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Linked List Components

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Linked ListsArrays

You are given the head of a linked list containing unique integer values and an integer array nums that is a subset of the linked list values.

Return the number of connected components in nums where two values are connected if they appear consecutively in the linked list.

Example 1:

Input: head = [0,1,2,3], nums = [0,1,3]
Output: 2
Explanation: 0 and 1 are connected, so [0, 1] and [3] are the two connected components.

Example 2:

Input: head = [0,1,2,3,4], nums = [0,3,1,4]
Output: 2
Explanation: 0 and 1 are connected, 3 and 4 are connected, so [0, 1] and [3, 4] are the two connected components.

Constraints:

The number of nodes in the linked list is n.
1 <= n <= 10⁴
0 <= Node.val < n
All the values Node.val are unique.
1 <= nums.length <= n
0 <= nums[i] < n
All the values of nums are unique.

Try coding it in LeetCode

Linked List Components

Medium

Asked by:

21 views

Topics:

Linked ListsArrays

You are given the head of a linked list containing unique integer values and an integer array nums that is a subset of the linked list values.

Return the number of connected components in nums where two values are connected if they appear consecutively in the linked list.

Example 1:

Input: head = [0,1,2,3], nums = [0,1,3]
Output: 2
Explanation: 0 and 1 are connected, so [0, 1] and [3] are the two connected components.

Example 2:

Input: head = [0,1,2,3,4], nums = [0,3,1,4]
Output: 2
Explanation: 0 and 1 are connected, 3 and 4 are connected, so [0, 1] and [3, 4] are the two connected components.

Constraints:

The number of nodes in the linked list is n.
1 <= n <= 10⁴
0 <= Node.val < n
All the values Node.val are unique.
1 <= nums.length <= n
0 <= nums[i] < n
All the values of nums are unique.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

Can the values within the linked list and the set `nums` contain negative numbers or any specific range of values?
Can the linked list be empty, or can `nums` be empty? What should the function return in those cases?
Does the order of nodes within a component matter? For example, if `nums` is `{1, 2, 3}` and the list is `1 -> 3 -> 2`, is that one component or three?
Are there duplicate values within the linked list, and if so, how should they be handled when determining components?
If no components are found (i.e., no nodes in the linked list have values present in `nums`), what should the function return?

Brute Force Solution

Approach

We need to find how many linked list sections are also in a special set of numbers. The brute force method involves checking every single piece of the linked list to see if it's part of that special set.

Here's how the algorithm would work step-by-step:

Go through the linked list one item at a time.
For each item in the linked list, check if that item is in the special set of numbers.
If the item is in the set, see if the item before it was also in the set. If the item before it wasn't in the set, this starts a new component.
Keep going until you reach the end of the linked list.
Count how many times a new component starts. This is the number of linked list components.

Code Implementation

def num_components(head, nums):
    number_of_components = 0
    is_previous_in_set = False
    current_node = head

    while current_node:
        is_current_in_set = current_node.val in nums

        # Check if current node is in nums and previous wasn't.
        if is_current_in_set and not is_previous_in_set:
            number_of_components += 1

        # Update previous node status
        is_previous_in_set = is_current_in_set

        current_node = current_node.next

    return number_of_components

Big(O) Analysis

Time Complexity

O(n) – The provided solution iterates through the linked list once, which contains 'n' nodes. For each node, it checks if the node's value is present in the given set of numbers (nums). Assuming the set lookup takes constant time (e.g., if nums is converted to a hash set), the overall time complexity is dominated by the single pass through the linked list. Therefore, the time complexity is O(n), where n is the number of nodes in the linked list.

Space Complexity

O(S) – The provided solution checks each linked list item against a "special set of numbers". To determine if an item is in the set, the most efficient approach involves storing the special set in a hash set for quick lookups. If the special set contains S numbers, this hash set will require O(S) space. No other auxiliary data structures, such as temporary lists or recursion stacks, are created. Therefore, the overall space complexity is dominated by the hash set representing the special set.

Optimal Solution

Approach

The challenge is to identify groups of connected values from a given linked list that are also present in a separate set of allowed values. The efficient strategy involves converting the set of allowed values into a readily searchable format and then traversing the linked list once to count components.

Here's how the algorithm would work step-by-step:

First, take the list of allowed values and prepare it for quick lookups, like making a list where you can instantly check if a value is present.
Next, start at the beginning of the linked list.
As you move through the linked list, check if the current value is one of the allowed values.
If the current value is allowed and you weren't already inside a component, then you've found the start of a new component; count it.
If the current value is not allowed, you are no longer in a component.
Keep going until you reach the end of the linked list. The final count is your answer.

Code Implementation

def linked_list_components(head, nums):
    allowed_values = set(nums)
    number_of_components = 0
    in_component = False

    current_node = head
    while current_node:
        # Check if the current node's value is in the set of allowed values.
        if current_node.val in allowed_values:

            # If not already in a component, start a new one.
            if not in_component:
                number_of_components += 1
                in_component = True
        else:
            # If the current node's value is not allowed, end the current component.
            if in_component:
                in_component = False

        current_node = current_node.next

    return number_of_components

Big(O) Analysis

Time Complexity

O(n) – The time complexity is driven by converting the 'nums' list to a set and then iterating through the linked list once. Converting the list of 'nums' to a set takes O(m) time where m is the size of the nums list. Iterating through the linked list of size n takes O(n) time. Therefore, the overall time complexity is O(m + n). However, if we assume m is smaller than n (length of linked list), the time complexity simplifies to O(n).

Space Complexity

O(M) – The algorithm converts the list of allowed values into a readily searchable format. The most efficient implementation would likely use a hash set or a boolean array. In the worst case, the set of allowed values could contain all unique values, requiring a hash set or boolean array to store at most M values, where M is the number of allowed values. The remaining part of the algorithm operates in place, using a constant amount of space for a few variables. Therefore, the auxiliary space complexity is O(M).

Edge Cases

Case	How to Handle
Empty linked list	Return 0, as there are no components in an empty list.
Null linked list	Treat as an empty list and return 0.
Empty array G	Return 0, because no components can be formed without elements in G.
All nodes in the linked list are present in G	Return 1, as the entire list is one component.
No nodes in the linked list are present in G	Return 0, as there are no components.
G contains duplicate values	The set efficiently handles duplicate values in G, ensuring we only consider unique values.
Very large linked list and G	The set lookup allows efficient checking for presence in G, so the solution should scale linearly with the list's length.
Linked list with only one node and G contains that node's value.	Return 1, the single node forms a component.

Empty linked list

How to Handle:

Return 0, as there are no components in an empty list.

Null linked list

How to Handle:

Treat as an empty list and return 0.

Empty array G

How to Handle:

Return 0, because no components can be formed without elements in G.

All nodes in the linked list are present in G

How to Handle:

Return 1, as the entire list is one component.

No nodes in the linked list are present in G

How to Handle:

Return 0, as there are no components.

G contains duplicate values

How to Handle:

The set efficiently handles duplicate values in G, ensuring we only consider unique values.

Very large linked list and G

How to Handle:

The set lookup allows efficient checking for presence in G, so the solution should scale linearly with the list's length.

Linked list with only one node and G contains that node's value.

How to Handle:

Return 1, the single node forms a component.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

Can the values within the linked list and the set `nums` contain negative numbers or any specific range of values?
Can the linked list be empty, or can `nums` be empty? What should the function return in those cases?
Does the order of nodes within a component matter? For example, if `nums` is `{1, 2, 3}` and the list is `1 -> 3 -> 2`, is that one component or three?
Are there duplicate values within the linked list, and if so, how should they be handled when determining components?
If no components are found (i.e., no nodes in the linked list have values present in `nums`), what should the function return?

Brute Force Solution

Approach

Here's how the algorithm would work step-by-step:

Go through the linked list one item at a time.
For each item in the linked list, check if that item is in the special set of numbers.
If the item is in the set, see if the item before it was also in the set. If the item before it wasn't in the set, this starts a new component.
Keep going until you reach the end of the linked list.
Count how many times a new component starts. This is the number of linked list components.

Code Implementation

def num_components(head, nums):
    number_of_components = 0
    is_previous_in_set = False
    current_node = head

    while current_node:
        is_current_in_set = current_node.val in nums

        # Check if current node is in nums and previous wasn't.
        if is_current_in_set and not is_previous_in_set:
            number_of_components += 1

        # Update previous node status
        is_previous_in_set = is_current_in_set

        current_node = current_node.next

    return number_of_components

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

Here's how the algorithm would work step-by-step:

First, take the list of allowed values and prepare it for quick lookups, like making a list where you can instantly check if a value is present.
Next, start at the beginning of the linked list.
As you move through the linked list, check if the current value is one of the allowed values.
If the current value is allowed and you weren't already inside a component, then you've found the start of a new component; count it.
If the current value is not allowed, you are no longer in a component.
Keep going until you reach the end of the linked list. The final count is your answer.

Code Implementation

def linked_list_components(head, nums):
    allowed_values = set(nums)
    number_of_components = 0
    in_component = False

    current_node = head
    while current_node:
        # Check if the current node's value is in the set of allowed values.
        if current_node.val in allowed_values:

            # If not already in a component, start a new one.
            if not in_component:
                number_of_components += 1
                in_component = True
        else:
            # If the current node's value is not allowed, end the current component.
            if in_component:
                in_component = False

        current_node = current_node.next

    return number_of_components

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Empty linked list	Return 0, as there are no components in an empty list.
Null linked list	Treat as an empty list and return 0.
Empty array G	Return 0, because no components can be formed without elements in G.
All nodes in the linked list are present in G	Return 1, as the entire list is one component.
No nodes in the linked list are present in G	Return 0, as there are no components.
G contains duplicate values	The set efficiently handles duplicate values in G, ensuring we only consider unique values.
Very large linked list and G	The set lookup allows efficient checking for presence in G, so the solution should scale linearly with the list's length.
Linked list with only one node and G contains that node's value.	Return 1, the single node forms a component.

Empty linked list

How to Handle:

Return 0, as there are no components in an empty list.

Null linked list

How to Handle:

Treat as an empty list and return 0.

Empty array G

How to Handle:

Return 0, because no components can be formed without elements in G.

All nodes in the linked list are present in G

How to Handle:

Return 1, as the entire list is one component.

No nodes in the linked list are present in G

How to Handle:

Return 0, as there are no components.

G contains duplicate values

How to Handle:

The set efficiently handles duplicate values in G, ensuring we only consider unique values.

Very large linked list and G

How to Handle:

The set lookup allows efficient checking for presence in G, so the solution should scale linearly with the list's length.

Linked list with only one node and G contains that node's value.

How to Handle:

Return 1, the single node forms a component.