15. Pascal's Triangle

Q: Why is each row's first and last entry 1?

C(n, 0) and C(n, n) both equal 1 — there's exactly one way to choose 0 items or all items.

Q: How does this relate to window functions?

Window functions like LAG(x) OVER (ORDER BY ts) produce a value for each row based on previous rows. The recurrence pattern — current depends on previous — is identical to Pascal's row construction.

easyAsked at Snowflake

Generate the first numRows of Pascal's triangle. Snowflake uses this to test row-at-a-time DP building — the same shape as windowing operations over a column where the next row depends on the previous one.

By Alex Chen, Founder, InterviewChamp.AI · Last verified 2026-05-20

Source citations

Public interview reports confirming this problem appears in Snowflake loops.

Glassdoor (2025-Q4)— Snowflake intern phone screen used this as a DP warm-up.
LeetCode Discuss (2025-08)— Reported at Snowflake new-grad screens.

Problem

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.

Constraints

1 <= numRows <= 30

Examples

Example 1

Input

numRows = 5

Output

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

Example 2

Input

numRows = 1

Output

[[1]]

Approaches

1. Compute via factorials

Each entry = C(n, k) computed with factorials.

Time: O(n^2)
Space: O(n^2)

function generate(numRows) {
  function fact(n) { let r = 1n; for (let i = 2n; i <= BigInt(n); i++) r *= i; return r; }
  const triangle = [];
  for (let n = 0; n < numRows; n++) {
    const row = [];
    for (let k = 0; k <= n; k++) {
      row.push(Number(fact(n) / (fact(k) * fact(n - k))));
    }
    triangle.push(row);
  }
  return triangle;
}

Tradeoff: Recomputes factorials, expensive. Don't ship.

2. Build row from previous row (optimal)

Each row begins and ends with 1. Interior entries are sum of adjacent entries in the previous row.

Time: O(numRows^2)
Space: O(numRows^2)

function generate(numRows) {
  const triangle = [];
  for (let i = 0; i < numRows; i++) {
    const row = new Array(i + 1).fill(1);
    for (let j = 1; j < i; j++) {
      row[j] = triangle[i - 1][j - 1] + triangle[i - 1][j];
    }
    triangle.push(row);
  }
  return triangle;
}

Tradeoff: Linear in output size. Each row depends only on the previous — classic 'window of size 1' DP.

Snowflake-specific tips

Snowflake interviewers want the row-at-a-time DP framing, not the factorial formula. Bonus signal: connect to window functions in SQL — Pascal's triangle is the canonical 'each row's output depends on the previous row's output', which is how Snowflake implements LAG/LEAD windowing.

Common mistakes

Computing each cell with factorials, which is O(n^2 * n) = O(n^3).
Off-by-one in the row length (row i has i+1 elements, not i).
Modifying the previous row in place — corrupts later reads.

Follow-up questions

An interviewer at Snowflake may pivot to one of these next:

Pascal's Triangle II (LC 119) — return only the kth row in O(k) space.
Count paths in a grid (related to C(n, k)).
How does Snowflake compute LAG over a partition?

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Output

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FAQ

Why is each row's first and last entry 1?

C(n, 0) and C(n, n) both equal 1 — there's exactly one way to choose 0 items or all items.

How does this relate to window functions?

Window functions like LAG(x) OVER (ORDER BY ts) produce a value for each row based on previous rows. The recurrence pattern — current depends on previous — is identical to Pascal's row construction.

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