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Pascal's Triangle II

Easy

Amazon

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Topics:

Arrays

Given an integer rowIndex, return the rowIndex^th (0-indexed) row of the Pascal's triangle.

In Pascal's triangle, each number is the sum of the two numbers directly above it as shown:

Example 1:

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

Example 2:

Input: rowIndex = 0
Output: [1]

Example 3:

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

Constraints:

0 <= rowIndex <= 33

Follow up: Could you optimize your algorithm to use only O(rowIndex) extra space?

Pascal's Triangle II

Easy

Amazon

1 view

Topics:

Arrays

Given an integer rowIndex, return the rowIndex^th (0-indexed) row of the Pascal's triangle.

In Pascal's triangle, each number is the sum of the two numbers directly above it as shown:

Example 1:

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

Example 2:

Input: rowIndex = 0
Output: [1]

Example 3:

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

Constraints:

0 <= rowIndex <= 33

Follow up: Could you optimize your algorithm to use only O(rowIndex) extra space?

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

Is the `rowIndex` 0-indexed or 1-indexed?
What is the maximum possible value for `rowIndex`? Is there an upper bound, and should I be concerned about integer overflow?
Should I return an empty list if `rowIndex` is negative?
Can I assume `rowIndex` will always be a valid integer?
Can I use extra space, or should I try to optimize for space complexity as much as possible?

Brute Force Solution

Approach

The brute force approach to Pascal's Triangle II focuses on building the entire triangle row by row. We start from the top and compute each element based on the elements in the previous row, ultimately arriving at the desired row. We recalculate every value even if we've computed it before.

Here's how the algorithm would work step-by-step:

Start by creating the first row of the triangle, which is just the number 1.
Then, to build the next row, look at the row above it.
For each number in the new row, calculate its value by adding the two numbers directly above it in the previous row.
If a number is at the edge of the row, just consider the missing number as zero.
Repeat this process, generating each row one at a time, until you reach the row that you're looking for.
Once you've generated the row you wanted, that's your answer.

Code Implementation

def get_pascal_row_brute_force(row_index):
    pascal_triangle = []

    # The first row is always [1]
    first_row = [1]
    pascal_triangle.append(first_row)

    for current_row_index in range(1, row_index + 1):
        previous_row = pascal_triangle[current_row_index - 1]
        current_row = []

        # First element of each row is always 1
        current_row.append(1)

        # Calculate intermediate elements by summing the two elements above
        for element_index in range(1, current_row_index):

            # Sum the two numbers above in previous row
            number_from_previous_row = previous_row[element_index - 1] + previous_row[element_index]
            current_row.append(number_from_previous_row)

        # Last element of each row is always 1
        current_row.append(1)
        pascal_triangle.append(current_row)

    # Return the requested row of Pascal's Triangle
    return pascal_triangle[row_index]

Big(O) Analysis

Time Complexity

O(n²) – The algorithm constructs Pascal's triangle row by row up to the target row 'n'. Generating each row requires iterating through its elements. The first row takes constant time, the second row takes time proportional to 2, the third row takes time proportional to 3, and so on, until the nth row takes time proportional to n. Therefore, the total time complexity is the sum of the integers from 1 to n, which is n*(n+1)/2. This simplifies to O(n²).

Space Complexity

O(N) – The brute force approach outlined builds Pascal's Triangle row by row until the target row (rowIndex) is reached. Each row generated is stored in memory. Since we build all rows from 0 up to rowIndex, we need to store at most two rows at a time: the previous row and the current row being built, each containing up to N elements where N is rowIndex + 1. Thus, the auxiliary space grows linearly with N, giving a space complexity of O(N).

Optimal Solution

Approach

To efficiently generate a specific row of Pascal's Triangle, we avoid calculating all the previous rows. Instead, we build the desired row directly, using the properties of Pascal's Triangle to update values in place.

Here's how the algorithm would work step-by-step:

Imagine you have a list to store the row you want to create.
Start by filling the list with ones. Every row in Pascal's Triangle begins and ends with one.
Now, moving from left to right, update each number in the list based on the number before it. You can do this using a simple formula that relies on the previous number in the row.
Repeat this process until you've reached the end of the row. This will correctly compute the next number in sequence using the previous ones.
After updating the entire list, you'll have the numbers for the requested row of Pascal's Triangle.

Code Implementation

def get_row(row_index):
    current_row = [1] * (row_index + 1)

    # We calculate elements based on previous ones
    for i in range(1, row_index + 1):
        for j in range(i - 1, 0, -1):

            # Update the current element based on its predecessor
            current_row[j] = current_row[j] + current_row[j - 1]

    return current_row

Big(O) Analysis

Time Complexity

O(n²) – We initialize a list of size n+1 with ones. Then, we iterate from the second row up to the target row, which involves updating the values in the list. The outer loop runs n times, and for each row, the inner loop iterates from the second element up to the second to last element, which is another loop that runs up to n times in the worst case. Therefore, the time complexity is approximately proportional to n * n, giving us O(n²).

Space Complexity

O(rowIndex) – The algorithm uses a list to store the row being generated. The size of this list is directly proportional to the input rowIndex, as it represents the number of elements in the desired row. Therefore, the auxiliary space required grows linearly with rowIndex. Consequently, the space complexity is O(rowIndex).

Edge Cases

Case	How to Handle
rowIndex is negative	Return an empty list since Pascal's Triangle is not defined for negative rows.
rowIndex is 0	Return a list containing only [1] since that's the first row of Pascal's Triangle.
rowIndex is a large positive integer that may cause integer overflow in calculating binomial coefficients.	Use long integer type or consider an alternative method (dynamic programming) to avoid potential integer overflow issues.
rowIndex is 1	Return a list containing [1, 1] as the second row of Pascal's triangle.
rowIndex close to the maximum representable integer size	Ensure the algorithm gracefully handles maximum value limits by choosing suitable data types and preventing any overflow conditions during calculations.
Algorithm involves iterative calculation of elements, ensure each value is derived properly from the previous row	Use Dynamic Programming and only store one row at a time to avoid creating unnecessarily large data structures.
Space Complexity of generated triangle row	Optimize space by only storing and overwriting the last row.
Time complexity of calculating elements	Ensure generation of pascal's triangle is O(k^2) where k is rowIndex through use of optimal iterative approach.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

Is the `rowIndex` 0-indexed or 1-indexed?
What is the maximum possible value for `rowIndex`? Is there an upper bound, and should I be concerned about integer overflow?
Should I return an empty list if `rowIndex` is negative?
Can I assume `rowIndex` will always be a valid integer?
Can I use extra space, or should I try to optimize for space complexity as much as possible?

Brute Force Solution

Approach

Here's how the algorithm would work step-by-step:

Start by creating the first row of the triangle, which is just the number 1.
Then, to build the next row, look at the row above it.
For each number in the new row, calculate its value by adding the two numbers directly above it in the previous row.
If a number is at the edge of the row, just consider the missing number as zero.
Repeat this process, generating each row one at a time, until you reach the row that you're looking for.
Once you've generated the row you wanted, that's your answer.

Code Implementation

def get_pascal_row_brute_force(row_index):
    pascal_triangle = []

    # The first row is always [1]
    first_row = [1]
    pascal_triangle.append(first_row)

    for current_row_index in range(1, row_index + 1):
        previous_row = pascal_triangle[current_row_index - 1]
        current_row = []

        # First element of each row is always 1
        current_row.append(1)

        # Calculate intermediate elements by summing the two elements above
        for element_index in range(1, current_row_index):

            # Sum the two numbers above in previous row
            number_from_previous_row = previous_row[element_index - 1] + previous_row[element_index]
            current_row.append(number_from_previous_row)

        # Last element of each row is always 1
        current_row.append(1)
        pascal_triangle.append(current_row)

    # Return the requested row of Pascal's Triangle
    return pascal_triangle[row_index]

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

Here's how the algorithm would work step-by-step:

Imagine you have a list to store the row you want to create.
Start by filling the list with ones. Every row in Pascal's Triangle begins and ends with one.
Now, moving from left to right, update each number in the list based on the number before it. You can do this using a simple formula that relies on the previous number in the row.
Repeat this process until you've reached the end of the row. This will correctly compute the next number in sequence using the previous ones.
After updating the entire list, you'll have the numbers for the requested row of Pascal's Triangle.

Code Implementation

def get_row(row_index):
    current_row = [1] * (row_index + 1)

    # We calculate elements based on previous ones
    for i in range(1, row_index + 1):
        for j in range(i - 1, 0, -1):

            # Update the current element based on its predecessor
            current_row[j] = current_row[j] + current_row[j - 1]

    return current_row

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
rowIndex is negative	Return an empty list since Pascal's Triangle is not defined for negative rows.
rowIndex is 0	Return a list containing only [1] since that's the first row of Pascal's Triangle.
rowIndex is a large positive integer that may cause integer overflow in calculating binomial coefficients.	Use long integer type or consider an alternative method (dynamic programming) to avoid potential integer overflow issues.
rowIndex is 1	Return a list containing [1, 1] as the second row of Pascal's triangle.
rowIndex close to the maximum representable integer size	Ensure the algorithm gracefully handles maximum value limits by choosing suitable data types and preventing any overflow conditions during calculations.
Algorithm involves iterative calculation of elements, ensure each value is derived properly from the previous row	Use Dynamic Programming and only store one row at a time to avoid creating unnecessarily large data structures.
Space Complexity of generated triangle row	Optimize space by only storing and overwriting the last row.
Time complexity of calculating elements	Ensure generation of pascal's triangle is O(k^2) where k is rowIndex through use of optimal iterative approach.