Pascal Triangle In Java Leetcode Solution

Pascal's Triangle is a triangular array of binomial coefficients that finds applications in combinatorics, probability theory, and algebra. Generating this pattern efficiently is a common challenge in programming.

In this article, you will learn how to construct Pascal's Triangle in Java, focusing on an iterative approach suitable for LeetCode-style problems.

Problem Statement

The core problem is to generate the first numRows of Pascal's Triangle. Each number in the triangle is the sum of the two numbers directly above it. The edges of the triangle are always 1.

For example, if numRows is 5, the output should represent the following structure:

1
   1 1
  1 2 1
 1 3 3 1
1 4 6 4 1

Example

For an input of numRows = 5, the expected output represented as a list of lists would be:

[
     [1],
    [1,1],
   [1,2,1],
  [1,3,3,1],
 [1,4,6,4,1]
]

Background & Knowledge Prerequisites

To understand the solution, readers should be familiar with:

• Java Basics: Variables, loops (for loops), conditional statements.
• ArrayList and List Interface: How to create, add elements to, and access elements from dynamic arrays in Java. Specifically, List> for the overall triangle and List for individual rows.

There are no external libraries or complex setups required for this problem.

Use Cases or Case Studies

Pascal's Triangle is more than just a mathematical curiosity; it has practical applications:

• Binomial Expansion: The numbers in each row correspond to the coefficients in the binomial expansion of (x + y)^n. For instance, row 4 (1 4 6 4 1) corresponds to (x+y)^4 = 1x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + 1y^4.
• Combinatorics: Each number C(n, k) (the k-th element in the n-th row, starting count from 0) represents the number of ways to choose k items from a set of n items.
• Probability Theory: It appears in probability calculations, especially for scenarios involving "n choose k" events.
• Computer Graphics: Used in algorithms for Bézier curves and surfaces due to its connection to binomial coefficients.

Solution Approaches

The most common and efficient approach for constructing Pascal's Triangle is iterative, building each row based on the previously generated row.

Approach 1: Iterative Row-by-Row Construction

This approach builds the triangle one row at a time. Each new row starts and ends with 1, and its inner elements are calculated by summing adjacent elements from the previous row.

One-line summary

Generate each row by adding adjacent elements from the previous row, starting and ending each new row with 1.

Code example

// Pascal's Triangle Generator
import java.util.ArrayList;
import java.util.List;

// Main class containing the entry point of the program
public class Main {

    // LeetCode-style solution method
    public static List<List<Integer>> generate(int numRows) {
        List<List<Integer>> triangle = new ArrayList<>();

        if (numRows == 0) {
            return triangle;
        }

        // The first row is always [1]
        List<Integer> firstRow = new ArrayList<>();
        firstRow.add(1);
        triangle.add(firstRow);

        // Iterate from the second row up to numRows
        for (int i = 1; i < numRows; i++) {
            List<Integer> prevRow = triangle.get(i - 1); // Get the previous row
            List<Integer> currentRow = new ArrayList<>();

            // The first element of every row is always 1
            currentRow.add(1);

            // Calculate inner elements based on the previous row
            for (int j = 1; j < i; j++) {
                // Each element is the sum of the two elements directly above it
                currentRow.add(prevRow.get(j - 1) + prevRow.get(j));
            }

            // The last element of every row is always 1
            currentRow.add(1);

            // Add the newly formed row to the triangle
            triangle.add(currentRow);
        }

        return triangle;
    }

    public static void main(String[] args) {
        // Test cases
        System.out.println("Pascal's Triangle for numRows = 1:");
        System.out.println(generate(1)); // Expected: [[1]]
        System.out.println("\\nPascal's Triangle for numRows = 3:");
        System.out.println(generate(3)); // Expected: [[1], [1, 1], [1, 2, 1]]
        System.out.println("\\nPascal's Triangle for numRows = 5:");
        System.out.println(generate(5));
        /* Expected:
        [
             [1],
            [1,1],
           [1,2,1],
          [1,3,3,1],
         [1,4,6,4,1]
        ]
        */
    }
}

Run

Sample output

Running the main method of the above Java code would produce:

Pascal's Triangle for numRows = 1:
[[1]]

Pascal's Triangle for numRows = 3:
[[1], [1, 1], [1, 2, 1]]

Pascal's Triangle for numRows = 5:
[[1], [1, 1], [1, 2, 1], [1, 3, 3, 1], [1, 4, 6, 4, 1]]

Stepwise explanation

1. Initialize the Triangle: Create an empty List> called triangle to store all the rows. Handle the edge case where numRows is 0 by returning an empty list.
2. Handle the First Row: If numRows is greater than 0, the very first row is always [1]. Create a List for this and add it to triangle.
3. Iterate for Subsequent Rows: Start a loop from i = 1 (representing the second row) up to numRows - 1. In each iteration i:

• Get Previous Row: Retrieve the prevRow from triangle using triangle.get(i - 1).
• Create Current Row: Initialize an empty List called currentRow.
• Add First Element: The first element of every row is always 1. Add 1 to currentRow.
• Calculate Inner Elements: Loop from j = 1 up to i - 1. For each j, calculate the element by summing prevRow.get(j - 1) and prevRow.get(j). Add this sum to currentRow. This covers all elements between the starting and ending 1s.
• Add Last Element: The last element of every row is always 1. Add 1 to currentRow.
• Add to Triangle: Add the currentRow to the triangle list.

1. Return Result: After the loop completes, triangle will contain all the generated rows. Return triangle.

This approach ensures that each row is correctly derived from the one immediately preceding it, building the triangle step by step.

Conclusion

Generating Pascal's Triangle is a classic problem that tests understanding of iterative patterns and list manipulation. The row-by-row iterative approach provides a clear and efficient solution by leveraging the relationship between consecutive rows. This method is straightforward to implement and offers good performance for typical constraints.

Summary

• Pascal's Triangle is a triangular array where each number is the sum of the two directly above it.
• The edges of the triangle always consist of 1s.
• The most common solution involves an iterative process, building the triangle row by row.
• Each new row starts and ends with 1.
• Inner elements of a new row are calculated by summing adjacent elements from the previous row.
• Java's List> and ArrayList are suitable data structures for representing the triangle and its rows.