What is the time complexity of Pascal's Triangle?

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

What is the space complexity of Pascal's Triangle?

The space complexity is O(n^2).

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

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?

**Direct Translation of the Summation Rule:** The core mathematical property `P(row, col) = P(row-1, col-1) + P(row-1, col)` is directly implemented. Each element at an inner position `j` in the current row `i` is precisely the sum of elements at `j-1` and `j` from the *previous* row `i-1`. **Leveraging Prior Computations (Dynamic Programming):** The solution explicitly uses the previously generated row (`triangle[i - 1]`) to construct the current one. This avoids redundant calculations, as each row (a subproblem) is computed once and stored, then directly accessed for subsequent rows.

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Pascal's Triangle - LeetCode 118 Solution

Q: How to solve Pascal's Triangle?

The provided solution embraces the inherent dynamic programming nature of Pascal's Triangle. It constructs the triangle row by row, building upon the results of the preceding row. We start by initializing an empty list, `triangle`, which will eventually hold all our rows. For each row index `i` from `0` up to `numRows - 1`: A new `row` is initialized. Crucially, it's created as a list of `i + 1` ones (e.g., `[1] * (i + 1)`). This cleverly handles the fixed '1's at the beginning and end of every row for all `i`. For rows beyond the second one (i.e., `i > 1`), we then iterate with `j` from `1` to `i - 1`. This range specifically targets the elements *between* the leading and trailing '1's that we just pre-filled. Inside this loop, the core dynamic programming step occurs: `row[j] = triangle[i - 1][j - 1] + triangle[i - 1][j]`. This directly translates the rule 'each number is the sum of the two numbers directly above it', using the already computed `(i - 1)`-th row. Once the current `row` is fully constructed, it's appended to the `triangle` list. This iterative process ensures that by the time we need to calculate `row i`, `row i-1` (which it depends on) is already present and fully correct in `triangle`.

Q: Which companies ask LeetCode 118. Pascal's Triangle in interviews?

This problem is frequently asked at Accenture, Cisco, Deloitte, Goldman Sachs, Google and 10 more companies.

Pascal's Triangle - Complete Solution Guide

Pascal's Triangle is LeetCode problem 118, 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 numRows , return the first numRows of Pascal's triangle . In Pascal's triangle , each number is the sum of the two numbers directly above it as shown: Example 1: Input: numRows = 5 Output: [[1],[1,1],[1,2,1],[1,3,3,1],[1,4,6,4,1]] Example 2: Input: numRows = 1 Output: [[1]] Constraints: 1 <= numRows <= 30

Detailed Explanation

Pascal's Triangle presents a visually elegant pattern that's surprisingly straightforward to generate algorithmically. At its core, the problem asks us to construct the first `numRows` of this famous mathematical triangle. The defining characteristic is simple: every number inside the triangle (i.e., not at the very edge) is the sum of the two numbers directly above it. The edges of each row are always `1`. For instance, if `numRows = 5`, the third row `[1,2,1]` is derived from the second row `[1,1]` where the '2' comes from `1 + 1`. The fourth row `[1,3,3,1]` similarly uses the third row `[1,2,1]` to calculate its inner '3's (first '3' is `1+2`, second '3' is `2+1`).

Solution Approach

The provided solution embraces the inherent dynamic programming nature of Pascal's Triangle. It constructs the triangle row by row, building upon the results of the preceding row. We start by initializing an empty list, `triangle`, which will eventually hold all our rows. For each row index `i` from `0` up to `numRows - 1`: A new `row` is initialized. Crucially, it's created as a list of `i + 1` ones (e.g., `[1] * (i + 1)`). This cleverly handles the fixed '1's at the beginning and end of every row for all `i`. For rows beyond the second one (i.e., `i > 1`), we then iterate with `j` from `1` to `i - 1`. This range specifically targets the elements *between* the leading and trailing '1's that we just pre-filled. Inside this loop, the core dynamic programming step occurs: `row[j] = triangle[i - 1][j - 1] + triangle[i - 1][j]`. This directly translates the rule 'each number is the sum of the two numbers directly above it', using the already computed `(i - 1)`-th row. Once the current `row` is fully constructed, it's appended to the `triangle` list. This iterative process ensures that by the time we need to calculate `row i`, `row i-1` (which it depends on) is already present and fully correct in `triangle`.

Step-by-Step Algorithm

Step 1: Initialize an empty list (or 2D array) to store the Pascal's Triangle.
Step 2: Iterate from 0 to `numRows - 1` (representing each row).
Step 3: For each row, create a new row list (or array) of size `i + 1` and initialize all elements to 1.
Step 4: If the row index `i` is greater than 1, iterate from 1 to `i - 1` to calculate the intermediate elements of the row using the formula: `row[j] = triangle[i - 1][j - 1] + triangle[i - 1][j]` (sum of elements from the previous row).
Step 5: Append the newly created row to the Pascal's Triangle list.
Step 6: After the loop finishes, return the Pascal's Triangle list.

Key Insights

**Direct Translation of the Summation Rule:** The core mathematical property `P(row, col) = P(row-1, col-1) + P(row-1, col)` is directly implemented. Each element at an inner position `j` in the current row `i` is precisely the sum of elements at `j-1` and `j` from the *previous* row `i-1`.
**Leveraging Prior Computations (Dynamic Programming):** The solution explicitly uses the previously generated row (`triangle[i - 1]`) to construct the current one. This avoids redundant calculations, as each row (a subproblem) is computed once and stored, then directly accessed for subsequent rows.
**Efficient Boundary Handling:** Instead of special-casing for the leading and trailing '1's in each row with separate conditional logic, the solution initializes `row = [1] * (i + 1)`. This simple step efficiently sets up the correct length and automatically populates the boundary '1's, making the subsequent inner loop solely responsible for computing the sums of the interior elements.

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(n^2)

Topics

This problem involves: Array, Dynamic Programming.

Companies

Asked at: Accenture, Cisco, Deloitte, Goldman Sachs, Google, HSBC, J.P. Morgan, Mitsogo, Oracle, Uber, Virtusa, Wix, X, Yahoo, tcs.

Pascal's Triangle | LeetCode 118

Easy

Array Dynamic Programming

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

Solution Code

Available in:

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

class Solution {
    public List<List<Integer>> generate(int numRows) {
        List<List<Integer>> triangle = new ArrayList<>();

        for (int i = 0; i < numRows; i++) {
            List<Integer> row = new ArrayList<>();
            for (int j = 0; j <= i; j++) {
                if (j == 0 || j == i) {
                    row.add(1);
                } else {
                    row.add(triangle.get(i - 1).get(j - 1) + triangle.get(i - 1).get(j));
                }
            }
            triangle.add(row);
        }

        return triangle;
    }
}

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Problem Info

Description

Problem 118: Pascal's Triangle - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming

Companies

AccentureCiscoDeloitteGoldman SachsGoogleHSBCJ.P. MorganMitsogo

Stats

Difficulty:Easy

Languages:4

Companies:15

Problem #:118

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 empty list (or 2D array) to store the Pascal's Triangle.
2
Step 2: Iterate from 0 to `numRows - 1` (representing each row).
3
Step 3: For each row, create a new row list (or array) of size `i + 1` and initialize all elements to 1.
4
Step 4: If the row index `i` is greater than 1, iterate from 1 to `i - 1` to calculate the intermediate elements of the row using the formula: `row[j] = triangle[i - 1][j - 1] + triangle[i - 1][j]` (sum of elements from the previous row).
5
Step 5: Append the newly created row to the Pascal's Triangle list.
6
Step 6: After the loop finishes, return the Pascal's Triangle list.

💡

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^2)

📚 Quadratic space - Uses lots of memory!

📝

Why?

The outer loop iterates `numRows` (n) times. The inner loop, in the worst case, iterates up to `numRows - 1` times. Therefore, the time complexity is approximately proportional to the sum of integers from 1 to n, which is O(n^2). The space complexity is also O(n^2) because we store the entire Pascal's Triangle, which has approximately n^2 elements.

📖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

A recursive solution can be implemented by defining a recursive function that calculates each element based on its position. It would directly implement the mathematical definition of Pascal's triangle.

Time: O(n^2)Space: O(n^2) due to recursive call stack, though potentially higher due to repeated computations if not memoized.

Trade-offs: While conceptually simpler, the recursive approach can be less efficient due to repeated calculations and potential stack overflow for large inputs compared to the iterative dynamic programming approach.

Combinatorial Approach

Each element in Pascal's triangle can be calculated using the binomial coefficient formula (n choose k), where n is the row number and k is the element position. This can be computed using factorials.

Time: O(n^2)Space: O(1) if we avoid using a large pre-computed factorial table (we can compute factorials on the fly).

Trade-offs: This approach can be more efficient in terms of space complexity than the dynamic programming approach if factorials are not pre-computed but might be slightly slower due to factorial calculations.

✨

Key Insights

1**Direct Translation of the Summation Rule:** The core mathematical property `P(row, col) = P(row-1, col-1) + P(row-1, col)` is directly implemented. Each element at an inner position `j` in the current row `i` is precisely the sum of elements at `j-1` and `j` from the *previous* row `i-1`.
2**Leveraging Prior Computations (Dynamic Programming):** The solution explicitly uses the previously generated row (`triangle[i - 1]`) to construct the current one. This avoids redundant calculations, as each row (a subproblem) is computed once and stored, then directly accessed for subsequent rows.
3**Efficient Boundary Handling:** Instead of special-casing for the leading and trailing '1's in each row with separate conditional logic, the solution initializes `row = [1] * (i + 1)`. This simple step efficiently sets up the correct length and automatically populates the boundary '1's, making the subsequent inner loop solely responsible for computing the sums of the interior elements.

📚

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: `numRows = 1`. The solution correctly handles this by returning `[[1]]`.
Edge case 2: `numRows = 0`. While not explicitly stated in the constraints, a robust solution should handle this gracefully by returning an empty list `[]`.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly calculating elements in the inner loop. Off-by-one errors in indices are very common when implementing this algorithm.

✗Common mistake 2: Not handling the base cases correctly, particularly the first row or rows.

✅

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 is directly related to binomial coefficients, which are fundamental in probability theory for calculating combinations and probabilities of events.

🏦Real-world scenario 2: Combinatorics and counting problems: Pascal's Triangle can be used to efficiently solve various combinatorics problems, such as counting the number of ways to choose a certain number of items from a set.

💡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

Use this problem to demonstrate clean code practices. Variable naming and code organization matter.

⚠️ 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 27, 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

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 14

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

@TechLeadJul 7

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I identify if a problem needs dynamic programming?

DP problems typically have: 1) Overlapping subproblems (same calculation repeated), 2) Optimal substructure (optimal solution uses optimal solutions to subproblems). Look for keywords like "minimum/maximum," "count ways," "is it possible," or "longest/shortest."

How should I approach easy-level algorithm problems?

Easy problems build foundational skills. Focus on: 1) Understanding the problem completely before coding, 2) Writing clean, readable code, 3) Handling all edge cases (empty input, single element, etc.), 4) Analyzing time and space complexity. Don't skip complexity analysis even for simple solutions.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

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

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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