15. Pascal's Triangle

easyAsked at Indeed

Generate the first numRows rows of Pascal's triangle — tests iterative 2D array construction that mirrors Indeed's incremental ranking coefficient tables.

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

Problem

Given an integer numRows, return the first numRows of Pascal's triangle. Each row is built so that 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 from scratch using binomial coefficients.

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

function generate(numRows) {
  const res = [];
  for (let i = 0; i < numRows; i++) {
    const row = [1];
    for (let j = 1; j < i; j++) row.push(res[i-1][j-1] + res[i-1][j]);
    if (i > 0) row.push(1);
    res.push(row);
  }
  return res;
}

Tradeoff:

2. Previous-row reference

Each row starts and ends with 1; interior values derive from the previous row, giving clean O(n^2) time with no unnecessary recomputation.

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

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

Tradeoff:

Indeed-specific tips

Indeed appreciates clean off-by-one handling on the first and last element of each row — state the boundary invariant aloud before coding.

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Output

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