optimized and corrected longest increasing subsequence problem

data_structures/binary_tree/floor_ceil_in_bst.py

@@ -0,0 +1,82 @@
+"""
+The floor of a key 'k' in a BST is the maximum
+value that is smaller than or equal to 'k'.
+
+The ceiling of a key 'k' in a BST is the minimum
+value that is greater than or equal to 'k'.
+
+Reference:
+https://bit.ly/46uB0a2
+
+Author : Arunkumar
+Date : 14th October 2023
+"""
+
+
+class TreeNode:
+    def __init__(self, key: int) -> None:
+        """
+        Initialize a TreeNode with the given key.
+
+        Args:
+            key (int): The key value for the node.
+        """
+        self.key = key
+        self.left: TreeNode | None = None
+        self.right: TreeNode | None = None
+
+
+def floor_ceiling(root: TreeNode | None, key: int) -> tuple[int | None, int | None]:
+    """
+    Find the floor and ceiling values for a given key in a Binary Search Tree (BST).
+
+    Args:
+        root (TreeNode): The root of the BST.
+        key (int): The key for which to find the floor and ceiling.
+
+    Returns:
+        tuple[int | None, int | None]:
+        A tuple containing the floor and ceiling values, respectively.
+
+    Examples:
+        >>> root = TreeNode(10)
+        >>> root.left = TreeNode(5)
+        >>> root.right = TreeNode(20)
+        >>> root.left.left = TreeNode(3)
+        >>> root.left.right = TreeNode(7)
+        >>> root.right.left = TreeNode(15)
+        >>> root.right.right = TreeNode(25)
+        >>> floor, ceiling = floor_ceiling(root, 8)
+        >>> floor
+        7
+        >>> ceiling
+        10
+        >>> floor, ceiling = floor_ceiling(root, 14)
+        >>> floor
+        10
+        >>> ceiling
+        15
+    """
+    floor_val = None
+    ceiling_val = None
+
+    while root is not None:
+        if root.key == key:
+            floor_val = root.key
+            ceiling_val = root.key
+            break
+
+        if key < root.key:
+            ceiling_val = root.key
+            root = root.left
+        else:
+            floor_val = root.key
+            root = root.right
+
+    return floor_val, ceiling_val
+
+
+if __name__ == "__main__":
+    import doctest
+
+    doctest.testmod()

dynamic_programming/longest_increasing_subsequence.py

@@ -1,5 +1,6 @@
 """
 Author  : Mehdi ALAOUI
+Optimized : Arunkumar
 
 This is a pure Python implementation of Dynamic Programming solution to the longest
 increasing subsequence of a given sequence.
@@ -13,46 +14,54 @@
 from __future__ import annotations
 
 
-def longest_subsequence(array: list[int]) -> list[int]:  # This function is recursive
+def longest_subsequence(array: list[int]) -> list[int]:
     """
-    Some examples
+    Find the longest increasing subsequence in the given array
+    using dynamic programming.
+
+    Args:
+        array (list[int]): The input array.
+
+    Returns:
+        list[int]: The longest increasing subsequence.
+
+    Examples:
     >>> longest_subsequence([10, 22, 9, 33, 21, 50, 41, 60, 80])
     [10, 22, 33, 41, 60, 80]
     >>> longest_subsequence([4, 8, 7, 5, 1, 12, 2, 3, 9])
     [1, 2, 3, 9]
     >>> longest_subsequence([9, 8, 7, 6, 5, 7])
-    [8]
+    [5, 7]
     >>> longest_subsequence([1, 1, 1])
-    [1, 1, 1]
+    [1]
     >>> longest_subsequence([])
     []
     """
-    array_length = len(array)
-    # If the array contains only one element, we return it (it's the stop condition of
-    # recursion)
-    if array_length <= 1:
-        return array
-        # Else
-    pivot = array[0]
-    is_found = False
-    i = 1
-    longest_subseq: list[int] = []
-    while not is_found and i < array_length:
-        if array[i] < pivot:
-            is_found = True
-            temp_array = [element for element in array[i:] if element >= array[i]]
-            temp_array = longest_subsequence(temp_array)
-            if len(temp_array) > len(longest_subseq):
-                longest_subseq = temp_array
-        else:
-            i += 1
-
-    temp_array = [element for element in array[1:] if element >= pivot]
-    temp_array = [pivot, *longest_subsequence(temp_array)]
-    if len(temp_array) > len(longest_subseq):
-        return temp_array
-    else:
-        return longest_subseq
+    if not array:
+        return []
+
+    n = len(array)
+    # Initialize an array to store the length of the longest
+    # increasing subsequence ending at each position.
+    lis_lengths = [1] * n
+
+    for i in range(1, n):
+        for j in range(i):
+            if array[i] > array[j]:
+                lis_lengths[i] = max(lis_lengths[i], lis_lengths[j] + 1)
+
+    # Find the maximum length of the increasing subsequence.
+    max_length = max(lis_lengths)
+
+    # Reconstruct the longest subsequence in reverse order.
+    subsequence = []
+    current_length = max_length
+    for i in range(n - 1, -1, -1):
+        if lis_lengths[i] == current_length:
+            subsequence.append(array[i])
+            current_length -= 1
+
+    return subsequence[::-1]  # Reverse the subsequence to get the correct order.
 
 
 if __name__ == "__main__":

scheduling/shortest_deadline_first.py

@@ -0,0 +1,70 @@
+"""
+Earliest Deadline First (EDF) Scheduling Algorithm
+
+This code implements the Earliest Deadline First (EDF)
+scheduling algorithm, which schedules processes based on their deadlines.
+If a process cannot meet its deadline, it is marked as "Idle."
+
+Reference:
+https://www.geeksforgeeks.org/
+earliest-deadline-first-edf-cpu-scheduling-algorithm/
+
+Author: Arunkumar
+Date: 14th October 2023
+"""
+
+
+def earliest_deadline_first_scheduling(
+    processes: list[tuple[str, int, int, int]]
+) -> list[str]:
+    """
+    Perform Earliest Deadline First (EDF) scheduling.
+
+    Args:
+        processes (List[Tuple[str, int, int, int]]): A list of
+        processes with their names,
+        arrival times, deadlines, and execution times.
+
+    Returns:
+        List[str]: A list of process names in the order they are executed.
+
+    Examples:
+        >>> processes = [("A", 1, 5, 2), ("B", 2, 8, 3), ("C", 3, 4, 1)]
+        >>> execution_order = earliest_deadline_first_scheduling(processes)
+        >>> execution_order
+        ['Idle', 'A', 'C', 'B']
+
+    """
+    result = []
+    current_time = 0
+
+    while processes:
+        available_processes = [
+            process for process in processes if process[1] <= current_time
+        ]
+
+        if not available_processes:
+            result.append("Idle")
+            current_time += 1
+        else:
+            next_process = min(
+                available_processes, key=lambda tuple_values: tuple_values[2]
+            )
+            name, _, deadline, execution_time = next_process
+
+            if current_time + execution_time <= deadline:
+                result.append(name)
+                current_time += execution_time
+                processes.remove(next_process)
+            else:
+                result.append("Idle")
+                current_time += 1
+
+    return result
+
+
+if __name__ == "__main__":
+    processes = [("A", 1, 5, 2), ("B", 2, 8, 3), ("C", 3, 4, 1)]
+    execution_order = earliest_deadline_first_scheduling(processes)
+    for i, process in enumerate(execution_order):
+        print(f"Time {i}: Executing process {process}")