Pascal's Triangle

Problem Overview

LeetCode Problem: Pascal's Triangle

Difficulty: Easy

URL: Pascal's Triangle - LeetCode

The problem statement is straightforward: given a non-negative integer numRows, generate the first numRows of Pascal's Triangle. In Pascal's Triangle, each number is the sum of the two numbers directly above it.

Understanding the Problem

Pascal's Triangle is a triangular array of binomial coefficients. The task is to construct the triangle up to a given number of rows (numRows). Each row represents the coefficients of the binomial expansion.

For example, if numRows is 5, the output should be:

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

Key DSA Concepts and Theory

Arrays

• Basics: An array is a collection of items stored at contiguous memory locations. It is used to store elements of the same data type.
• Multi-dimensional Arrays: Used to store data in a tabular form (like Pascal’s Triangle).

Mathematical Background: Pascal's Triangle

• Construction Rule: Each element of a row can be constructed by adding the elements in the previous row that are directly above and to the left above the current element (except for the first and last element of each row, which are always 1).

Solution Approach

The approach involves iterating over the number of rows, and for each row, constructing its elements based on the elements from the previous row.

Detailed Step-by-Step Explanation

1. Initialize Result Array: Create a 2D array or a list to hold all the rows of the Pascal's Triangle.
2. Iterate Through Rows: For each row up to numRows:
   • Initialize a new row with 1.
   • For each position in the current row (except the first and last position), compute the sum of the values from the previous row.
   • Add the computed row to the result list.
3. Return the Result: Once all rows are computed, return the 2D list containing all rows of the Pascal’s Triangle.

C++ Code

#include <iostream>
#include <vector>

std::vector<std::vector<int>> generate(int numRows) {
    std::vector<std::vector<int>> triangle(numRows);

    for (int i = 0; i < numRows; i++) {
        triangle[i].resize(i + 1);
        triangle[i][0] = triangle[i][i] = 1;

        for (int j = 1; j < i; j++) {
            triangle[i][j] = triangle[i - 1][j - 1] + triangle[i - 1][j];
        }
    }

    return triangle;
}

int main() {
    int numRows = 5;
    std::vector<std::vector<int>> result = generate(numRows);

    for (const auto& row : result) {
        for (int val : row) {
            std::cout << val << " ";
        }
        std::cout << std::endl;
    }

    return 0;
}

Java Code

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

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

        for (int i = 0; i < numRows; i++) {
            List<Integer> row = new ArrayList<>();
            row.add(1);

            for (int j = 1; j < i; j++) {
                int previousValue = triangle.get(i - 1).get(j - 1) + triangle.get(i - 1).get(j);
                row.add(previousValue);
            }

            if (i > 0) {
                row.add(1);
            }

            triangle.add(row);
        }

        return triangle;
    }

    public static void main(String[] args) {
        int numRows = 5;
        List<List<Integer>> triangle = generate(numRows);

        for (List<Integer> row : triangle) {
            System.out.println(row);
        }
    }
}

Time and Space Complexity Analysis

• Time Complexity: O(n²), where n is numRows. This is because we have to build each of the n levels, and each level requires building up to n elements.

• Space Complexity: O(n²). The space complexity is also O(n²) because we store all the elements of the triangle up to row n.

Common Mistakes to Avoid

1. Index Out of Bounds: Ensure to handle indices carefully when accessing the elements of the previous row.
2. Initializing Inner Arrays: Proper initialization of arrays/lists is crucial before updating or accessing elements.

Similar Problems on LeetCode

1. Pascal's Triangle II (LeetCode 119): Here, you're required to compute only a single row in Pascal's Triangle without storing the entire triangle.
2. Triangle (LeetCode 120): This involves finding the minimum path sum from the top to the bottom in a triangle.

Additional Resources and References

• Introduction to Arrays - GeeksforGeeks
• Understanding Pascal's Triangle - Khan Academy
• Dynamic Programming and Pascal's Triangle - Wikipedia

By understanding these concepts and practicing similar problems, one can enhance their problem-solving skills related to arrays and combinatorial mathematics.