How do you solve Pascal's Triangle II on LeetCode?

Pascal's Triangle II (LeetCode #119) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Each element in a row is the sum of the two elements directly above it.

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

Pascal's Triangle II 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 II?

Always clarify the problem requirements before jumping into coding. Think about edge cases, like rowIndex = 0 or rowIndex = 1. Explain your thought process clearly while coding.

What are common mistakes when solving Pascal's Triangle II?

Not understanding how to build the triangle row by row. Confusing the indices when accessing elements from the previous row.

Pascal's Triangle II — LeetCode #119 (Easy)

Tags: Array, Dynamic Programming

Related patterns: Dynamic Programming, Array

Brute Force approach

Time complexity: O(n²). Space complexity: O(1).

We can generate Pascal's Triangle row by row until we reach the desired rowIndex. Each row is built using the values from the previous row, making it straightforward but not the most efficient.

The time complexity is O(n²) because we build each row using the previous row, which takes linear time for each row, and we do this for n rows. The space complexity is O(1) since we only store the current row.

Step 1: Initialize a list to hold the first row of Pascal's Triangle, which is [1].
Step 2: For each row from 1 to rowIndex, create a new list starting with 1.
Step 3: For each element in the new row (except the first and last), calculate its value by summing the two elements directly above it from the previous row.
Step 4: Append 1 to the end of the new row and replace the previous row with the new row.
Step 5: Repeat until the desired rowIndex is reached.

For rowIndex = 3: 1. Start with row = [1] 2. i = 1: new_row = [1, 1] → row = [1, 1] 3. i = 2: new_row = [1]; j = 1: new_row = [1, 2]; new_row = [1, 2, 1] → row = [1, 2, 1] 4. i = 3: new_row = [1]; j = 1: new_row = [1, 3]; j = 2: new_row = [1, 3, 3]; new_row = [1, 3, 3, 1] → row = [1, 3, 3, 1]

Optimal Solution approach

Time complexity: O(n). Space complexity: O(n).

We can compute the row directly using the properties of binomial coefficients, which allows us to build the row in a single pass without needing to store all previous rows.

The time complexity is O(n) because we compute each element of the row in constant time. The space complexity is O(n) as we store the entire row.

Step 1: Initialize a list with size rowIndex + 1, filled with 0.
Step 2: Set the first element to 1, as the first element of any row in Pascal's Triangle is always 1.
Step 3: For each index from 1 to rowIndex, calculate the value using the formula: row[j] = row[j - 1] * (rowIndex - j + 1) / j.
Step 4: Return the constructed row.

For rowIndex = 3: 1. Start with row = [1, 0, 0, 0] 2. j = 1: row[1] = 1 * (3 - 1 + 1) / 1 = 3 → row = [1, 3, 0, 0] 3. j = 2: row[2] = 3 * (3 - 2 + 1) / 2 = 3 → row = [1, 3, 3, 0] 4. j = 3: row[3] = 3 * (3 - 3 + 1) / 3 = 1 → row = [1, 3, 3, 1]

Key Insights

Each element in a row is the sum of the two elements directly above it.
The nth row can be computed directly using binomial coefficients.

Common Mistakes

Not understanding how to build the triangle row by row.
Confusing the indices when accessing elements from the previous row.

Interview Tips

Always clarify the problem requirements before jumping into coding.
Think about edge cases, like rowIndex = 0 or rowIndex = 1.
Explain your thought process clearly while coding.

#119

Pascal's Triangle II

Easy

Array↗Dynamic Programming↗Dynamic Programming↗Array↗

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

We can compute the row directly using the properties of binomial coefficients, which allows us to build the row in a single pass without needing to store all previous rows.

⚙️

Algorithm

4 steps

1Step 1: Initialize a list with size rowIndex + 1, filled with 0.
2Step 2: Set the first element to 1, as the first element of any row in Pascal's Triangle is always 1.
3Step 3: For each index from 1 to rowIndex, calculate the value using the formula: row[j] = row[j - 1] * (rowIndex - j + 1) / j.
4Step 4: Return the constructed row.

solution.py6 lines

1def getRow(rowIndex):
2    row = [0] * (rowIndex + 1)
3    row[0] = 1
4    for j in range(1, rowIndex + 1):
5        row[j] = row[j - 1] * (rowIndex - j + 1) // j
6    return row

ℹ

Complexity note: The time complexity is O(n) because we compute each element of the row in constant time. The space complexity is O(n) as we store the entire row.

1Each element in a row is the sum of the two elements directly above it.
2The nth row can be computed directly using binomial coefficients.

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