Solution Explanation for Pascal's Triangle II

This problem asks to generate the rowIndex-th row of Pascal's Triangle. Pascal's Triangle is a triangular array where each number is the sum of the two numbers directly above it. The solution presented uses a dynamic programming approach for efficiency.

Approach: Dynamic Programming with Space Optimization

The core idea is to iteratively build each row based on the previous row. We don't need to store the entire Pascal's triangle; we only need to maintain the current row. This significantly reduces space complexity.

Algorithm:

1. Initialization: Create an array f of size rowIndex + 1 and initialize all elements to 1. This represents the first row (rowIndex = 0) which contains only 1.

2. Iteration: Iterate from the second row (i = 2) up to the rowIndex. For each row i, update the array f from right to left. The value at index j becomes the sum of the values at indices j and j-1 from the previous row. This efficiently calculates each element in the current row using the previous row's values.

3. Return: After the iterations are complete, the array f contains the elements of the rowIndex-th row of Pascal's Triangle. Return the array.

Time and Space Complexity Analysis

• Time Complexity: The outer loop iterates rowIndex times. The inner loop iterates at most rowIndex times in the worst case (last row). Therefore, the overall time complexity is O(rowIndex²). However, it's closer to O(rowIndex * (rowIndex /2)) because the inner loop's iterations reduce on each outer iteration. For the purpose of analysis we can simplify to O(rowIndex²)

• Space Complexity: The algorithm uses an array f of size rowIndex + 1 to store the current row. Thus, the space complexity is O(rowIndex). This is an optimization compared to storing the entire triangle which would require O(rowIndex²) space.

Code Examples (with slight variations for clarity)

The provided code in various languages implements this optimized dynamic programming approach. Here's a slightly modified version in Python for better readability:

class Solution:
    def getRow(self, rowIndex: int) -> List[int]:
        if rowIndex == 0:
            return [1]
 
        row = [1]  # Initialize the first row
        for i in range(1, rowIndex + 1):
            new_row = [1]  # Each row starts with 1
            for j in range(1, i):
                new_row.append(row[j - 1] + row[j])  # Calculate intermediate elements
            new_row.append(1)  # Each row ends with 1
            row = new_row  # Update the row for the next iteration
 
        return row

This version explicitly shows the building up of each row and improves readability compared to the in-place update used in the original solution. The time and space complexity remain the same. All other provided languages implement essentially the same algorithm with minor syntax differences.