Introduction

If you only need one row of Pascal's triangle, generating every row above it is wasted work. The efficient solution is to compute each value from the previous one using the binomial relationship between neighboring coefficients, which gives the whole row in linear time.

Use the Relationship Between Neighboring Coefficients

The values in row n are the binomial coefficients C(n, 0) through C(n, n). A naive approach uses factorials for every position, but that repeats a lot of expensive arithmetic.

The better recurrence is:

C(n, k) = C(n, k - 1) x (n - k + 1) / k

That means once you know one coefficient, you can derive the next one directly.

Start with:

• 'C(n, 0) = 1'

Then compute each next value from the one before it.

Build the Row Iteratively

Here is a direct Python implementation:

1def pascal_row(n: int) -> list[int]:
2    row = [1]
3    current = 1
4
5    for k in range(1, n + 1):
6        current = current * (n - k + 1) // k
7        row.append(current)
8
9    return row
10
11print(pascal_row(0))
12print(pascal_row(1))
13print(pascal_row(5))

Output:

[1]
[1, 1]
[1, 5, 10, 10, 5, 1]

This method computes exactly the row you need, and nothing more.

Why This Is Better Than Building the Full Triangle

If you generate all rows from 0 through n, the total work is O(n²). That is fine when you truly need the whole structure, but not when the problem asks for only one row.

The iterative row-only method:

• runs in O(n) time
• uses only constant extra space beyond the output list
• avoids repeated factorial computation

It is also numerically clean because each division is exact when done in the recurrence order shown above.

You Can Exploit Symmetry Too

Pascal rows are symmetric. For example, row 6 is:

1, 6, 15, 20, 15, 6, 1

So if you want, you can compute only half the row and mirror it. That can save a little work in performance-sensitive code, although the simple linear recurrence is already fast enough for most uses.

1def pascal_row_symmetric(n: int) -> list[int]:
2    half = []
3    current = 1
4
5    for k in range(0, n // 2 + 1):
6        if k == 0:
7            current = 1
8        else:
9            current = current * (n - k + 1) // k
10        half.append(current)
11
12    if n % 2 == 0:
13        return half + half[-2::-1]
14    return half + half[::-1]
15
16print(pascal_row_symmetric(6))

The symmetry optimization is useful, but it is optional. The biggest efficiency gain already comes from avoiding full-triangle generation.

Think in Terms of Combinations

The row is not just a triangle pattern. It is a sequence of combinations. That is why the recurrence works so well. Each step moves from one combination count to the next without recomputing the entire combinatorial formula from scratch.

That interpretation also makes it easier to explain why the numbers stay integral even though there is a division in the recurrence.

Common Pitfalls

• Building the full triangle when the problem asks for only one row.
• Recomputing factorials for every position, which creates unnecessary work and very large intermediate values.
• Using floating-point division instead of exact integer arithmetic.
• Forgetting that many code examples index rows from zero rather than one.
• Overcomplicating the problem with symmetry before implementing the simple linear recurrence correctly.

Summary

• The efficient way to compute one Pascal row is to derive each coefficient from the previous one.
• This gives an O(n) solution instead of O(n²) full-triangle generation.
• Integer arithmetic keeps the coefficients exact.
• Symmetry can reduce work further, but the main win is using the neighbor recurrence.
• Think of the row as binomial coefficients, not only as a printed triangle pattern.