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}")

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}")

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}")

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}")

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}")

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}")

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}")