Pascal's Triangle

by atalaykutlay

Solution to "Pascal's Triangle" on Leetcode.

Pascal's Triangle

1class Solution:
2    def generate(self, numRows: int) -> List[List[int]]:
3    
4        results = []
5
6        for row in range(1,numRows+1):
7            if row <= 2:
8                result = [1] * row
9                results.append(result)
10                continue
11            
12            result = results[-1] + [1]
13            new_row = result[:]
14            for i in range(1, row-1):
15                new_row[i] = result[i-1] + result[i]
16            
17            results.append(new_row)
18        
19        return results

✨ Code Explanation ✨

This code is a solution to the "Pascal's Triangle" problem. The function "generate" takes an integer input "numRows" and returns a 2D list representing Pascal's Triangle with "numRows" number of rows. The code starts by initializing an empty list called "results" to store each row of the triangle. It then starts a for loop that iterates "row" from 1 to "numRows". Inside the loop, there are three main scenarios: when "row" is 1 or 2, and when "row" is greater than 2. When "row" is 1 or 2, it creates a new list called "result" with all elements set to 1 using [1] * row. It then appends this "result" to the "results" list. When "row" is greater than 2, it retrieves the last row of "results" and adds an additional 1 at the end to form the new row. Then, it creates a copy of this new row called "new_row" to keep track of the current row being constructed. It then uses a nested for loop to iterate from index 1 to index row-1 of the "result" list. In each iteration, it assigns the sum of the previous row elements at index i-1 and i to the corresponding element of "new_row". After constructing "new_row", it appends it to the "results" list. Finally, it returns the "results" list representing Pascal's Triangle with "numRows" number of rows.