Coding problem · Easy · Basics

Pascal's Triangle Row

Easy Algorithms & data structures

Return row n (starting from 0) of Pascal's triangle. Each row starts and ends with 1, and every other value is the sum of the two values above it.

row 0:     1
row 1:    1 1
row 2:   1 2 1
row 3:  1 3 3 1

Example 1

Input: n = 3
Output: [1,3,3,1]

Example 2

Input: n = 0
Output: [1]

Hints

Hint 1

Build each row from the previous one.

Hint 2

New row: 1, then prev[i-1] + prev[i] for the middle, then 1.

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Solution and explanation

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Time O(n²) · Space O(n)

Start from [1] and build n times: each new row is 1, the pairwise sums of the previous row, then 1.

Reference solution

function pascalRow(n) {
  let row = [1];
  for (let i = 0; i < n; i++) {
    const next = [1];
    for (let j = 1; j < row.length; j++) next.push(row[j - 1] + row[j]);
    next.push(1);
    row = next;
  }
  return row;
}

function pascalRow(n: number): number[] {
  let row = [1];
  for (let i = 0; i < n; i++) {
    const next = [1];
    for (let j = 1; j < row.length; j++) next.push(row[j - 1] + row[j]);
    next.push(1);
    row = next;
  }
  return row;
}

def pascal_row(n: int) -> list[int]:
    row = [1]
    for _ in range(n):
        row = [1] + [row[j - 1] + row[j] for j in range(1, len(row))] + [1]
    return row

<?php

function pascalRow(int $n): array
{
    $row = [1];
    for ($i = 0; $i < $n; $i++) {
        $next = [1];
        for ($j = 1; $j < count($row); $j++) {
            $next[] = $row[$j - 1] + $row[$j];
        }
        $next[] = 1;
        $row = $next;
    }
    return $row;
}

class Solution {
    public int[] pascalRow(int n) {
        int[] row = new int[n + 1];
        row[0] = 1;
        for (int i = 1; i <= n; i++) for (int j = i; j > 0; j--) row[j] += row[j - 1];
        return row;
    }
}

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