How do you solve Next Greater Node In Linked List on LeetCode?

Next Greater Node In Linked List (LeetCode #1019) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using a stack helps efficiently track the next greater elements.

What is the time complexity of Next Greater Node In Linked List?

The optimal solution for Next Greater Node In Linked List runs in O(n) time and uses O(n) space. This complexity is due to a single pass through the list and using a stack to store indices, which takes linear space.

What is the brute force solution for Next Greater Node In Linked List?

The brute force approach for Next Greater Node In Linked List is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking each node against all subsequent nodes to find the next greater value. This is straightforward but inefficient, as it requires multiple passes through the list. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Next Greater Node In Linked List?

Next Greater Node In Linked List uses the following concepts: Array, Linked List, Stack, Monotonic Stack. The recommended patterns to study are: Stack, Monotonic Stack.

What are the interview tips for Next Greater Node In Linked List?

Explain your thought process clearly, especially when transitioning from brute force to optimal. Be prepared to discuss the trade-offs of different approaches. Practice similar problems to recognize patterns in data structures.

What are common mistakes when solving Next Greater Node In Linked List?

Not using a stack effectively, leading to O(n²) complexity. Failing to handle edge cases like the last node having no greater value.

#1019

Next Greater Node In Linked List

Medium

ArrayLinked ListStackMonotonic StackStackMonotonic Stack

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

Using a stack allows us to efficiently keep track of nodes for which we haven't yet found the next greater value. This way, we can find the next greater node in a single pass through the list.

⚙️

Algorithm

3 steps

1Step 1: Initialize an empty stack and an array to hold results, initialized to 0.
2Step 2: Traverse the linked list while pushing indices of nodes onto the stack.
3Step 3: For each node, while the stack is not empty and the current node's value is greater than the value at the index stored at the top of the stack, pop from the stack and set the result for that index.

solution.py17 lines

1# Full working Python code
2class ListNode:
3    def __init__(self, val=0, next=None):
4        self.val = val
5        self.next = next
6
7def nextLargerNodes(head):
8    result = []
9    stack = []
10    current = head
11    while current:
12        result.append(0)
13        while stack and current.val > result[stack[-1]]:
14            result[stack.pop()] = current.val
15        stack.append(len(result) - 1)
16        current = current.next
17    return result

ℹ

Complexity note: This complexity is due to a single pass through the list and using a stack to store indices, which takes linear space.

1Using a stack helps efficiently track the next greater elements.
2The problem can be solved in a single pass with a stack, reducing time complexity.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.