15. Pascal's Triangle

easyAsked at CircleCI

Generate Pascal's triangle up to numRows, testing your ability to build iterative tabular structures common in test-result matrix generation at CircleCI.

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

Problem

Given an integer numRows, return the first numRows of Pascal's triangle, where 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. Brute force

Build each row by summing adjacent elements from the previous row.

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

function generate(numRows) {
  const res = [[1]];
  for (let i = 1; i < numRows; i++) {
    const prev = res[i - 1];
    const row = [1];
    for (let j = 1; j < prev.length; j++) row.push(prev[j-1] + prev[j]);
    row.push(1);
    res.push(row);
  }
  return res;
}

Tradeoff:

2. In-place row construction

Same iterative approach but emphasizes clarity of boundary conditions, demonstrating clean code under interview pressure.

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

function generate(numRows) {
  const triangle = [];
  for (let i = 0; i < numRows; i++) {
    const row = new Array(i + 2).fill(1);
    for (let j = 1; j < i; j++) {
      row[j] = triangle[i-1][j-1] + triangle[i-1][j];
    }
    row.pop();
    triangle.push(row.slice(0, i + 1));
  }
  return triangle;
}

Tradeoff:

CircleCI-specific tips

CircleCI may ask this as a warmup before pivoting to 2D DP — be prepared to extend to Pascal's triangle row access in O(k) space.

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Output

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