Get started with two-pointers to solve linear data structure problems efficiently.

Coding

The Two Pointers pattern uses two coordinated pointers or iterators to traverse data efficiently. By adjusting their movement according to problem-specific conditions, it uncovers meaningful connections within any data structure, reducing redundant work and improving performance.

Two Pointers

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Behavioral

Both lists contain nodes with the same values in the same order, but each node is a separate object in memory.

The two lists share a common tail sequence starting from the same node object in memory.

The last node of $listA$ has the same value as the last node of $listB$.

Both lists have the same length and identical values at every position.

Which scenarios describe a *true* intersection between two singly linked lists, listA and listB?

Which scenarios describe a true intersection between two singly linked lists, listA and listB?

What is the correct output for the input given below?

```
List A: 60 → 50 → 80 → 30 → 10 → 90

List B: 55 → 90 → 80 → 67 → 21 → 66
```
 

What is the correct output for the input given below?
<pre><code>List A: 60 → 50 → 80 → 30 → 10 → 90

List B: 55 → 90 → 80 → 67 → 21 → 66
</code></pre>

What is the intersecting point if the following lists are given as input?

```
listA:   2 → 1 ↘
                 3 → 9 → 21
listB:   1 → 2 ↗
```

What is the intersecting point if the following lists are given as input?
<pre><code>listA: 2 → 1 ↘
 3 → 9 → 21
listB: 1 → 2 ↗
</code></pre>

# Input Format (As handled by the judge. Your program is not given these inputs )
# listA – The first linked list.
# listB – The second linked list.
# skipA – The number of nodes to skip in listA (starting from the head) to reach the intersection node.
# skipB – The number of nodes to skip in listB (starting from the head) to reach the intersection node.
# intersectVal – The value of the node where the intersection begins. Set to 0 if the two linked lists do not intersect.
# Using these inputs, the judge constructs the linked lists and connects them at the intersection point, if one exists.
# The two list heads, headA and headB, are then passed to your function.
# If your function correctly identifies and returns the intersecting node, the solution is accepted.
def getIntersectionNode(headA, headB):

    # Replace this placeholder return statement with your code
    return None

"# Quick Fix \nYou are Edward, a Senior Software Engineer who specializes in coding interview patterns, data structures and algorithms, and is great at debugging issues in code. \n- Edward, please keep in mind that you do not have to print the data of the placeholders. Also, do not mention anything about yourself. \nEdward, you are given the following data in the placeholders: \n- Function name: `getIntersectionNode` \n- User’s submitted solution code: USER_CODE \n- Error: ERROR_DESCRIPTION \n- Failed test case: FAILED_TEST_CASE \n- Coding language: Python \n\n- Edward, your task is to understand the type of error that is occurring in the coding language provided. \n- Once you understand the error using the error’s description, and why the test case failed. You will suggest how the learner can improve their solution code to fix the error. \n- Edward, you will not give the corrected code, just a hint on what the error tells, and how to fix it. \n- Edward, please limit your response strictly to 100 words. No more words than the limit."

"# Review Code \nEdward, your current task is to review the `getIntersectionNode` function(s) and the other function(s) invoked in it/them.  \n-Take your time and review the user’s solution code. \n- Do not review any extra/helper files or functions not invoked in the solution function `getIntersectionNode`. \n- Do not review the code of data structure objects instantiated like a binary tree, linked list, trie, union find, etc., in the solution functions. Just review the solution/algorithm function of the problem.  \n- Also, if the solution code doesn’t use the coding pattern stated above, then please encourage the user to reattempt the problem, to correctly use the coding pattern."

"# Review time complexity \n- Please provide a brief time complexity analysis of the function(s) `getIntersectionNode`.  \n- While calculating time complexity, include the time complexity of built-in Python functions invoked by the code.  \n- The answer should be to the point and in a few lines."

"# Review space complexity \n- Please provide a brief space complexity analysis of the function(s) `getIntersectionNode`.  \n- While calculating space complexity, include the space complexity of built-in Python functions invoked by the code.  \n- Do not include the space used for the input parameters and the space used for just returning an output (not used as a scratch space).  \n- The answer should be to the point and in a few lines."

def evaluate_results(self) -> object:
 """
 self.__inputs is the list of inputs that are passed as input to function
 self.__actual_output is a list of outputs that is produced by the learner's code
 return : self.__edu__result object
 """
 def getIntersectionNode(headA, headB):
 ptrA = headA
 ptrB = headB
 
 while ptrA != ptrB:
 ptrA = headB if ptrA is None else ptrA.next
 ptrB = headA if ptrB is None else ptrB.next

 return ptrA
 class LinkedListNode:
 # __init__ will be used to make a LinkedListNode type object.
 def __init__(self, data, next=None):
 self.data = data
 self.next = next
 
 # Template for the linked list
 class LinkedList:
 # __init__ will be used to make a LinkedList type object.
 def __init__(self):
 self.head = None
 
 # insert_node_at_head method will insert a LinkedListNode at head
 # of a linked list.
 def insert_node_at_head(self, node):
 if self.head:
 node.next = self.head
 self.head = node
 else:
 self.head = node
 
 # create_linked_list method will create the linked list using the
 # given integer array with the help of InsertAthead method.
 def create_linked_list(self, lst):
 for x in reversed(lst):
 new_node = LinkedListNode(x)
 self.insert_node_at_head(new_node)
 
 # returns the number of nodes in the linked list
 def get_length(self, head):
 temp = head
 length = 0
 while(temp):
 length+=1
 temp = temp.next
 return length
 
 # returns the node at the specified position(index) of the linked list
 def get_node(self, head, pos):
 if pos != -1:
 p = 0
 ptr = head
 while p < pos:
 ptr = ptr.next
 p += 1
 return ptr
 
 # __str__(self) method will display the elements of linked list.
 def __str__(self):
 result = ""
 temp = self.head
 while temp:
 result += str(temp.data)
 temp = temp.next
 if temp:
 result += ", "
 result += ""
 return result
 
 def main(list1, list2, skipA, skipB,intersect_va):
 listA = LinkedList()
 listA.create_linked_list(list1)
 listB = LinkedList()
 listB.create_linked_list(list2)
 
 # Create intersection if valid skip indices exist
 if skipA == 0 and skipB == 0:
 listB = listA
 elif skipA < len(list1) and skipB < len(list2):
 nodeA = listA.head
 for _ in range(skipA):
 nodeA = nodeA.next
 
 nodeB = listB.head
 for _ in range(skipB-1):
 nodeB = nodeB.next
 
 # Link listB's node to listA's intersection node
 nodeB.next = nodeA
 
 # Find intersection node
 result = getIntersectionNode(listA.head, listB.head)
 if result:
 return(f"Intersected at: {result.data}")
 else:
 return(f"No intersection")
 
 expected = main(self.__inputs[0] , self.__inputs[1], self.__inputs[2], self.__inputs[3], self.__inputs[4])
 # actual output (from user's code)
 actual = self.__actual_outputs[0]
 expected_outputs = []
 # store the 1st expected output that should be displayed to the user 

 if expected == actual:
 expected_outputs.append(expected)
 self.__edu_result.store_result(expected_outputs, TestCaseResult.PASS)
 else:
 expected_outputs.append(expected)
 self.__edu_result.store_result(expected_outputs, TestCaseResult.FAIL)

 return self.__edu_result

from Solution import *

def get_length(head):
 temp = head
 length = 0
 while(temp):
 length+=1
 temp = temp.next
 return length

def main(list1, list2, skipA, skipB,intersect_va):
 listA = list1
 # listA.create_linked_list(list1)
 listB = list2
 # listB.create_linked_list(list2)

 # Create intersection if valid skip indices exist
 if skipA == 0 and skipB == 0:
 listB = listA
 elif skipA < get_length(list1) and skipB < get_length(list2):
 nodeA = listA
 for _ in range(skipA):
 nodeA = nodeA.next

 nodeB = listB
 for _ in range(skipB-1):
 nodeB = nodeB.next

 # Link listB's node to listA's intersection node
 nodeB.next = nodeA

 # Find intersection node
 result = getIntersectionNode(listA, listB)
 if result:
 return(f"Intersected at: {result.val}")
 else:
 return(f"No intersection")

intersection-of-two-linked-lists.yaml

testcases:
  - name: "testcase 1"
    inputs:
    - 1 : [4, 1, 8, 4, 5]
    - 2 : [5, 6, 6, 8, 4, 5]
    - 3 : 2
    - 4 : 3
    - 5 : 8
    output: 
    - 1 : "Intersected at: 8"
  - name: "testcase 2"
    inputs:
    - 1 : [1, 2, 3]
    - 2 : [4, 5, 6]
    - 3 : 3
    - 4 : 3
    - 5 : 0
    output: 
    - 1 : "No intersection"
  - name: "testcase 3"
    inputs:
    - 1 : [1, 2, 3]
    - 2 : [1, 2, 3]
    - 3 : 0
    - 4 : 0
    - 5 : 1
    output: 
    - 1 : "Intersected at: 1"
  - name: "testcase 4"
    inputs:
    - 1 : [2, 6, 4]
    - 2 : [1, 5,4]
    - 3 : 2
    - 4 : 2
    - 5 : 4
    output: 
    - 1 : "Intersected at: 4"
  - name: "testcase 5"
    inputs:
    - 1 : [1]
    - 2 : [1]
    - 3 : 0
    - 4 : 0
    - 5 : 1
    output: 
    - 1 : "Intersected at: 1"
  - name: "testcase 6"
    inputs:
    - 1 : [2,3,6,21,5]
    - 2 : [7,21,5]
    - 3 : 3
    - 4 : 1
    - 5 : 21
    output: 
    - 1 : "Intersected at: 21"

def sum(input_1, input_2):
    return input_1+input_2

def validate_inputs(self) -> None:
 """
 Validates the linked list intersection problem inputs based on the given constraints.
 self.__inputs:
 [0] -> listA (list of integers)
 [1] -> listB (list of integers)
 [2] -> optional intersection index or other parameter
 """
 def is_valid_1d_array(arr):
 if not isinstance(arr, list):
 return False
 return all(isinstance(elem, int) and not isinstance(elem, bool) for elem in arr)
 
 # Validate listA
 if not is_valid_1d_array(self.__inputs[0]):
 self.store_result("Invalid input: listA must be a 1D list of integers.", ValidationFunction.FAIL)
 return
 elif not (1 <= len(self.__inputs[0]) <= 1000):
 self.store_result("The number of nodes in listA must be between 1 and 1000.", ValidationFunction.FAIL)
 return
 elif any(x < 1 or x > 100000 for x in self.__inputs[0]):
 self.store_result("Each node value in listA must be between 1 and 100000.", ValidationFunction.FAIL)
 return
 
 # Validate listB
 if not is_valid_1d_array(self.__inputs[1]):
 self.store_result("Invalid input: listB must be a 1D list of integers.", ValidationFunction.FAIL)
 return
 elif not (1 <= len(self.__inputs[1]) <= 1000):
 self.store_result("The number of nodes in listB must be between 1 and 1000.", ValidationFunction.FAIL)
 return
 elif any(x < 1 or x > 100000 for x in self.__inputs[1]):
 self.store_result("Each node value in listB must be between 1 and 100000.", ValidationFunction.FAIL)
 return
 
 # Optional: if intersection index is provided
 if len(self.__inputs) > 2:
 inter_index = self.__inputs[2]
 if not isinstance(inter_index, int) or inter_index < -1:
 self.store_result("Intersection index must be an integer greater than or equal to -1.", ValidationFunction.FAIL)
 return
 
 # If all checks pass
 self.store_result("", ValidationFunction.PASS)

Try to solve the Intersection of Two Linked Lists problem.

Problem: Intersection of Two Linked Lists

Statement

Examples

Understand the problem

Figure it out!