: nums = [9, 3, 7, 1, -2, 6, 8]
+Output: [8, 7, 9]
+# or [7, 9, 8] or [9, 7, 8] ...
+```
+
+## Solution
+
+Here's how we can solve this problem using a min-heap:
+
+- Create a min-heap that stores the first 3 elements of the array. These represent the 3 largest elements we have seen so 
+  far, with the smallest of the 3 at the root of the heap.
+  ![Solution 1](./images/solutions/top_k_largest_element_in_array_solution_1.png)
+
+- Iterate through the remaining elements in the array.
+  - If the current element is larger than the root of the heap, pop the root and push the current element into the heap.
+  - Otherwise, continue to the next element.
+  
+  ![Solution 2](./images/solutions/top_k_largest_element_in_array_solution_2.png)
+  ![Solution 3](./images/solutions/top_k_largest_element_in_array_solution_3.png)
+  ![Solution 4](./images/solutions/top_k_largest_element_in_array_solution_4.png)
+  ![Solution 5](./images/solutions/top_k_largest_element_in_array_solution_5.png)
+
+After iterating through all the elements, the heap contains the 3 largest elements in the array.
+
+### Complexity Analysis
+
+#### Time Complexity Breakdown
+
+- The heapify function takes O(3) = O(1) time
+- We iterate through all elements in the array once: O(n) time
+- The heappop and heappush operations take O(log 3) = O(1) time each
+
+#### Space Complexity
+
+- We use a heap of size 3 to store the 3 largest elements: O(3) = O(1) space 
+- The algorithm uses constant space regardless of input size
+
+Note: The time and space complexity become more interesting when 3 is a variable number k.
+
+---
+
+# Kth Largest Element in an Array
+
+Write a function that takes an array of unsorted integers nums and an integer k, and returns the kth largest element in
+the array. This function should run in O(n log k) time, where n is the length of the array.
+
+## Examples
+
+```text
+Input:
+nums = [5, 3, 2, 1, 4]
+k = 2
+
+Output: 4
+```
+
+## Solutions
+
+- [Approach 1](#approach-1-sorting)
+- [Approach 2](#approach-2-min-heap)
+
+### Approach 1: Sorting
+
+The simplest approach is to sort the array in descending order and return the kth element. This approach has a time 
+complexity of O(n log n) where n is the number of elements in the array, and a space complexity of O(1).
+
+### Approach 2: Min Heap
+
+By using a min-heap, we can reduce the time complexity to O(n log k), where n is the number of elements in the array and
+k is the value of k.
+The idea behind this solution is to iterate over the elements in the array while storing the k largest elements we've
+seen so far in a min-heap. At each element, we check if it is greater than the smallest element (the root) of the heap.
+If it is, we pop the smallest element from the heap and push the current element into the heap. This way, the heap will
+always contain the k largest elements we've seen so far.
+After iterating over all the elements, the root of the heap will be the kth largest element in the array.
+
+![Solution 2.1](./images/solutions/kth_largest_element_in_array_solution_1.png)
+![Solution 2.2](./images/solutions/kth_largest_element_in_array_solution_2.png)
+![Solution 2.3](./images/solutions/kth_largest_element_in_array_solution_3.png)
+![Solution 2.4](./images/solutions/kth_largest_element_in_array_solution_4.png)
+![Solution 2.5](./images/solutions/kth_largest_element_in_array_solution_5.png)
+![Solution 2.6](./images/solutions/kth_largest_element_in_array_solution_6.png)
+![Solution 2.7](./images/solutions/kth_largest_element_in_array_solution_7.png)
+![Solution 2.8](./images/solutions/kth_largest_element_in_array_solution_8.png)
+![Solution 2.9](./images/solutions/kth_largest_element_in_array_solution_9.png)
+![Solution 2.10](./images/solutions/kth_largest_element_in_array_solution_10.png)
+![Solution 2.11](./images/solutions/kth_largest_element_in_array_solution_11.png)
+
+#### Complexity Analysis
+
+##### Time Complexity: O(n log k)
+
+Where n is the number of elements in the array and k is the input parameter. We iterate over
+all elements, and in the worst case, we both push and pop each element from the heap, which takes O(log k) time per
+element.
+
+##### Space Complexity: O(k)
+
+Where k is the input parameter. The space is used by the heap to store the k largest elements.
diff --git a/algorithms/heap/topklargest/__init__.py b/algorithms/heap/topklargest/__init__.py
new file mode 100644
index 00000000..382b1952
--- /dev/null
+++ b/algorithms/heap/topklargest/__init__.py
@@ -0,0 +1,83 @@
+from typing import List
+import heapq
+
+
+def k_largest(nums: List[int], k: int = 3) -> List[int]:
+    """
+    Finds the k largest elements in a given list. K is defaulted to three, but can be used to tweak the k largest
+     elements in the array
+    Args:
+        nums(list): list of elements to check for
+        k(int): number of elements to check, defaulted to 3
+    Returns:
+        list: top k largest elements
+    """
+    # input validation to ensure we don't get unexpected results
+    if not nums or k <= 0:
+        return []
+    
+    # Adjust k if it exceeds the length of nums
+    k = min(k, len(nums))
+    
+    # create a minimum heap with the first k elements
+    min_heap = nums[:k]
+    heapq.heapify(min_heap)
+
+    # iterate through the remaining elements
+    for num in nums[k:]:
+        # if the current number is greater than the element at the top of the heap
+        if num > min_heap[0]:
+            # Remove it and add this element
+            heapq.heappushpop(min_heap, num)
+
+    # return the top k elements
+    return min_heap
+
+
+def kth_largest(nums: List[int], k: int) -> int:
+    """
+    Finds the kth largest element in a given list
+    Args:
+        nums(list): list of elements to check for
+        k(int): the kth largest element to return
+    Returns:
+        int: the kth largest element
+    """
+    # input validation to ensure we don't get unexpected results
+    if not nums or k <= 0 or k > len(nums):
+        return -1
+
+    # create a minimum heap with the first k elements
+    min_heap = []
+
+    # iterate through the remaining elements
+    for num in nums:
+        if len(min_heap) < k:
+            heapq.heappush(min_heap, num)
+        # if the current number is greater than the element at the top of the heap
+        elif num > min_heap[0]:
+            # Remove it and add this element
+            heapq.heappushpop(min_heap, num)
+
+    # return the top kth element
+    return min_heap[0]
+
+
+def kth_largest_sorting(nums: List[int], k: int) -> int:
+    """
+    Finds the kth largest element in a given list using sorting
+    Args:
+        nums(list): list of elements to check for
+        k(int): the kth largest element to return
+    Returns:
+        int: the kth largest element
+    """
+    # input validation to ensure we don't get unexpected results
+    if not nums or k <= 0 or k > len(nums):
+        return -1
+
+    # Sort the list which incurs a time complexity cost of O(n log(n)). Space complexity is O(n) due to creating
+    # a new sorted list
+    sorted_nums = sorted(nums, reverse=True)
+    # Return the kth largest element
+    return sorted_nums[k - 1]
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diff --git a/algorithms/heap/topklargest/test_top_k_largest_elements.py b/algorithms/heap/topklargest/test_top_k_largest_elements.py
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--- /dev/null
+++ b/algorithms/heap/topklargest/test_top_k_largest_elements.py
@@ -0,0 +1,45 @@
+import unittest
+from typing import List
+from parameterized import parameterized
+from algorithms.heap.topklargest import k_largest, kth_largest, kth_largest_sorting
+
+K_LARGEST_ELEMENTS_TEST_CASES = [
+    ([9, 3, 7, 1, -2, 6, 8], 3, [8, 7, 9]),
+    ([5, 3, 2, 1, 4], 2, [5, 4]),
+]
+
+
+class TopKLargestElementsTestCase(unittest.TestCase):
+    @parameterized.expand(K_LARGEST_ELEMENTS_TEST_CASES)
+    def test_top_k_largest_elements(self, nums: List[int], k: int, expected: List[int]):
+        actual = k_largest(nums, k)
+        self.assertListEqual(sorted(expected), sorted(actual))
+
+
+KTH_LARGEST_ELEMENTS_TEST_CASES = [
+    ([3, 2, 1, 5, 6, 4], 2, 5),
+    ([3, 2, 3, 1, 2, 4, 5, 5, 6], 4, 4),
+    ([1], 1, 1),
+    ([7, 10, 4, 3, 20, 15], 3, 10),
+    ([7, 10, 4, 3, 20, 15], 4, 7),
+    ([7, 10, 4, 3, 20, 15], 5, 4),
+    ([5, 3, 2, 1, 4], 2, 4),
+]
+
+
+class TopKthLargestElementsTestCase(unittest.TestCase):
+    @parameterized.expand(KTH_LARGEST_ELEMENTS_TEST_CASES)
+    def test_top_kth_largest_element(self, nums: List[int], k: int, expected: int):
+        actual = kth_largest(nums, k)
+        self.assertEqual(expected, actual)
+
+    @parameterized.expand(KTH_LARGEST_ELEMENTS_TEST_CASES)
+    def test_top_kth_largest_element_sorting(
+        self, nums: List[int], k: int, expected: int
+    ):
+        actual = kth_largest_sorting(nums, k)
+        self.assertEqual(expected, actual)
+
+
+if __name__ == "__main__":
+    unittest.main()