The Pascal Row

An easy Python interview practice problem on DataDriven. Write the Python and run it against test cases, with instant feedback.

Domain
Python
Difficulty
easy
Seniority
L3
Asked in
Citadel

Problem

An analyst only ever asks for one row of Pascal's triangle at a time: what are the entries of row n (0-indexed), computed directly from the binomial coefficients rather than by building every row above it?

Each number is the sum of two numbers above it.

Example 1

Input:

n = 3

Output:

[1,3,3,1]

Example 2

Input:

n = 0

Output:

[1]

Worked solution and explanation

What this really is

This is a binomial-coefficient recurrence dressed up as a triangle puzzle. Everyone knows the additive build where each entry is the sum of two above it, and that is exactly what the prompt rules out: it materializes every prior row. The skill is carrying C(n, k) forward from C(n, k-1) with one multiply and one divide. The trap is the order of those two operations: divide first and the entries silently truncate to wrong integers.

Walking to the answer

Step 1: Seed the row with C(n, 0)

Every row starts at 1, so row = [1] is both the seed and the full answer for n = 0. No special case needed.

Step 2: Derive each term from the previous one

C(n, k) = C(n, k-1) * (n - k + 1) / k. You already hold C(n, k-1) in row[k - 1], so each new entry costs constant work and no factorials.

Step 3: Multiply first, then floor-divide

The product row[k - 1] * (n - k + 1) is always divisible by k, but row[k - 1] alone usually is not. For n = 5, k = 2: 5 // 2 * 4 gives 8, while 5 * 4 // 2 gives the correct 10.

The solution

One row, one pass
def pascal_row(n):
    row = [1]
    for k in range(1, n + 1):
        val = row[k - 1] * (n - k + 1) // k
        row.append(val)
    return row

Common follow-up questions