Pascal's Triangle II




Problem Classification
Mathematical Problem: This problem requires knowledge about Pascal’s Triangle, which is a mathematical concept used in algebra and combinatorics.
Array/List Manipulation: The task involves generating a specific row in Pascal’s Triangle as a list, which involves operations on lists such as accessing elements and modifying them.
Recursion: The problem requires an understanding of recursive functions, which is a method where the solution to a problem depends on solutions to smaller instances of the same problem.
Dynamic Programming: Similar to the previous problem about generating Pascal’s triangle, this problem can be also considered a dynamic programming problem because the value of each element in the row depends on previously computed values (the row above it).
Language Agnostic Coding Drills
Understanding Pascal’s Triangle: This problem involves generating a specific row of Pascal’s triangle. The basic understanding of how Pascal’s triangle is structured and its properties is essential for this problem.
List Manipulation: The problem involves manipulating lists, adding elements to them and accessing their elements.
Recursion: This solution uses recursion. Understanding the basic concept of recursion is needed to solve this problem.
Targeted Drills in Python
 Understanding Pascal’s Triangle:
This drill involves explaining and understanding the properties of Pascal’s Triangle. For example, each row of the triangle starts and ends with 1, and each other number in the row is the sum of the two numbers directly above it.
 List Manipulation:


 Recursion:


The final integrated drill would be to write a function that generates a specific row of Pascal’s triangle, which is essentially the solution to this problem.
Here is one way to implement getting a Pascal’s triangle row in Python:


The key steps are:
Initialize
row
to [1]Iterate up to the target
rowIndex
Append next value based on previous value and index
Return final
row
This dynamically builds up each row, using the previous value and current index to calculate the next Pascal’s triangle value.
We leverage the mathematical relationship in Pascal’s triangle to populate each row efficiently without having to store the entire triangle.
Dynamic Programming
Stage 1: Recursive Formulation
Here is a precise English description of the problem:
Given an integer variable rowIndex, the objective is to return the rowIndex’th row of Pascal’s triangle as a list of integers.
Pascal’s triangle is a triangular array of integers such that each value in the triangle is the sum of the two values directly above it.
Specifically, the 0th row of Pascal’s triangle contains the single value 1. The 1st row contains the values 1 and 1. Subsequent rows are generated by taking the previous row, prepend a 1, append a 1, and fill the remaining values by summing the two values directly above from the previous row.
For example, if rowIndex is 3, then the 3rd row [1,3,3,1] should be returned, since this is the 4th row of Pascal’s triangle. If rowIndex is 0, then [1] should be returned representing just the first value.
The key constraints are:
 rowIndex will be an integer in the range 0 to 33 inclusive, representing the target row.
 The return value should be a list of integers representing the specified row of Pascal’s triangle.
Here are some of the key subproblems that contribute to solving this overall problem of generating a Pascal’s triangle row:
Calculating an individual element in the row
 Requires summing the two elements directly above it from the previous row
Generating the previous row
 Needed to compute the current row values by summing above elements
Prepending a 1 to the row
 Every Pascal’s triangle row starts with a 1
Appending a 1 to the row
 Every Pascal’s triangle row ends with a 1
Initializing the 0th and 1st rows
 Base cases to build up from
Iterating through row indexes
 Generate each row value incrementally
By finding solutions for the subproblems of calculating individual elements, constructing the previous row, prepending/appending 1’s, handling base cases, and iterating indexes, we can build up a solution for generating any Pascal’s triangle row.
Here is one way to write a recursive formula expressing the solution to generating a Pascal’s triangle row in terms of smaller subproblems:
Let’s define:
p(r, c) = the element at row r and column c of Pascal’s triangle
row(n) = the full nth row of Pascal’s triangle
Then the recursive relationships are:
Base cases:
row(0) = [1] row(1) = [1, 1]
Recursive formula:
p(r, c) = p(r1, c1) + p(r1, c) for c = 1 to r1
row(n) = [1] + [p(n, c) for c = 1 to n1] + [1]
This expresses each element p(r,c) in terms of the two elements directly above it from row r1.
The full row(n) concatenates the beginning 1, the recursively computed p(r,c) values, and ending 1.
So we build up each row by leveraging solutions to subproblems of computing individual elements from the previous row.
Specification
The problem of generating a Pascal’s triangle row can be specified in terms of input and output as:
Input:
 rowIndex  An integer representing the index of the desired Pascal’s triangle row, ranging from 0 to 33
Output:
 A list of integers representing the Pascal’s triangle row at the specified rowIndex.
More formally:
Given:
rowIndex  An integer between 0 and 33 inclusive
Compute:
row  A list of integers such that:
row[0] = 1
row[len(row)1] = 1
row[c] = row[c1] + row[c] for c in range(1, len(row)1)
Where row represents the rowIndex’th row of Pascal’s triangle based on the recursive definition.
Return:
row
So in summary, the input is an integer row index, and the output is the corresponding row of Pascal’s triangle expressed as a list of integers.
The boundary conditions or base cases for the recursive formulation of generating a Pascal’s triangle row are:
 row(0) = [1]
 row(1) = [1, 1]
Where row(n) represents the nth row of Pascal’s triangle.
These base cases establish that:
The 0th row just contains the single element 1.
The 1st row contains two 1s.
All other rows are built up from these initial rows using the recursive formula.
The base cases are necessary because:
The 0th row provides an initialization point to start the recursion.
The 1st row provides the elements needed to generate the 2nd row.
Rows 2 onwards depend on the structure of the rows before them.
So the base cases rows(0) and rows(1) anchor the recurrence relation and allow recursively generating any higher numbered row in Pascal’s triangle based on the previous rows.
Solution
The full recursive formula for generating the nth row of Pascal’s triangle is:
p(r, c) = element at row r, column c
Base cases:
row(0) = [1]
row(1) = [1, 1]
Recursive formula:
p(r, c) = p(r1, c1) + p(r1, c) for c = 1 to r1
row(n) = [1] + [p(n, c) for c = 1 to n1] + [1]
Overall solution:
row(rowIndex)
Where:
The base cases define the first two rows
The recursive formula generates element p(r,c) based on the two elements directly above it
Concatenating the beginning 1, recursively computed p(r,c) values, and ending 1 builds row(n)
The final solution is the row() corresponding to the input index
So by leveraging the predefined base cases and recursive relationships, we can build up any desired row of Pascal’s triangle.
Yes, the recursive formula for generating a Pascal’s triangle row can be broken down into these simpler subproblems:
The full recurrence is:
p(r, c) = p(r1, c1) + p(r1, c)
This can be decomposed into two simpler subproblems:
 Get left element from previous row:
left(r, c) = p(r1, c1)
 Get right element from previous row:
right(r, c) = p(r1, c)
 Compute sum:
p(r, c) = left(r, c) + right(r, c)
The overall solution combines these:
row(n) = [1] + [p(r, c) for c in 1..n1] + [1]
Where each p(r, c) is computed by summing the corresponding left and right elements from the previous row.
By breaking down into subproblems of getting left/right elements and computing their sum, we simplify the recurrence into more modular pieces.
Stage 2: Bottomup Solution
Based on the recursive formula, here are all the possible subproblems that would be called by this algorithm to generate a Pascal’s triangle row:
p(0, 0)  The single element in row 0
 Base case
p(1, 0) and p(1, 1)  The first two elements in row 1
 Base cases
p(2, 0) through p(2, 1)  Elements in 2nd row
 Depend on p(1, 0) and p(1, 1)
p(3, 0) through p(3, 2)  Elements in 3rd row
 Depend on p(2, 0) through p(2, 1)
…
p(r, 0) through p(r, r1)  Elements in row r
 Depend on rows 0 through r1
…
p(n, 0) through p(n, n1)  Elements in target nth row
 Depend on rows 0 through n1
So in summary, the subproblems are the elements p(r, c) from row 0 through row n1 that collectively build up the target nth row, with base cases p(0,0) and p(1,0) to p(1,1).
For memoizing and storing the results of the subproblems for Pascal’s triangle, the most suitable data structure would be a 2D array (nested lists in Python).
We can create a 2D array dp where dp[r][c] stores the computed value for element p(r, c) in the triangle.
A 2D array is a natural fit because:
The subproblem values have a clear 2D matrix structure matching the layout of Pascal’s triangle.
We can directly map the row and column indices to array indices for O(1) access.
Iterating through the rows and columns is straightforward with nested loops.
Arrays provide fast lookup and storage for sequential tablebased data.
A dictionary could also work, but a 2D array directly mirrors the tabular structure of Pascal’s triangle without unnecessary mappings.
So in summary, a 2D array indexed by [row][col] is an ideal memoization structure for this problem, providing fast lookup and storage aligned to the triangular data layout.
Identify Dependencies
Here is a dependency graph showing how the subproblems in generating a Pascal’s triangle row rely on each other:
p(0,0)

p(1,0) p(1,1)
\ /
p(2,0)

p(2,1)
\ /
p(3,0)

p(3,1)
\ /
p(3,2)
Where:
 p(r, c) represents the element at row r, column c
 An arrow from p(r, c) to p(s, t) indicates p(s, t) depends directly on p(r, c).
Key observations:
Each element p(r, c) depends directly on the two elements p(r1, c1) and p(r1, c) above it.
The base cases p(0,0), p(1,0) and p(1,1) have no dependencies.
All other elements form a dependency chain leading back to the base cases.
This dependency graph helps visualize the interdependencies and build order requirements of the elements to generate any Pascal’s triangle row.