119. Pascal’s Triangle II §

Date: 2026-03-16 Difficulty: Easy Topics: Array, Dynamic Programming, Recursion Link: https://leetcode.com/problems/pascals-triangle-ii/

Final Solution §

class Solution:
    def getRow(self, rowIndex: int) -> List[int]:
        if rowIndex == 0:
            return [1]
        prevRow = self.getRow(rowIndex - 1)
        newRow = [1] * (rowIndex + 1)
        for i in range(1, rowIndex):
            newRow[i] = prevRow[i - 1] + prevRow[i]
        return newRow

Initial Issues §

• First attempt stored all rows (list of lists) — unnecessary since we only need the final row
• Treated return type inconsistently (flat list in base case, list of lists in recursive case)
• return tree[rowIndex - 1] worked by accident due to negative indexing cancellation
• Included unnecessary rowIndex == 1 base case — rowIndex == 0 alone is sufficient since range(1, 1) is empty

Key Learnings §

• Only need to track previous row to build the next — O(rowIndex) space optimization
• Recursive function returns a single row, not all rows — each frame builds next row from what it receives
• Fewer base cases = cleaner code. Always check if an extra base case is actually needed by tracing the next value through.

Nuances to Remember §

• range(1, 1) is empty — so row 1 [1, 1] builds correctly from row 0 [1] without special-casing
• Space: O(rowIndex) for the row + O(rowIndex) for call stack
• Time: O(rowIndex^2) — building each row takes O(row length)

119. Pascal’s Triangle II §

Date: 2026-03-17 Difficulty: Easy Topics: Array, Dynamic Programming, Recursion Link: https://leetcode.com/problems/pascals-triangle-ii/

Final Solution §

class Solution:
    def getRow(self, rowIndex: int) -> List[int]:
        if rowIndex == 0:
            return [1]
        prevRow = self.getRow(rowIndex - 1)
        newRow = [1] * (rowIndex + 1)
        for i in range(1, rowIndex):
            newRow[i] = prevRow[i - 1] + prevRow[i]
        return newRow

Initial Issues §

• First attempt used a 2D array and helper function with buildPascal returning all rows — unnecessary since we only need the final row
• Had type mismatch: base cases returned flat lists but recursive case treated return value as list of lists
• Included unnecessary rowIndex == 1 base case — rowIndex == 0 is sufficient since range(1, 1) is empty

Key Learnings §

• Optimized from storing all rows to only keeping previous row — O(rowIndex) space
• Each frame only needs the previous row to build the next, so return a single list instead of list of lists
• rowIndex == 0 returning [1] is the only base case needed — when rowIndex == 1, the loop range(1, 1) is empty so newRow stays as [1, 1]

Nuances to Remember §

• Row i has i + 1 elements
• Middle elements: prevRow[i-1] + prevRow[i]
• Inner loop range is range(1, rowIndex) — skips first and last (always 1)
• Space: O(rowIndex) for the row + O(rowIndex) for call stack