How do you solve Pascal's Triangle on LeetCode?

Pascal's Triangle (LeetCode #118) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n²) time complexity and O(n) space complexity. Key insight: Each element in Pascal's triangle is the sum of the two elements directly above it.

What is the brute force solution for Pascal's Triangle?

The brute force approach for Pascal's Triangle is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach constructs each row of Pascal's triangle from scratch using the properties of the triangle. Each element is computed by summing the two elements directly above it in the previous row. The optimal Optimal Solution approach improves this to O(n²).

What algorithm or data structure is used to solve Pascal's Triangle?

Pascal's Triangle uses the following concepts: Array, Dynamic Programming. The recommended patterns to study are: Dynamic Programming, Array.

What are the interview tips for Pascal's Triangle?

Explain your thought process clearly while coding. Consider edge cases, such as numRows = 1. Practice building the triangle visually to understand the relationships.

What are common mistakes when solving Pascal's Triangle?

Forgetting to handle the edges of the triangle (the 1s). Not initializing the row correctly before filling it.

#118

Pascal's Triangle

Easy

ArrayDynamic ProgrammingDynamic ProgrammingArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n²)
Space	O(1)	O(n)

💡

Intuition

Time O(n²)Space O(n)

The optimal solution builds each row based on the previous row, reusing the list to minimize space usage. This approach is efficient in both time and space, as it only requires storing the current and previous rows.

⚙️

Algorithm

5 steps

1Step 1: Initialize a list to hold the first row as [1].
2Step 2: For each subsequent row, create a new list starting with 1.
3Step 3: Calculate each element in the row using the last row, and append it to the new row.
4Step 4: Append 1 at the end of the new row.
5Step 5: Replace the previous row with the new row and repeat until all rows are generated.

solution.py13 lines

1# Full working Python code
2
3def generate(numRows):
4    triangle = []
5    for row_num in range(numRows):
6        row = [1]  # Start with 1
7        if triangle:
8            last_row = triangle[-1]
9            for j in range(1, len(last_row)):
10                row.append(last_row[j - 1] + last_row[j])
11            row.append(1)  # End with 1
12        triangle.append(row)
13    return triangle

ℹ

Complexity note: The time complexity remains O(n²) since we still generate n rows with each row containing up to n elements. The space complexity is O(n) due to storing the triangle in a list.

1Each element in Pascal's triangle is the sum of the two elements directly above it.
2The triangle is symmetric, meaning row n is the same as row n reversed.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.