Introduction

In this blog post, we will explore how to solve the LeetCode problem "119" 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

Given an integer rowIndex, return the rowIndexth (0-indexed) row of the Pascal's triangle. In Pascal's triangle, each number is the sum of the two numbers directly above it as shown: Example 1: Input: rowIndex = 3 Output: [1,3,3,1] Example 2: Input: rowIndex = 0 Output: [1] Example 3: Input: rowIndex = 1 Output: [1,1] Constraints: 0 <= rowIndex <= 33 Follow up: Could you optimize your algorithm to use only O(rowIndex) extra space?

Explanation

Here's an efficient solution to generate the rowIndex-th row of Pascal's triangle:

• Iterative Row Generation: Build the Pascal's triangle row by row, starting from the first row [1].
• In-place Update: Instead of storing all previous rows, update the current row in-place based on the previous row's values, significantly reducing space complexity. Specifically, generate each element based on the formula: row[j] = row[j-1] + row[j].

• Optimization: We will compute the row from right to left. If we compute it from left to right, we would overwrite the values needed in the later computations.

• Runtime Complexity: O(rowIndex²), Storage Complexity: O(rowIndex)

Code

    def getRow(rowIndex: int) -> list[int]:
    """
    Given an integer rowIndex, return the rowIndexth (0-indexed) row of the Pascal's triangle.
    """
    row = [1] * (rowIndex + 1)  # Initialize the row with 1s

    for i in range(1, rowIndex + 1):
        for j in range(i - 1, 0, -1):  # Iterate from right to left
            row[j] = row[j-1] + row[j]
    return row