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Solution Approach

Solution Overview

There are 3 main approaches to solve this problem:

  1. Iterative Row-by-Row Construction - Complex approach
  2. Mathematical Formula (Binomial Coefficients) - Complex approach
  3. Dynamic Programming Approach - Complex approach

Approach 1: Iterative Row-by-Row Construction

Let's start by understanding the structure of Pascal's Triangle. Each row begins and ends with 1, and every other element is the sum of the two elements directly above it.

Thinking Process: We can build the triangle row by row, starting from the first row which is always [1]. For each subsequent row, we know:

  • The first and last elements are always 1
  • Each middle element at position j is the sum of elements at positions j-1 and j from the previous row

Intuition: This approach directly follows the definition of Pascal's Triangle, making it intuitive and easy to understand.

Algorithm:

  1. Initialize the result array with the first row [1].
  2. For each subsequent row i from 1 to numRows-1:
  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. Add the row to the result array.
  8. Return the result array.

Time Complexity:

O(n²)

Where n is the number of rows. We need to calculate n rows, and each row i has i elements, leading to a total of n(n+1)/2 elements, which is O(n²).

Space Complexity:

O(n²)

We store all elements of the triangle, which is n(n+1)/2 elements, resulting in O(n²) space complexity.

Approach 2: Mathematical Formula (Binomial Coefficients)

Pascal's Triangle can also be generated using the mathematical formula for binomial coefficients. The element at row n and position k is C(n, k) = n! / (k! * (n-k)!).

Thinking Process: Instead of building each row based on the previous row, we can directly calculate each element using the binomial coefficient formula. 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 the result array.
  2. For each row i from 0 to numRows-1:
  3. a. Create a new row.
  4. b. The first element of the row is always 1.
  5. c. For each position j from 1 to i:
  6. - Calculate C(i, j) = C(i, j-1) * (i-j+1) / j
  7. - Add the calculated value to the row.
  8. d. Add the row to the result array.
  9. Return the result array.

Time Complexity:

O(n²)

We still need to calculate n rows with a total of n(n+1)/2 elements, resulting in O(n²) time complexity.

Space Complexity:

O(n²)

We store all elements of the triangle, which is n(n+1)/2 elements, resulting in O(n²) space complexity.

Approach 3: Dynamic Programming Approach

We can view this problem as a dynamic programming problem, where each element depends on elements from the previous row.

Thinking Process: Let's define dp[i][j] as the value at row i and position j in Pascal's Triangle. The recurrence relation is:

  • dp[i][0] = dp[i][i] = 1 (first and last elements of each row are 1)
  • dp[i][j] = dp[i-1][j-1] + dp[i-1][j] for 1 ≤ j ≤ i-1

Intuition: This approach explicitly uses dynamic programming principles to build the triangle, making the dependencies between elements clear.

Algorithm:

  1. Initialize a 2D array dp of size numRows × numRows.
  2. Set dp[0][0] = 1 (the first row has only one element, which is 1).
  3. For each row i from 1 to numRows-1:
  4. a. Set dp[i][0] = dp[i][i] = 1 (first and last elements are 1).
  5. b. For each position j from 1 to i-1:
  6. - Set dp[i][j] = dp[i-1][j-1] + dp[i-1][j].
  7. Convert the 2D array to the required format and return.

Time Complexity:

O(n²)

We need to fill a triangular portion of a 2D array with n rows, resulting in O(n²) time complexity.

Space Complexity:

O(n²)

We use a 2D array of size numRows × numRows, resulting in O(n²) space complexity.

Pseudocode

solution.pseudo
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function generate(numRows):
    // Initialize result with the first row
    result = [[1]]
    
    // Generate each row based on the previous row
    for i from 1 to numRows-1:
        // Start with 1
        row = [1]
        
        // Calculate middle elements
        for j from 1 to i-1:
            row.append(result[i-1][j-1] + result[i-1][j])
        
        // End with 1
        row.append(1)
        
        // Add row to result
        result.append(row)
    
    return result
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