What is the time complexity of Pascal's Triangle II?

The optimal solution has a time complexity of O(n^2).

What data structures are used to solve Pascal's Triangle II?

This problem uses Array, Dynamic Programming. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Pascal's Triangle II?

**In-place Calculation with Backward Iteration:** The core trick is updating the row in-place. The backward loop (`j` from `i` down to `1`) is essential. If `j` were to iterate forwards, `row[j-1]` would be updated before `row[j]` could use its *original* value, corrupting the sum. By going backwards, `row[j-1]` remains stable until `row[j]` has correctly utilized it. **Optimal Space Complexity (`O(k)`):** This solution uses only a single list of size `rowIndex + 1` to store the computed row. This means its space complexity is `O(rowIndex)`, directly satisfying the follow-up question for an optimized space usage. We avoid storing all preceding rows, which would incur `O(rowIndex^2)` space.

Which companies ask LeetCode 119. Pascal's Triangle II in interviews?

This problem is frequently asked at Goldman Sachs, Yahoo.

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Pascal's Triangle II - LeetCode 119 Solution

Q: How to solve Pascal's Triangle II?

The provided solution elegantly computes the `rowIndex`th row using an in-place dynamic programming approach, adhering to an `O(rowIndex)` space complexity. It begins by initializing a list `row` of size `rowIndex + 1` with all ones. This clever initialization sets up the boundary elements correctly (the first and last numbers in any row of Pascal's Triangle are always 1) and provides a mutable structure for subsequent calculations. The algorithm then iterates with an outer loop that conceptually represents building the triangle row by row, up to our target. The inner loop, `for j in range(i, 0, -1)`, is crucial. It iterates *backwards* from `i` down to `1`, updating `row[j]` by adding `row[j-1]` to its current value. This backward iteration ensures that when `row[j]` is updated, `row[j-1]` still holds its value from the *previous conceptual row*, preventing us from using an already-updated value from the *current* iteration which would lead to incorrect sums. By working this way, the `row` list incrementally transforms, ultimately holding the correct values for the `rowIndex`th row.

Pascal's Triangle II - Complete Solution Guide

Pascal's Triangle II is LeetCode problem 119, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

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?

Detailed Explanation

This problem challenges us to extract a specific `rowIndex` (0-indexed) from Pascal's Triangle. Instead of generating the entire triangle, which can be memory-intensive for higher row indices, we're asked to return just that single target row as a list of integers. For instance, if `rowIndex` is 0, we expect `[1]`; if `rowIndex` is 1, `[1,1]`; and for `rowIndex = 3`, the output should be `[1,3,3,1]`. The core rule governing Pascal's Triangle is that each number is the sum of the two numbers directly above it, with the edges always being 1s.

Solution Approach

The provided solution elegantly computes the `rowIndex`th row using an in-place dynamic programming approach, adhering to an `O(rowIndex)` space complexity. It begins by initializing a list `row` of size `rowIndex + 1` with all ones. This clever initialization sets up the boundary elements correctly (the first and last numbers in any row of Pascal's Triangle are always 1) and provides a mutable structure for subsequent calculations. The algorithm then iterates with an outer loop that conceptually represents building the triangle row by row, up to our target. The inner loop, `for j in range(i, 0, -1)`, is crucial. It iterates *backwards* from `i` down to `1`, updating `row[j]` by adding `row[j-1]` to its current value. This backward iteration ensures that when `row[j]` is updated, `row[j-1]` still holds its value from the *previous conceptual row*, preventing us from using an already-updated value from the *current* iteration which would lead to incorrect sums. By working this way, the `row` list incrementally transforms, ultimately holding the correct values for the `rowIndex`th row.

Step-by-Step Algorithm

Step 1: Initialize an array/list `row` of size `rowIndex + 1` with all elements set to 1.
Step 2: Iterate from `i = 1` to `rowIndex - 1`. This loop represents each element's calculation in the row beyond the first and last which are already 1.
Step 3: Inner loop iterates from `j = i` down to `1`. This iterates backward through the row calculating each element using the previous element in the array.
Step 4: Inside the inner loop, add the previous element `row[j - 1]` to the current element `row[j]`. This updates the array in place.
Step 5: After each row is generated, the last element (which is always 1) is added to the array.
Step 6: Return the array `row` which holds the generated `rowIndex`-th row of Pascal's triangle.

Key Insights

**In-place Calculation with Backward Iteration:** The core trick is updating the row in-place. The backward loop (`j` from `i` down to `1`) is essential. If `j` were to iterate forwards, `row[j-1]` would be updated before `row[j]` could use its *original* value, corrupting the sum. By going backwards, `row[j-1]` remains stable until `row[j]` has correctly utilized it.
**Optimal Space Complexity (`O(k)`):** This solution uses only a single list of size `rowIndex + 1` to store the computed row. This means its space complexity is `O(rowIndex)`, directly satisfying the follow-up question for an optimized space usage. We avoid storing all preceding rows, which would incur `O(rowIndex^2)` space.
**Smart Initialization:** Starting the `row` list with all ones (`[1] * (rowIndex + 1)`) is more than just a placeholder. It correctly sets the fixed '1's at the beginning and end of every row and provides base values for the intermediate sums. For `rowIndex` values of 0 or 1, this initialization alone is sufficient, as the loops won't even execute or will execute minimally to yield the correct result.

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(n)

Topics

This problem involves: Array, Dynamic Programming.

Companies

Asked at: Goldman Sachs, Yahoo.

Pascal's Triangle II | LeetCode 119

Easy

Array Dynamic Programming

Time: O(n^2)Space: O(n)

Solution Code

Available in:

import java.util.ArrayList;
import java.util.List;

class Solution {
    public List<Integer> getRow(int rowIndex) {
        List<Integer> row = new ArrayList<>(rowIndex + 1);
        row.add(1);
        for (int i = 1; i <= rowIndex; i++) {
            for (int j = i - 1; j >= 1; j--) {
                row.set(j, row.get(j) + row.get(j - 1));
            }
            row.add(1);
        }
        return row;
    }
}

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Problem Info

Description

Problem 119: Pascal's Triangle II - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming

Companies

Goldman SachsYahoo

Stats

Difficulty:Easy

Languages:4

Companies:2

Problem #:119

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

6 steps

1
Step 1: Initialize an array/list `row` of size `rowIndex + 1` with all elements set to 1.
2
Step 2: Iterate from `i = 1` to `rowIndex - 1`. This loop represents each element's calculation in the row beyond the first and last which are already 1.
3
Step 3: Inner loop iterates from `j = i` down to `1`. This iterates backward through the row calculating each element using the previous element in the array.
4
Step 4: Inside the inner loop, add the previous element `row[j - 1]` to the current element `row[j]`. This updates the array in place.
5
Step 5: After each row is generated, the last element (which is always 1) is added to the array.
6
Step 6: Return the array `row` which holds the generated `rowIndex`-th row of Pascal's triangle.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n^2)

🐢 Quadratic - Gets slow with large inputs. Consider optimization!

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is O(rowIndex^2) because of the nested loops. The outer loop runs `rowIndex - 1` times, and the inner loop runs a maximum of `rowIndex -1` times in the last iteration. The space complexity is O(rowIndex) because we only store the current row in an array of size `rowIndex + 1`.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Recursive Approach

Recursively calculate each element based on Pascal's identity: C(n, k) = C(n-1, k-1) + C(n-1, k), where C(n, k) is the k-th element of the n-th row.

Time: O(rowIndex^2)Space: O(rowIndex^2) due to recursive calls and implicit stack space

Trade-offs: Less efficient than the iterative approach in terms of space complexity due to the overhead of recursive calls. Easier to understand conceptually but less optimal for large values of `rowIndex`.

Combinatorial Approach

Use the combinations formula n! / (k! * (n-k)!) to directly calculate the elements of the row, where n is the row index and k is the element index.

Time: O(rowIndex^2)Space: O(rowIndex)

Trade-offs: Mathematically elegant. However, calculating factorials can lead to integer overflow issues for larger values of `rowIndex` without using specialized libraries for arbitrary-precision arithmetic.

✨

Key Insights

1**In-place Calculation with Backward Iteration:** The core trick is updating the row in-place. The backward loop (`j` from `i` down to `1`) is essential. If `j` were to iterate forwards, `row[j-1]` would be updated before `row[j]` could use its *original* value, corrupting the sum. By going backwards, `row[j-1]` remains stable until `row[j]` has correctly utilized it.
2**Optimal Space Complexity (`O(k)`):** This solution uses only a single list of size `rowIndex + 1` to store the computed row. This means its space complexity is `O(rowIndex)`, directly satisfying the follow-up question for an optimized space usage. We avoid storing all preceding rows, which would incur `O(rowIndex^2)` space.
3**Smart Initialization:** Starting the `row` list with all ones (`[1] * (rowIndex + 1)`) is more than just a placeholder. It correctly sets the fixed '1's at the beginning and end of every row and provides base values for the intermediate sums. For `rowIndex` values of 0 or 1, this initialization alone is sufficient, as the loops won't even execute or will execute minimally to yield the correct result.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: `rowIndex` = 0. The solution correctly handles this by returning `[1]`.
Edge case 2: `rowIndex` = 1. The solution correctly returns `[1, 1]`.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly handling the base cases (rowIndex = 0 and rowIndex = 1). Ensure you handle these cases correctly without relying on the loops, otherwise you can have array index out of bounds errors.

✗Common mistake 2: Not utilizing space optimization. Many beginners will generate all previous rows unnecessarily, leading to O(rowIndex^2) space complexity.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Probability calculations. Pascal's Triangle provides binomial coefficients, which are fundamental in probability theory for calculating combinations.

🏦Real-world scenario 2: Combinatorics problems. Pascal's Triangle helps in solving many counting problems, such as arranging items, selecting subsets, etc.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Don't rush. Explain your thought process clearly. Easy problems are often about communication.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: September 28, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 13

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 6

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

Should I use top-down (memoization) or bottom-up (tabulation)?

Start with top-down (recursion + memoization) as it directly maps to the recursive thinking and handles only needed subproblems. Convert to bottom-up for better space efficiency and to avoid stack overflow. Bottom-up is often faster due to no recursion overhead.

What should I learn from easy problems?

Easy problems introduce core patterns that appear in harder problems. Master basic operations (iteration, conditionals, simple data structures), recognize common patterns (counting, searching, basic transformations), and practice explaining your thought process clearly.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Pascal's Triangle II"?

For this easy-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether dynamic programming techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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