Introduction

In this blog post, we will explore how to solve the LeetCode problem "1019" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You are given the head of a linked list with n nodes. For each node in the list, find the value of the next greater node. That is, for each node, find the value of the first node that is next to it and has a strictly larger value than it. Return an integer array answer where answer[i] is the value of the next greater node of the ith node (1-indexed). If the ith node does not have a next greater node, set answer[i] = 0. Example 1: Input: head = [2,1,5] Output: [5,5,0] Example 2: Input: head = [2,7,4,3,5] Output: [7,0,5,5,0] Constraints: The number of nodes in the list is n. 1 <= n <= 104 1 <= Node.val <= 109

Explanation

Here's the breakdown of the approach and the Python code:

• Convert to Array: Transform the linked list into a regular Python list. This allows for easier indexing and manipulation.
• Stack-Based Approach: Use a stack to keep track of indices of elements for which we haven't found a next greater element yet.

• Iterate and Compare: Iterate through the list. If the current element is greater than the element at the index on top of the stack, it means we've found the next greater element for that index. Pop the stack until this is no longer true. Then, push the current element's index onto the stack.

• Runtime Complexity: O(N), where N is the number of nodes in the linked list.

• Space Complexity: O(N)

Code

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

def nextLargerNodes(head):
    """
    Finds the next greater node for each node in a linked list.

    Args:
        head: The head of the linked list.

    Returns:
        A list of integers where each element is the value of the next greater node
        or 0 if there is no next greater node.
    """

    # Convert linked list to a list
    nodes = []
    curr = head
    while curr:
        nodes.append(curr.val)
        curr = curr.next

    n = len(nodes)
    result = [0] * n
    stack = []  # Stack to store indices

    for i in range(n):
        while stack and nodes[i] > nodes[stack[-1]]:
            result[stack[-1]] = nodes[i]
            stack.pop()
        stack.append(i)

    return result