You are given the head of a linked list.
Remove every node which has a node with a greater value anywhere to the right side of it.
Return the head of the modified linked list.
Example 1:
Input: head = [5,2,13,3,8] Output: [13,8] Explanation: The nodes that should be removed are 5, 2 and 3. - Node 13 is to the right of node 5. - Node 13 is to the right of node 2. - Node 8 is to the right of node 3.
Example 2:
Input: head = [1,1,1,1] Output: [1,1,1,1] Explanation: Every node has value 1, so no nodes are removed.
Constraints:
[1, 105].1 <= Node.val <= 105When 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:
The brute force way to remove nodes from a linked list involves checking every possible combination of nodes to remove. This means we'll explore each possible sublist, deciding for each node whether to keep it or delete it, and then checking if the remaining list meets our requirements.
Here's how the algorithm would work step-by-step:
def remove_nodes_brute_force(head, requirement_function, comparison_function):
all_possible_sublists = []
def generate_sublists(current_node, current_sublist):
if current_node is None:
all_possible_sublists.append(current_sublist)
return
# Explore the possibility of excluding
# current node from the sublist.
generate_sublists(current_node.next, current_sublist.copy())
# Explore the possibility of including
# current node in the sublist.
current_sublist.append(current_node.val)
generate_sublists(current_node.next, current_sublist.copy())
generate_sublists(head, [])
valid_sublists = []
for sublist in all_possible_sublists:
# We convert the list of values to a linked list.
# It is convenient for the later requirement check.
dummy_head = ListNode(0)
current = dummy_head
for value in sublist:
current.next = ListNode(value)
current = current.next
list_head = dummy_head.next
if requirement_function(list_head):
valid_sublists.append(sublist)
best_sublist = None
for sublist in valid_sublists:
# Choose the 'best' valid sublist based on
# the provided comparison function
if best_sublist is None:
best_sublist = sublist
elif comparison_function(sublist, best_sublist):
best_sublist = sublist
if best_sublist is None:
return None
# We convert the best sublist to a linked list.
dummy_head = ListNode(0)
current = dummy_head
for value in best_sublist:
current.next = ListNode(value)
current = current.next
return dummy_head.next
class ListNode:
def __init__(self, val=0, next=None):
self.val = val
self.next = nextThe key to efficiently removing nodes from a linked list lies in traversing the list and making on-the-spot decisions. Instead of revisiting nodes, we'll maintain knowledge of the maximum value encountered so far to determine which nodes need removal as we go. This prevents extra work and makes the process much faster.
Here's how the algorithm would work step-by-step:
class ListNode:
def __init__(self, value=0, next_node=None):
self.value = value
self.next_node = next_node
def remove_nodes(head):
if not head:
return None
maximum_value_so_far = head.value
current_node = head.next_node
previous_node = head
while current_node:
if current_node.value < maximum_value_so_far:
# Remove the current node by updating the previous node's next.
previous_node.next_node = current_node.next_node
else:
# Update the maximum value because we found a larger value
maximum_value_so_far = current_node.value
previous_node = current_node
current_node = current_node.next_node
# Check if the head node needs to be removed
if head.value < 0:
return head.next_node
else:
return head| Case | How to Handle |
|---|---|
| Null or Empty Linked List Input | Return null immediately; no nodes to process. |
| All nodes have values to be removed | The list should become empty after processing; correctly handle head changes. |
| No nodes need to be removed | The original list should be returned without modifications; verify no accidental changes. |
| Consecutive nodes need to be removed | Properly link the remaining nodes after removing the consecutive nodes. |
| The head node needs to be removed | Update the head pointer correctly; handle cases where the new head also needs removal. |
| Linked list with a single node | Check if the single node needs removal, and return null if so, otherwise the original list. |
| Large linked list (potential memory/performance issues) | Iterative approach avoids stack overflow issues associated with recursion. |
| All node values are the same | The algorithm should correctly identify and remove the nodes based on the given criteria, even if the values are identical. |