# Leetcode – Pascal’s Triangle II in Python

## Introduction

Pascal’s Triangle is an intriguing mathematical concept that’s been studied for centuries. It holds numerous applications, from solving problems in probability to understanding algebraic expansions. In LeetCode, the “Pascal’s Triangle II” problem presents an interesting twist on generating Pascal’s Triangle – instead of generating the entire triangle, you are asked to return a specific row without actually generating the whole triangle. This article takes you on a deep dive into the problem, its mathematical underpinnings, and a variety of solutions in Python.

## Understanding the Problem

In the “Pascal’s Triangle II” problem, we are given an index rowIndex and we need to return the rowIndex-th row of the Pascal’s Triangle. The rowIndex is 0-indexed, meaning the first row is 0, the second row is 1, and so on.

For example, given rowIndex = 3, the function should return [1, 3, 3, 1] as it is the fourth row of Pascal’s Triangle.

## Mathematical Background

Pascal’s Triangle has several fascinating mathematical properties. Each row represents the coefficients in the binomial expansion of (a + b)^n. Additionally, the element at the k-th position of the n-th row (0-indexed) can be calculated directly using the binomial coefficient formula, also known as “n choose k”:

Value = n! / (k! * (n-k)!)

## Approach 1: Binomial Coefficients with Optimization

An optimized way to calculate the binomial coefficient for each element of the rowIndex-th row is to use the following relation:

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

This allows us to calculate each element in the row iteratively, without having to calculate factorials, which can be expensive.

Here is the Python code for this approach:

def getRow(rowIndex):
row = [1]
value = 1
for k in range(1, rowIndex + 1):
value = value * (rowIndex - k + 1) // k
row.append(value)
return row

This approach has a time complexity of O(N), where N is the rowIndex.

## Approach 2: Dynamic Programming

Another approach to solving this problem is using dynamic programming. We can keep a single array and calculate the next row based on the current row, in-place.

Here is the Python code for the dynamic programming approach:

def getRow(rowIndex):
row = [1]

for i in range(1, rowIndex + 1):
newRow = [1]
for j in range(1, i):
newRow.append(row[j-1] + row[j])
newRow.append(1)
row = newRow

return row

This approach also has a time complexity of O(N^2), but the constant factor might be slightly higher than the binomial coefficient approach.

## Approach 3: Using the Inverse of Modular Multiplicative

We can optimize the binomial coefficient approach further by calculating the inverse of the modular multiplicative, which can be more efficient than division.

def getRow(rowIndex):
row = [1]
value = 1
for k in range(1, rowIndex + 1):
value = value * pow(k, -1, rowIndex + 1) * (rowIndex - k + 1)
row.append(value)
return row

## Key Takeaways

1. Understanding the mathematical properties of Pascal’s Triangle is crucial for solving this problem efficiently.
2. Dynamic programming can often be an intuitive way to approach problems that have overlapping subproblems.
3. Optimizations, like using the inverse of modular multiplicative, can sometimes be used to improve efficiency in certain algorithms.

## Conclusion

The Pascalâ€™s Triangle II problem is an excellent example of how mathematical insights can greatly simplify a problem. It not only tests your knowledge of Pascal’s Triangle but also challenges you to think about optimization and efficiency. In this article, we discussed three approaches, including an optimized binomial coefficient approach, dynamic programming, and an approach using the inverse of modular multiplicative. Each approach has its own merits and understanding when to use which approach is key to becoming proficient in problem-solving.