added algorithm to remove duplicates from a linked list

data_structures/linked_list/remove_duplicates.py

@@ -0,0 +1,109 @@
+from __future__ import annotations
+
+from dataclasses import dataclass
+
+
+@dataclass
+class Node:
+    data: int
+    next_node: Node | None = None
+
+
+def print_linked_list(head: Node | None) -> None:
+    """
+    Print the entire linked list iteratively.
+
+    >>> head = insert_node(None, 0)
+    >>> head = insert_node(head, 2)
+    >>> head = insert_node(head, 1)
+    >>> print_linked_list(head)
+    0->2->1
+    >>> head = insert_node(head, 4)
+    >>> head = insert_node(head, 5)
+    >>> print_linked_list(head)
+    0->2->1->4->5
+    """
+    if head is None:
+        return
+    while head.next_node is not None:
+        print(head.data, end="->")
+        head = head.next_node
+    print(head.data)
+
+
+def insert_node(head: Node | None, data: int) -> Node | None:
+    """
+    Insert a new node at the end of a linked list
+    and return the new head.
+
+    >>> head = insert_node(None, 10)
+    >>> head = insert_node(head, 9)
+    >>> head = insert_node(head, 8)
+    >>> print_linked_list(head)
+    10->9->8
+    """
+    new_node = Node(data)
+    if head is None:
+        return new_node
+
+    temp_node = head
+    while temp_node.next_node:
+        temp_node = temp_node.next_node
+    temp_node.next_node = new_node
+    return head
+
+
+def remove_duplicates(head: Node | None) -> Node | None:
+    """
+    Remove nodes with duplicate data
+
+    >>> head=insert_node(None,1)
+    >>> head=insert_node(head,1)
+    >>> head=insert_node(head,2)
+    >>> head=insert_node(head,3)
+    >>> head=insert_node(head,3)
+    >>> head=insert_node(head,4)
+    >>> head=insert_node(head,5)
+    >>> head=insert_node(head,5)
+    >>> head=insert_node(head,5)
+    >>> new_head= remove_duplicates(head)
+    >>> print_linked_list(new_head)
+    1->2->3->4->5
+    """
+    if head is None or head.next_node is None:
+        return head
+
+    has_occurred = {}
+
+    new_head = head
+    last_node = head
+    has_occurred[head.data] = True
+    current_node = None
+    if head.next_node:
+        current_node = head.next_node
+    while current_node is not None:
+        if current_node.data not in has_occurred:
+            last_node.next_node = current_node
+            last_node = current_node
+            has_occurred[current_node.data] = True
+        current_node = current_node.next_node
+    last_node.next_node = None
+    return new_head
+
+
+if __name__ == "__main__":
+    import doctest
+
+    doctest.testmod()
+
+    head = insert_node(None, 1)
+    head = insert_node(head, 1)
+    head = insert_node(head, 2)
+    head = insert_node(head, 3)
+    head = insert_node(head, 3)
+    head = insert_node(head, 4)
+    head = insert_node(head, 5)
+    head = insert_node(head, 5)
+
+    new_head = remove_duplicates(head)
+    print_linked_list(new_head)

dynamic_programming/distinct_subsequences.py

@@ -0,0 +1,51 @@
+from future import __annotations__
+
+
+def subsequenceCounting(s1: str, s2: str, n: int, m: int) -> int:
+    """
+    Uses bottom-up dynamic programming/tabulation
+    to count the number of distinct
+    subsequences of string s2 in string s1
+
+    >>> s1 = "babgbag"
+    >>> s2 = "bag"
+    >>> subsequenceCounting(s1, s2, len(s1), len(s2))
+    """
+    # Initialize a DP table to store the count of distinct subsequences
+    dp = [[0 for i in range(m + 1)] for j in range(n + 1)]
+
+    # Base case: There is exactly one subsequence of an empty string s2 in s1
+    for i in range(n + 1):
+        dp[i][0] = 1
+
+    # Initialize dp[0][i] to 0 for i > 0 since an empty s1 cannot have a non-empty subsequence of s2
+    for i in range(1, m + 1):
+        dp[0][i] = 0
+
+    # Fill in the DP table using dynamic programming
+    for i in range(1, n + 1):
+        for j in range(1, m + 1):
+            # If the current characters match, we have two choices:
+            # 1. Include the current character in both s1 and s2 (dp[i-1][j-1])
+            # 2. Skip the current character in s1 (dp[i-1][j])
+            if s1[i - 1] == s2[j - 1]:
+                dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j]
+            else:
+                dp[i - 1][j]
+
+    # The final value in dp[n][m] is the count of distinct subsequences
+    return dp[n][m]
+
+
+if __name__ == "__main__":
+    import doctest
+
+    doctest.testmod()
+    s1 = "babgbag"
+    s2 = "bag"
+
+    # Find the number of distinct subsequences of string s2 in string s1
+    print(
+        "The Count of Distinct Subsequences is",
+        subsequenceCounting(s1, s2, len(s1), len(s2)),
+    )