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# Pascal's Triangle II

Simulation Array

## Problem

Given an integer `rowIndex`, return the `rowIndexth` row of the Pascal's triangle.

Notice that the row index starts from **0**.

{% hint style="info" %}
For example:

```
Input: rowIndex = 3
Output: [1,3,3,1]
```

{% endhint %}

### Thought Process

* All we have to do is generate all rows up to the row that is given. We can do this by applying the algorithm we used in *Pascal Triangle I*

## Solution

```
class Solution:
    def getRow(self, rowIndex: int) -> List[int]:
        
        if rowIndex < 0:
            return []
        
        matrix = [[1]]
        
        for i in range(1,rowIndex+1):
            row = [1]
            for j in range(1,i):
                row.append(matrix[i-1][j-1] + matrix[i-1][j])
            row.append(1)
            matrix.append(row)
        return matrix[rowIndex]
```

## Key Facts

* Where this problem is different from *Pascal's Triangle I* is in the **input**. The input here is indicating which row to generate, and this is **0 based** i.e. k=0 corresponds to row 1.
* Since we are generating the matrix up until the row that we want, we need the first for-loop's range to go to *rowIndex + 1* since this is 0 based.

## Time Complexity

* **Time:**$$O(m\*n)$$because we have to generate every row and column up until rowIndex
* **Space:**$$O(m\*n)$$ because we a matrix of rows and columns.