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Given an integer numRows, return the first numRows of Pascal’s triangle. In Pascal’s triangle, each number is the sum of the two numbers directly above it.

Clarifying Questions

  1. What are the constraints on numRows?
    • Typically 1 <= numRows <= 30 based on common test cases.
  2. What should be returned if numRows is 0?
    • Since Pascal’s triangle with 0 rows trivially has no content, it would be an empty list [].
  3. Should the function handle invalid inputs like negative numbers or non-integer values?
    • Generally, let’s assume the input will be a non-negative integer as per the constraints.

Strategy

The problem requires generating Pascal’s triangle up to a given number of rows. Each row in Pascal’s triangle starts and ends with the number 1, and each interior number is the sum of the two numbers above it from the previous row.

Steps to achieve this:

  1. Initialize the triangle with the first row [1].
  2. Iterate from the second row to the numRows:
    • Start each row with [1].
    • For each new element, sum the two numbers from the previous row.
    • End each row with [1].
  3. Append each generated row to the results list.

Code

def generate(numRows):
    if numRows == 0:
        return []
    
    triangle = []
    
    for row_num in range(numRows):
        # Start with a row of `1`s
        row = [1] * (row_num + 1)
        
        # Calculate the interior values
        for j in range(1, row_num):
            row[j] = triangle[row_num - 1][j - 1] + triangle[row_num - 1][j]
        
        triangle.append(row)
    
    return triangle

# Example usage
print(generate(5))

Time Complexity

The time complexity of this approach is (O(n^2)), where n is numRows. This is because:

  1. You are looping through each row (n iterations).
  2. Within each row, you’re filling values (k iterations, where (k \leq n)).

Thus, the overall complexity is driven by the nested loops iterating through all the elements of the triangle.

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