118. Pascal's Triangle

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YouTube Video https://youtu.be/hfN4VONP4HQ [A1] - Brute Force Time: O(n^2) Space: O(1) class Solution: def twoSum(self, nums: List[int], target: int) -> List[int]: for i in range(len(nums)): for j in range(i+1, len(nums)): ...

LeetCode's Valid Anagram problem checks if two strings, `s` and `t`, are anagrams by comparing character frequencies.

LeetCode's Contains Duplicate problem checks if there are any duplicate elements in a given array, often solved using hash sets or sorting.

Time: O(n^2)
Space: O(n^2)
class Solution:
def generate(self, numRows: int) -> List[List[int]]:
# Initialize the triangle with the first row
pascal_triangle = []
for i in range(numRows):
# Start each row with a list
row = [1] * (i + 1) # Every row starts with 1s
# Fill in the interior values
for j in range(1, i):
row[j] = pascal_triangle[i - 1][j - 1] + pascal_triangle[i - 1][j]
# Append the completed row to the triangle
pascal_triangle.append(row)
return pascal_triangle
Solution class: It defines a class that co ntains a method for generating Pascal’s Triangle.
generate method: This method takes an integer numRows as input, representing the number of rows in Pascal’s Triangle to generate.
Initializes an empty list called pascal_triangle that will hold all rows of Pascal’s Triangle.
Uses a loop to iterate over the range of numRows to generate each row of the triangle.
Each row is initialized with 1s using [1] * (i + 1).
For rows beyond the second, the interior values (non-1s) are calculated using values from the previous row: pascal_triangle[i - 1][j - 1] + pascal_triangle[i - 1][j].
After completing a row, it appends the row to the pascal_triangle list.
Finally, the pascal_triangle is returned.