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Solution Approach

Solution Overview

There are 3 main approaches to solve this problem:

  1. Generate Full Triangle - Complex approach
  2. Iterative In-Place Calculation - Complex approach
  3. Mathematical Formula (Binomial Coefficients) - Complex approach

Approach 1: Generate Full Triangle

One straightforward approach is to generate the entire Pascal's Triangle up to the rowIndex-th row, and then return the last row.

Thinking Process: We can use the same approach as in the "Pascal's Triangle" problem to generate all rows up to the rowIndex-th row. Then, we simply return the last row.

Intuition: This approach is simple and directly follows the definition of Pascal's Triangle. However, it doesn't meet the follow-up requirement of using only O(rowIndex) extra space, as we're storing all rows of the triangle.

Algorithm:

  1. Initialize the result array with the first row [1].
  2. For each subsequent row i from 1 to rowIndex:
  3. a. Create a new row starting with 1.
  4. b. For each position j from 1 to i-1:
  5. - Calculate the value as the sum of elements at positions j-1 and j from the previous row.
  6. c. Append 1 to the end of the row.
  7. d. Replace the result array with this new row.
  8. Return the result array.

Time Complexity:

O(rowIndex²)

We need to calculate rowIndex rows, and each row i has i+1 elements, leading to a total of (rowIndex+1)(rowIndex+2)/2 elements, which is O(rowIndex²).

Space Complexity:

O(rowIndex)

We only need to store the current row, which has rowIndex+1 elements, resulting in O(rowIndex) space complexity.

Approach 2: Iterative In-Place Calculation

We can optimize the space complexity by calculating the row in-place, updating the array from right to left.

Thinking Process: If we update the array from left to right, we would overwrite values that we still need for subsequent calculations. However, if we update from right to left, we can ensure that we don't overwrite values before we use them.

Intuition: By updating the array from right to left, we can calculate the rowIndex-th row using only O(rowIndex) extra space, meeting the follow-up requirement.

Algorithm:

  1. Initialize an array of size rowIndex+1 with all 1s.
  2. For each row i from 2 to rowIndex:
  3. a. For each position j from i-1 down to 1:
  4. - Update the value at position j by adding the value at position j-1.
  5. Return the array.

Time Complexity:

O(rowIndex²)

We still need to perform calculations for each element in the triangle up to the rowIndex-th row, resulting in O(rowIndex²) time complexity.

Space Complexity:

O(rowIndex)

We only use a single array of size rowIndex+1, resulting in O(rowIndex) space complexity.

Approach 3: Mathematical Formula (Binomial Coefficients)

We can directly calculate the rowIndex-th row using the mathematical formula for binomial coefficients.

Thinking Process: The element at position j in the rowIndex-th row is C(rowIndex, j) = rowIndex! / (j! * (rowIndex-j)!). However, computing factorials can lead to overflow for large numbers, so we need to use an optimized approach.

Intuition: We can use the property that C(n, k) = C(n, k-1) * (n-k+1) / k, which allows us to calculate each element based on the previous element in the same row, avoiding factorial calculations.

Algorithm:

  1. Initialize an array of size rowIndex+1.
  2. Set the first element to 1.
  3. For each position j from 1 to rowIndex:
  4. a. Calculate C(rowIndex, j) = C(rowIndex, j-1) * (rowIndex-j+1) / j
  5. b. Set the value at position j to the calculated value.
  6. Return the array.

Time Complexity:

O(rowIndex)

We calculate each element in the row once, resulting in O(rowIndex) time complexity.

Space Complexity:

O(rowIndex)

We use a single array of size rowIndex+1, resulting in O(rowIndex) space complexity.

Pseudocode

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function getRow(rowIndex):
    // Initialize result with the first row
    result = [1]
    
    // Generate each row based on the previous row
    for i from 1 to rowIndex:
        // Create a new row
        row = [1]
        
        // Calculate middle elements
        for j from 1 to i-1:
            row.append(result[j-1] + result[j])
        
        // End with 1
        row.append(1)
        
        // Replace result with the new row
        result = row
    
    return result
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