Pascal's Triangle II

class Solution {
public:
    vector<int> generateRow(int rowIndex) {
        vector<int> result = {1};

        for (int i = 1; i <= rowIndex; i++)
            result.push_back((int)((result.back() * (long long)(rowIndex - i + 1)) / i));

        return result;
    }
};

This solution involves generating a specific row of Pascal's Triangle given the row index. The implementation is in C++ and utilizes a simple but effective approach with the generateRow function.

Key details of the implementation:

• The function generateRow accepts an integer rowIndex which specifies the zero-based index of the row in Pascal's Triangle that you want to generate.
• It initializes the result vector with the first element being 1, which is consistent across all rows in Pascal's Triangle.
• A loop then runs from 1 to rowIndex, utilizing a formula to compute each element of the row based on the previous element (result.back()). This formula takes advantage of the combinatorial properties of Pascal's Triangle: Each element is the result of multiplying the previous element by a fraction that involves the current index and the total number of elements.
• The element is computed using (result.back() * (long long)(rowIndex - i + 1)) / i. This operation ensures that integer overflow is avoided by first casting to long long.
• The final result is a vector of integers representing the desired row in Pascal's Triangle.

Benefits of this approach:

• Space-efficient since only one vector is continuously updated without building the entire triangle.
• Prevents integer overflow using type-casting to long long.
• The approach directly reaches the result without unnecessary computations of preceding rows.

Use this code when you need to retrieve a specific row directly from Pascal's Triangle efficiently, especially when dealing with large indices, as it maintains a predictable computational complexity.

class Solution {
    public List<Integer> generateRow(int rowIndex) {
        List<Integer> result = new ArrayList<Integer>() {
            {
                add(1);
            }
        };

        for (int i = 1; i <= rowIndex; i++) {
            result.add((int) ((result.get(result.size() - 1) * (long) (rowIndex - i + 1)) / i));
        }

        return result;
    }
}

The problem "Pascal's Triangle II" requires generating a specific row of Pascal's triangle, where the row index is provided as an input. The solution involves using a Java method generateRow that returns the desired row as a list of integers.

• Start by initializing a list named result containing a single element 1. This element corresponds to the first number in every row of Pascal's Triangle.
• Use a for loop to iterate from 1 to rowIndex. In each iteration, calculate the next number in the row by using the formula (result.get(result.size() - 1) * (long) (rowIndex - i + 1)) / i. This formula leverages the properties of Pascal's Triangle where each element is the sum of the two directly above it.
• The calculated value is then added to the result list.
• Finally, the fully formed result list representing the rowIndex-th row of Pascal's Triangle is returned.

This approach ensures efficient calculation using only one list and operates in O(n) time complexity, where n is the number of elements in the target row.

int* fetchPascalRow(int rowIndex, int* size) {
    *size = rowIndex + 1;
    int* result = malloc((rowIndex + 1) * sizeof(int));
    result[0] = 1;
    for (int i = 1; i <= rowIndex; i++)
        result[i] = (int)((result[i - 1] * (long long)(rowIndex - i + 1)) / i);
    return result;
}

The provided C function, fetchPascalRow, generates a specific row from Pascal's Triangle, given the row index. When you execute the function with an integer value representing the desired row index, it returns a pointer to an array containing that row's elements.

Here's how the function operates:

1. Begin by determining the number of elements in the desired row. This is equal to rowIndex + 1. The *size pointer parameter is set to this value to inform the caller of the array's length.

2. Allocate memory dynamically for an array result to hold the row's values. The size of memory allocated is (rowIndex + 1) * sizeof(int).

3. Initialize the first element of the array, result[0], to 1, since the first element of every row in Pascal's Triangle is 1.

4. Utilize a loop starting from i = 1 up to rowIndex to fill in the subsequent values of the array based on the properties of Pascal's Triangle. Each element is computed using the formula:

   [ \text{result}[i] = \left(\text{result}[i - 1] \times \frac{\text{rowIndex} - i + 1}{i}\right) ]

   This formula iteratively calculates each value based on the previous value, multiplying by the corresponding ratio derived from the row index and current position.

5. Return the result pointer, which now points to an array containing the appropriate values from the specified row of Pascal's Triangle.

By following this algorithm, the fetchPascalRow function efficiently computes and retrieves any desired row in Pascal's Triangle, leveraging dynamic memory allocation and iterative calculations.

var pascalRow = function(index) {
    let result = [1];
    for (let i = 1; i <= index; i++) {
        result.push(Math.floor((result[result.length - 1] * (index - i + 1)) / i));
    }
    return result;
};

This JavaScript function, pascalRow, generates a specific row of Pascal's Triangle based on the provided index. Here's a summarized breakdown of the code:

• The function starts by initializing an array, result, containing a single element [1]. This represents the first element of every Pascal's Triangle row.
• A for loop runs starting from 1 up to the specified index. In each iteration:
  • The next element of the triangle's row is calculated and appended to the result array. The element is computed using the formula: Math.floor((result[result.length - 1] * (index - i + 1)) / i). This formula uses the last element of the current result array to calculate the next element.
  • This formula factors in combinatorial calculations which are derived from the properties of Pascal's Triangle, where each element is the sum of the two directly above it from the previous row, in a sense here simplified to a product/division operation for efficiency.
• The function concludes by returning the fully constructed row as result.

This concise implementation efficiently computes the desired row of Pascal’s Triangle by leveraging the mathematical relationship between the elements, thus avoiding the need to build the entire triangle up to the requested row.

class Solution:
    def fetchRow(self, rowIndex):
        triangleRow = [1]
        for i in range(1, rowIndex + 1):
            triangleRow.append(int((triangleRow[-1] * (rowIndex - i + 1)) / i))
        return triangleRow

The provided Python code defines a single method fetchRow within the Solution class that efficiently computes a specific row in Pascal's Triangle. The method accepts an integer rowIndex and generates the row at that index directly using a list comprehension combined with a mathematical formula specifically derived from the properties of Pascal's Triangle.

• The computation starts with a base list initialised with [1].
• As successive items of the row are generated, they heavily utilize the previously computed values (dynamic programming approach).
• The relationship (triangleRow[-1] * (rowIndex - i + 1)) / i is used to compute the next item in the row, exploiting the factorial relationship inherent in the rows of Pascal's Triangle.

This approach results in a very efficient computational routine for fetching a specific row as it only requires O(n) time complexity and space where n represents the row index.