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Intersection of Two linked list using python hash-set

docs/linked-list/Intersection_Linked_list_python.md

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+---
+id: Intersection_Linked_lists_python
+sidebar_position: 1
+title: "Find the Intersection Point of Two Linked Lists in Python using hash set"
+description: "This tutorial explains how to find the intersection point of two singly linked lists using Python."
+sidebar_label: "Linked List Intersection"
+tags: [dsa, linked-lists, intersection,python,hash-set]
+---
+
+# Intersection of Two Linked Lists
+
+## Problem Statement
+
+In a linked list, the intersection of two lists occurs when two linked lists share a common node. Given two linked lists, this problem aims to find the node where the two lists intersect. If they do not intersect, the result should be `None`.
+
+### Example
+
+Consider the following two linked lists:
+
+- **List A**: 1 -> 2 -> 3
+- **List B**: 4 -> 5
+- Both lists intersect at the node with value `3`.
+
+The expected output in this case is:
+
+
+## Approach
+
+To solve this problem, we can use a hash set to store the nodes of the first linked list. As we traverse the second linked list, we can check if any node exists in the set. If we find a match, that node is the intersection point.
+
+### Steps:
+
+1. Traverse the first linked list and add each node to a set.
+2. Traverse the second linked list and check if any node is in the set.
+3. If a node is found in the set, return that node as the intersection.
+4. If no nodes match, return `None`.
+
+## Python Code
+
+Below is the implementation of the solution in Python:
+
+```python
+class ListNode:
+    def __init__(self, x):
+        self.val = x
+        self.next = None
+
+class Solution:
+    def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> ListNode:
+        # Using a set to store the nodes of the first linked list
+        seen_nodes = set()
+        curr = headA
+
+        while curr:
+            seen_nodes.add(curr)  # Add the current node to the set
+            curr = curr.next
+
+        curr2 = headB
+        while curr2:
+            if curr2 in seen_nodes:  # Check if current node is in the set
+                return curr2  # Intersection found
+            curr2 = curr2.next
+
+        return None  # No intersection
+
+# Helper function to create a linked list from a list
+def create_linked_list(values):
+    if not values:
+        return None
+    head = ListNode(values[0])
+    curr = head
+    for value in values[1:]:
+        curr.next = ListNode(value)
+        curr = curr.next
+    return head
+
+# Example usage
+if __name__ == "__main__":
+    # Create linked lists for the example
+    # List A: 1 -> 2 -> 3
+    # List B: 4 -> 5
+    # Intersection at node with value 3
+    intersection_node = ListNode(3)
+
+    headA = create_linked_list([1, 2])
+    headA.next.next = intersection_node  # Connect intersection
+    headB = create_linked_list([4, 5])
+    headB.next.next = intersection_node  # Connect intersection
+
+    solution = Solution()
+    intersection = solution.getIntersectionNode(headA, headB)
+
+    if intersection:
+        print(f"Intersection at node with value: {intersection.val}")
+    else:
+        print("No intersection")