118. Pascal's Triangle - Detailed Explanation

Problem Statement

Pascal's Triangle is a triangle of numbers where each number is the sum of the two numbers directly above it. You are given an integer numRows. Your task is to generate the first numRows of Pascal's triangle.

How Pascal's Triangle Works

• The first row is always [1].
• Every subsequent row starts and ends with 1.
• Each inner element of a row is computed by summing the two numbers from the previous row that are directly above it.

Example Inputs, Outputs, and Explanations

1. Example 1

   • Input: numRows = 5
   • Output:

     [
       [1],
       [1,1],
       [1,2,1],
       [1,3,3,1],
       [1,4,6,4,1]
     ]
     

   • Explanation:
     • Row 1: [1]
     • Row 2: [1, 1]
     • Row 3: [1, 2, 1] (2 is computed as 1 + 1 from the previous row)
     • Row 4: [1, 3, 3, 1] (3 is computed as 1+2 and 2+1 respectively)
     • Row 5: [1, 4, 6, 4, 1] (4 is computed as 1+3, 6 as 3+3, etc.)

2. Example 2

   • Input: numRows = 1
   • Output:

     [
       [1]
     ]
     

   • Explanation:
     Only the first row is needed.

Constraints

• 1 <= numRows <= 30

Hints to Get Started

• Hint 1:
  Remember that the first and last element of every row is always 1. Use this fact to simplify the construction of each row.

• Hint 2:
  Each element in the interior of the row is the sum of the two elements directly above it from the previous row. Think about iterating over the previous row to compute the new row.

Approaches to Solve the Problem

Approach 1: Iterative Row-by-Row Construction

Idea

• Build the triangle one row at a time.
• For each new row, initialize with a 1 at the beginning.
• Then compute each middle element as the sum of two adjacent numbers from the previous row.
• End the row with a 1.

Step-by-Step Process

1. Initialize an empty list (or array) to hold all rows.
2. Iterate from 0 to numRows - 1:
   • Create a new list for the current row.
   • Set the first element as 1.
   • For each position j (from 1 to current row length - 1), compute:
     • currentRow[j] = previousRow[j - 1] + previousRow[j]
   • Append a final 1 to the current row.
   • Add the current row to the triangle.
3. Return the triangle.

Visual Example (for numRows = 5)

• Row 1:
  [1]

• Row 2:
  Start with [1]
  End with [1] → [1, 1]

• Row 3:
  Start with [1]
  Compute middle: 1 (from row 2) + 1 (from row 2) → 2
  End with [1] → [1, 2, 1]

• Row 4:
  Start with [1]
  Compute first middle: 1 + 2 = 3
  Compute second middle: 2 + 1 = 3
  End with [1] → [1, 3, 3, 1]

• Row 5:
  Start with [1]
  Compute middles:

  • 1 + 3 = 4
  • 3 + 3 = 6
  • 3 + 1 = 4
    End with [1] → [1, 4, 6, 4, 1]

Approach 2: Mathematical Combination Formula (Alternative Variation)

Idea

• Each element at row n and position k can be directly computed using combinations:
  • C(n-1, k) = (n-1)! / (k! * ((n-1) - k)!)
• This method uses the mathematical formula to compute each element without iterating over previous rows.

Considerations

• Although this approach is mathematically elegant, it involves factorial computations which can be less efficient and may require handling of large numbers.
• The iterative approach is generally simpler to implement and understand for this problem.

Code Implementation

Python Implementation