Split Linked List in Parts

▶

Steps

setup

Initialize total_nodes and current pointer

Set total_nodes to 0 and current pointer to head of the list to start counting nodes.

💡 Counting nodes is essential to determine how to split the list evenly.

Line:

total_nodes = 0
current = head

💡 We start with no nodes counted and a pointer at the list head.

traverse

Count first node

Increment total_nodes to 1 because current points to the first node with value 1. Move current to next node.

💡 Each node visited increases the count to find total length.

Line:

total_nodes += 1
current = current.next

💡 We have counted one node and moved forward to continue counting.

traverse

Count second node

Increment total_nodes to 2 as current points to node with value 2. Move current to next node.

💡 Counting nodes one by one ensures accurate total length.

Line:

total_nodes += 1
current = current.next

💡 We have counted two nodes so far.

traverse

Count third node

Increment total_nodes to 3 as current points to node with value 3. Move current to next node which is null, ending count.

💡 Reaching null means we have counted all nodes.

Line:

total_nodes += 1
current = current.next

💡 Total nodes counted is 3.

setup

Calculate part size and remainder

Calculate base part size as total_nodes // k = 3 // 5 = 0 and remainder as total_nodes % k = 3 % 5 = 3.

💡 Base size and remainder determine how nodes are distributed among parts.

Line:

part_size = total_nodes // k
remainder = total_nodes % k

💡 Parts will have size 0 or 1 depending on remainder.

setup

Initialize parts array and reset current pointer

Create an array parts of size k with all None and reset current pointer to head to start splitting.

💡 Parts array will hold the head of each split part.

Line:

parts = [None] * k
current = head

💡 Ready to assign nodes to parts.

fill_row

Assign first part head and calculate size

Set parts[0] to current node (1). Calculate size as part_size + 1 because remainder > 0, so size = 1. Decrement remainder to 2.

💡 Parts with remainder get one extra node.

Line:

parts[i] = current
size = part_size + (1 if remainder > 0 else 0)
if remainder > 0:
    remainder -= 1

💡 First part will have one node.

traverse

Traverse to end of first part

Since size is 1, no inner loop traversal needed. Current points to node 1, which is the end of this part.

💡 For size 1, the part is just one node, no traversal inside part needed.

Line:

for j in range(size - 1):
    if current:
        current = current.next

💡 Current is at last node of part 0.

split

Break link to separate first part

Save next_part as current.next (node 2). Break link by setting current.next to None. Move current pointer to next_part (node 2).

💡 Breaking the link isolates the first part from the rest of the list.

Line:

if current:
    next_part = current.next
    current.next = None
    current = next_part

💡 First part is now a standalone list with one node.

fill_row

Assign second part head and calculate size

Set parts[1] to current node (2). Calculate size as part_size + 1 because remainder > 0, so size = 1. Decrement remainder to 1.

💡 Second part also gets one node due to remainder.

Line:

parts[i] = current
size = part_size + (1 if remainder > 0 else 0)
if remainder > 0:
    remainder -= 1

💡 Second part will have one node.

traverse

Traverse to end of second part

Since size is 1, no inner loop traversal needed. Current points to node 2, end of this part.

💡 No traversal needed for size 1 parts.

Line:

for j in range(size - 1):
    if current:
        current = current.next

💡 Current is at last node of part 1.

split

Break link to separate second part

Save next_part as current.next (node 3). Break link by setting current.next to None. Move current pointer to next_part (node 3).

💡 Breaking link isolates second part.

Line:

if current:
    next_part = current.next
    current.next = None
    current = next_part

💡 Second part is now isolated.

fill_row

Assign third part head and calculate size

Set parts[2] to current node (3). Calculate size as part_size + 1 because remainder > 0, so size = 1. Decrement remainder to 0.

💡 Third part also gets one node due to remainder.

Line:

parts[i] = current
size = part_size + (1 if remainder > 0 else 0)
if remainder > 0:
    remainder -= 1

💡 Third part will have one node.

traverse

Traverse to end of third part

Since size is 1, no inner loop traversal needed. Current points to node 3, end of this part.

💡 No traversal needed for size 1 parts.

Line:

for j in range(size - 1):
    if current:
        current = current.next

💡 Current is at last node of part 2.

split

Break link to separate third part

Save next_part as current.next (null). Break link by setting current.next to None (already None). Move current pointer to next_part (null).

💡 Breaking link isolates third part; current becomes null indicating no more nodes.

Line:

if current:
    next_part = current.next
    current.next = None
    current = next_part

💡 Third part is isolated; no more nodes remain.

fill_row

Assign empty fourth part

Set parts[3] to current which is null, indicating an empty part.

💡 When no nodes remain, parts are empty (None).

Line:

parts[i] = current

💡 Empty parts are represented by None.

fill_row

Assign empty fifth part

Set parts[4] to current which is null, indicating an empty part.

💡 Remaining parts with no nodes are empty.

Line:

parts[i] = current

💡 All parts are assigned, including empty ones.

reconstruct

Return parts array as result

Return the array parts containing heads of each split part, including empty parts represented by None.

💡 The final output is an array of linked list heads representing each part.

Line:

return parts

💡 The list is split into k parts as required.

class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next

def splitListToParts(head, k):
    total_nodes = 0  # STEP 1
    current = head  # STEP 1
    while current:  # STEP 2-4 loop
        total_nodes += 1  # STEP 2-4
        current = current.next  # STEP 2-4

    part_size = total_nodes // k  # STEP 5
    remainder = total_nodes % k  # STEP 5

    parts = [None] * k  # STEP 6
    current = head  # STEP 6

    for i in range(k):  # STEP 7-17 loop
        parts[i] = current  # STEP 7,10,13,16,17
        size = part_size + (1 if remainder > 0 else 0)  # STEP 7,10,13
        if remainder > 0:  # STEP 7,10,13
            remainder -= 1  # STEP 7,10,13

        for j in range(size - 1):  # STEP 8,11,14
            if current:
                current = current.next

        if current:  # STEP 9,12,15
            next_part = current.next
            current.next = None
            current = next_part

    return parts  # STEP 18

📊

Split Linked List in Parts - Watch the Algorithm Execute, Step by Step

Watching each pointer move and link break helps you understand how the list is divided evenly and why some parts may be empty.

advance

1

→

2

→

3

advance

1

→

2

→

3

Result: 1

advance

1

→

2

→

3

Result: 2

advance

1

→

2

→

3

Result: 3

advance

1

→

2

→

3

Result:

Part size:0

Remainder:3

advance

1

→

2

→

3

Result:

Parts:[null, null, null, null, null]

connect

1

→

2

→

3

Result:

Parts:[[1], null, null, null, null]

Remainder:2

advance

1

→

2

→

3

Result:

Parts:[[1], null, null, null, null]

detach

1

→

2

→

3

Result:

Parts:[[1], null, null, null, null]

connect

1

→

2

→

3

Result:

Parts:[[1], [2], null, null, null]

Remainder:1

advance

1

→

2

→

3

Result:

Parts:[[1], [2], null, null, null]

detach

1

→

2

→

3

Result:

Parts:[[1], [2], null, null, null]

connect

1

→

2

→

3

Result:

Parts:[[1], [2], [3], null, null]

Remainder:0

advance

1

→

2

→

3

Result:

Parts:[[1], [2], [3], null, null]

detach

1

→

2

→

3

Result:

Parts:[[1], [2], [3], null, null]

connect

1

→

2

→

3

Result:

Parts:[[1], [2], [3], null, null]

connect

1

→

2

→

3

Result:

Parts:[[1], [2], [3], null, null]

reconstruct

1

→

2

→

3

Result: [[1], [2], [3], null, null]

Key Takeaways

✓ The algorithm counts nodes first to determine exact part sizes and remainder for even distribution.

This counting step is crucial and often overlooked; without it, splitting evenly is impossible.

✓ Parts with remainder get one extra node, ensuring the first few parts are larger if nodes don't divide evenly.

Visualizing remainder distribution clarifies why some parts have one node and others are empty.

✓ Breaking links inline isolates parts without extra memory, showing efficient in-place list manipulation.

Seeing links broken step-by-step reveals how the list is physically split, which is hard to grasp from code alone.

Practice

1. You are given a circular array where each element represents the number of steps to move forward or backward. The goal is to determine if there exists a cycle where all moves are in the same direction and the cycle length is greater than 1. Which algorithmic approach guarantees an optimal O(n) time and O(1) space solution for this problem?

easy

A. Fast and slow pointer technique (Floyd's cycle detection) adapted for circular arrays with direction checks

B. Greedy approach that tries to jump as far as possible each time without revisiting indices

C. Dynamic programming to store reachable indices and cycle lengths

D. Brute force simulation with a visited set for each start index

Solution

1. Step 1: Understand problem constraints

   The problem requires detecting cycles in a circular array with direction consistency and cycle length > 1.

2. Step 2: Identify algorithm that efficiently detects cycles

   Fast and slow pointer (Floyd's cycle detection) can detect cycles in O(n) time and O(1) space, with added direction checks to ensure cycle validity.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Fast-slow pointer is classic for cycle detection in sequences [OK]

Hint: Cycle detection in sequences -> fast-slow pointer [OK]

Common Mistakes:

• Thinking greedy jumps detect cycles correctly
• Assuming DP applies here
• Using brute force is optimal

2. The following code attempts to detect a cycle in a circular array using the fast-slow pointer approach. Identify the line containing the subtle bug that causes incorrect cycle detection.

medium

A. Line with 'return true' inside while loop returns prematurely

B. Line with 'if slow == fast:' missing cycle length check

C. Line with 'direction = nums[i] > 0' incorrectly sets direction

D. Line with 'nums[marker] = 0' incorrectly marks visited elements

Solution

1. Step 1: Identify cycle detection condition

   The code returns true immediately when slow == fast, but does not check if cycle length > 1.

2. Step 2: Understand why self-loop is invalid

   Cycle of length 1 (self-loop) is invalid; must check if slow != next_index(slow) before returning true.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Missing cycle length check causes false positives [OK]

Hint: Check cycle length > 1 to avoid self-loop false positives [OK]

Common Mistakes:

• Returning true on self-loop cycles
• Mixing directions
• Not marking visited nodes

3. What is the time and space complexity of Floyd's cycle detection algorithm used to find the start of a cycle in a linked list of length n?

medium

A. Time: O(n), Space: O(1)

B. Time: O(n^2), Space: O(1)

C. Time: O(n), Space: O(n)

D. Time: O(n log n), Space: O(1)

Solution

1. Step 1: Analyze time complexity of pointer movements

   Fast pointer moves twice as fast as slow pointer, so they meet in O(n) steps, and locating cycle start also takes O(n) steps, total O(n).

2. Step 2: Analyze space complexity

   Only a fixed number of pointers are used, no extra data structures, so space is O(1).

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Linear time and constant space are standard for Floyd's algorithm [OK]

Hint: Two pointers traverse list linearly, no extra space needed [OK]

Common Mistakes:

• Confusing space with hash set approach
• Assuming quadratic time due to nested loops
• Mistaking recursion stack space

4. Identify the bug in the following code snippet for detecting and returning the cycle length in a linked list.

medium

A. Line 11: The length counting loop should start with length = 0 instead of 1.

B. Line 6: slow pointer should move two steps instead of one.

C. Line 7: fast pointer should move one step instead of two.

D. Line 4: The condition should check both fast and fast.next to avoid null pointer errors.

Solution

1. Step 1: Check loop condition for pointer safety

   The loop condition only checks if fast is not null, but fast.next may be null causing runtime error on fast.next.next.

2. Step 2: Confirm other lines are correct

   Slow moves one step, fast moves two steps correctly; length counting starts at 1 correctly.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   Missing fast.next check causes null pointer dereference [OK]

Hint: Always check fast and fast.next before advancing fast by two steps [OK]

Common Mistakes:

• Forgetting fast.next check
• Off-by-one in length counting
• Swapping slow and fast pointer steps

5. If the linked list nodes can be reused multiple times (i.e., the list is cyclic or can be traversed repeatedly), which modification is necessary to the optimal palindrome check algorithm?

hard

A. No modification needed; the current algorithm works as is.

B. Use a stack to store first half values instead of reversing to avoid modifying the list.

C. Convert the list to an array to handle multiple traversals safely.

D. Restore the reversed second half to original order after comparison to preserve list structure.

Solution

1. Step 1: Understand reuse implications

   If nodes are reused or list is cyclic, modifying it breaks future traversals.

2. Step 2: Restore list after palindrome check

   Reversing second half in-place must be undone to preserve original list structure.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   Restoring reversed half ensures list integrity for reuse [OK]

Hint: Always restore list after in-place reversal if list is reused [OK]

Common Mistakes:

• Ignoring list restoration causing side effects
• Switching to stack approach unnecessarily increasing space
• Assuming array conversion is always better