Pascal's Triangle

Count Primes

Unique Paths

Build the first numRows rows of Pascal's triangle where each value sums the two above it.

Given an integer `numRows`, return the first `numRows` rows of **Pascal's triangle**. In Pascal's triangle each number is the sum of the two numbers directly above it, and every row starts and ends with `1`.

**Input / output**
- Input: `numRows: integer`
- Output: a list of `numRows` lists, each row of the triangle

**Examples**
1. `numRows = 5` returns `[[1],[1,1],[1,2,1],[1,3,3,1],[1,4,6,4,1]]`.
2. `numRows = 1` returns `[[1]]`.
3. `numRows = 2` returns `[[1],[1,1]]`.

**Constraints**
- `1 <= numRows <= 30`

**Follow-up**
Row `k` (0-indexed) is exactly the binomial coefficients `C(k, 0..k)`. Can you generate each row from the previous one using only additions, without recomputing factorials?

- Row `0` is `[1]`.
- Every later row begins and ends with `1`.
- An interior value at position `j` equals the sum of positions `j-1` and `j` in the previous row.

Understand the Pattern

- Start the result with the first row `[1]`.
- For each new row, walk the previous row and add adjacent pairs.
- Wrap the sums with a leading and trailing `1`.

Build Row by Row

- You never need more than the previous row to build the next one.
- Derive the next row by zipping the previous row shifted by one.
- This keeps the logic to pure additions.

Rolling Previous Row

- Each of the `numRows` rows costs work proportional to its length.
- Total time is `O(numRows^2)`.
- Space is `O(numRows^2)` for the returned triangle.