Date: 2026-06-06
Time: 18:29
This file implements LeetCode 118 — Pascal's Triangle. It owns exactly one responsibility: given a row count numRows, produce the first numRows rows of Pascal's triangle as a list of lists. It's a standalone solution module following the repo's convention of one problem per directory with solution.py, test_solution.py, plan.md, and review.md.
generate(numRows: int) -> list[list[int]]The sole public function. Contract:
numRows in range [1, 30] (per LeetCode constraints — not validated in code).numRows lists, where row i (0-indexed) has i + 1 elements representing the i-th row of Pascal's triangle.Iterative bottom-up construction. Rather than using recursion or a math formula (C(n, k)), the solution builds each row from the previous one. This is the classic DP-style approach for Pascal's triangle:
1. Initialize each row as all 1s (correct for the boundary elements).
2. Overwrite the interior elements by summing adjacent elements from the prior row.
This avoids factorial computation and keeps the logic O(numRows^2) in both time and space — which matches the output size, so it's optimal.
Pre-fill then patch. The idiom row = [1] * (i + 1) sets both boundary values (first and last) to 1 upfront. The inner loop for j in range(1, i) only touches interior positions. When i < 2, range(1, i) is empty, so rows 0 and 1 ([1] and [1, 1]) are produced correctly without special-casing.
Imports: None. The function uses only built-in Python types and operations.
Imported by: The corresponding pascals-triangle/test_solution.py imports generate for testing. The massive "Imported By" list in the prompt is misleading — those are unrelated test files that happen to share a common test harness pattern, not actual importers of this module's generate function.
generate(5)
i=0: row = [1] → triangle = [[1]]
i=1: row = [1, 1] (inner loop empty) → triangle = [[1], [1,1]]
i=2: row = [1, 1, 1] → row[1] = 1+1=2 → triangle = [[1], [1,1], [1,2,1]]
i=3: row = [1,1,1,1] → row[1]=1+2=3, row[2]=2+1=3
→ [..., [1,3,3,1]]
i=4: row = [1,1,1,1,1] → row[1]=1+3=4, row[2]=3+3=6, row[3]=3+1=4
→ [..., [1,4,6,4,1]]
Each row reads from triangle[i-1] (the already-appended previous row) at positions j-1 and j, producing the standard Pascal's identity: C(n,k) = C(n-1,k-1) + C(n-1,k).
i has exactly i + 1 elements.row[0] and row[i] are always 1 (boundary condition, enforced by pre-fill).j in range(1, i) guarantees j-1 >= 0 and j <= i-1, so triangle[i-1][j-1] and triangle[i-1][j] are always valid lookups — no bounds checking needed.triangle grows by exactly one row per iteration, so triangle[i-1] is always the most recently appended row.None. The function trusts its caller to pass a valid numRows >= 1. Passing 0 returns [] (the loop doesn't execute). Passing a negative value also returns []. No exceptions are raised or caught.
pascals-triangle-ii/solution.py — The follow-up problem that returns only the k-th row, likely using O(k) space instead of O(n^2)pascals-triangle/test_solution.py — See which edge cases and row counts are testedpascals-triangle/plan.md — The planning document that preceded this implementationiterative-vs-combinatorial-pascals — Compare this DP approach to direct binomial coefficient computation C(n,k) = n! / (k!(n-k)!) for trade-offs in precision and speedclimbing-stairs/solution.py:climbStairs — Another bottom-up DP solution in the repo that follows the same "build from previous" patternpascal-generate-pure — generate is a pure function with no side effects, no imports, and no mutation of external statepascal-boundary-by-prefill — Boundary values (first and last element of each row = 1) are set by pre-filling with [1] * (i + 1), not by conditional logicpascal-inner-loop-safe — The inner loop range(1, i) guarantees all triangle[i-1] lookups are in-bounds without explicit bounds checkingpascal-zero-rows-returns-empty — Calling generate(0) returns [] since the outer loop range is empty, even though this is outside the stated LeetCode constraints