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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:
Input: rowIndex = 3
Output: [1, 3, 3, 1]

Constraints:
- 0 <= rowIndex <= 33

Clarifying Questions:

1. Input Constraints: Is rowIndex guaranteed to fall within the specified constraints?
   • Answer: Yes, rowIndex is guaranteed to be between 0 and 33.
2. Output Format: Should the output be a list of integers representing the specific row?
   • Answer: Yes, the output should be a list of integers.

Strategy:

To generate the rowIndexth row of Pascal’s Triangle, we can utilize the following properties:

• Each element in the Pascal’s Triangle can be generated iteratively using the previous row.
• The first and the last element of any row is always 1.
• Any other element can be calculated by summing the two elements from the previous row.

Here’s a step-by-step plan:

1. Start with the first row [1].
2. Iteratively generate each subsequent row up to rowIndex. For each row, generate new elements by summing appropriate elements from the previous row, with the rule that the first and the last elements are always 1.
3. Maintain only the required row at each iteration to save space.

Code:

def getRow(rowIndex):
    row = [1]
    for _ in range(rowIndex):
        row = [x + y for x, y in zip([0] + row, row + [0])]
    return row

# Example usage:
print(getRow(3))  # Output: [1, 3, 3, 1]

Time Complexity:

• Time Complexity: O(rowIndex^2). This is because each row generation involves summing operations, which takes linear time concerning the number of elements in the row. Summing up these operations for all rows leads to an overall quadratic time complexity.
• Space Complexity: O(rowIndex). We use a single list to store the current row, which grows up to rowIndex + 1 in size.

This approach efficiently generates the desired row while keeping space usage minimal.

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