Pascals Triangle Ii

                                            
class Solution {
public:
    vector<int> getRow(int rowIndex) {
        vector<int> res(rowIndex + 1, 0);
        res[0] = 1;
        for(int i = 1; i <= rowIndex; i++){
            for(int j = i; j >= 1; j--){
                res[j] += res[j-1];
            }
        }
        return res;
    }
};

                                            
class Solution {
    public List<Integer> getRow(int rowIndex) {
        List<Integer> row = new ArrayList<>();
        for (int i = 0; i <= rowIndex; i++) {
            row.add(0, 1);
            for (int j = 1; j < row.size() - 1; j++) {
                row.set(j, row.get(j) + row.get(j + 1));
            }
        }
        return row;
    }
}

                                            
class Solution(object):
    def getRow(self, rowIndex):
        """
        :type rowIndex: int
        :rtype: List[int]
        """
        row = [1] * (rowIndex + 1)
        for i in range(2, rowIndex + 1):
            for j in range(i - 1, 0, -1):
                row[j] += row[j - 1]
        return row

                                            
/**
 * Note: The returned array must be malloced, assume 
 * caller calls free().
 */
int* getRow(int rowIndex, int* returnSize){
    *returnSize = rowIndex + 1;
    int* row = (int*)malloc(*returnSize * sizeof(int));
    
    for(int i = 0; i <= rowIndex; i++){
        for(int j = i; j >= 0; j--){
            if(j == 0 || j == i){
                row[j] = 1;
            } else{
                row[j] += row[j-1];
            }
        }
    }
    
    return row;
}

                                            
public class Solution {
    public IList<int> GetRow(int rowIndex) {
        IList<int> result = new List<int>();
        for(int i = 0; i <= rowIndex; i++){
            result.Add(1);
            for(int j = i - 1; j > 0; j--){
                result[j] += result[j-1];
            }
        }
        return result;
    }
}

                                            
/**
 * @param {number} rowIndex
 * @return {number[]}
 */
const getRow = (rowIndex) => {
    const result = [];
    for (let i = 0; i <= rowIndex; i++) {
        result[i] = 1;
        for (let j = i - 1; j > 0; j--) {
            result[j] = result[j] + result[j - 1];
        }
    }
    return result;
};

                                            
function getRow(rowIndex: number): number[] {
    const row: number[] = [1];
    for (let i = 1; i <= rowIndex; i++) {
        for (let j = i - 1; j > 0; j--) {
            row[j] = row[j] + row[j - 1];
        }
        row.push(1);
    }
    return row;
};

                                            
class Solution {
    /**
     * @param Integer $rowIndex
     *
     * @return Integer[]
     */
    function getRow($rowIndex) {
        $row = [1];
        for ($i = 1; $i <= $rowIndex; $i++) {
            $nextRow = array_fill(0, $i + 1, 1);
            for ($j = 1; $j < $i; $j++) {
                $nextRow[$j] = $row[$j - 1] + $row[$j];
            }
            $row = $nextRow;
        }
        return $row;
    }
}

                                            
class Solution {
    func getRow(_ rowIndex: Int) -> [Int] {
        var result = Array(repeating: 1, count: rowIndex + 1)
        
        for i in 1..<result.count {
            for j in (1..<i).reversed() {
                result[j] += result[j - 1]
            }
        }
        
        return result
    }
}

                                            
class Solution {
    fun getRow(rowIndex: Int): List<Int> {
        val result = IntArray(rowIndex + 1) { 0 }
        result[0] = 1
        
        for (i in 1..rowIndex) {
            for (j in i downTo 1) {
                result[j] += result[j - 1]
            }
        }
        
        return result.asList()
    }
}

                                            
class Solution {
  List<int> getRow(int rowIndex) {
    List<int> row = [1];
    
    for (int i = 1; i <= rowIndex; i++) {
      List<int> newRow = [];
      for (int j = 0; j <= i; j++) {
        if (j == 0 || j == i) {
          newRow.add(1);
        } else {
          newRow.add(row[j - 1] + row[j]);
        }
      }
      row = newRow;
    }
    
    return row;
  }
}

                                            
func getRow(rowIndex int) []int {
    numRows := rowIndex + 1
    res := make([]int, numRows)
    res[0] = 1
    for i := 1; i < numRows; i++ {
        for j := i; j > 0; j-- {
            res[j] += res[j-1]
        }
    }
    return res
}

                                            
# @param {Integer} row_index
# @return {Integer[]}
def get_row(row_index)
    row = [1]
    (1..row_index).each do |i|
        row = [1] + row.each_cons(2).map { |a, b| a + b } + [1]
    end
    row
end

                                            
object Solution {
    def getRow(rowIndex: Int): List[Int] = {
        var row = List.fill(rowIndex + 1)(1)
        
        for (i <- 1 to rowIndex) {
            for (j <- i - 1 to 1 by -1) {
                row = row.updated(j, row(j) + row(j - 1))
            }
        }
        
        row
    }
}

To solve this coding challenge, we need to generate a specific row of Pascal's Triangle given a zero-based index. Pascal's Triangle is a triangular array of numbers where each number is the sum of the two numbers directly above it. In order to find the

rowIndex

-th row, we adopt an efficient technique that requires only O(rowIndex) extra space.

Explanation

In this challenge, we aim to generate the

rowIndex

-th row of Pascal's Triangle. The

0

-th row is simply

[1]

. From there, each subsequent row is generated by considering the values from the previous row.

Pascal's Triangle Fundamental Concept:

The edges of Pascal's Triangle are always
1

.
Any interior value at position
j

in row
i

can be calculated as the sum of the values at position
j

and
j-1

from row
i-1

.

Optimized Approach:

Given the constraints where

0 <= rowIndex <= 33

, an optimized solution is important to consider. Instead of generating the entire Pascal's Triangle up to the

rowIndex

, we only store the current row while iteratively updating it.

Detailed Steps in Pseudocode:

Initialization : Start with a list
row

filled with ones. The length of this list is
rowIndex + 1

.
Iteration and Update : Use nested loops to update the values of
row

. For each index
i

from 2 to
rowIndex

:

Update the elements of
row

from the end to the beginning, thus ensuring that no element is overwritten prematurely.

Pseudocode with Comments:

                                            
# Define function to get specific row of Pascal's Triangle
function getPascalRow(rowIndex):
# Step 1: Initialize a list 'row' of size rowIndex + 1 with all elements set to 1
row = [1] * (rowIndex + 1)

# Step 2: Iterate over each row from 2 to rowIndex
for i in range 2 to rowIndex + 1:
# Step 3: Update the row from right to left to avoid overwriting earlier values
for j in range i - 1 to 0 step -1:
# Each element is the sum of the two elements directly above it
row[j] = row[j] + row[j - 1]

# Return the final row which represents the rowIndex-th row of Pascal's Triangle
return row

Step-by-Step Explanation:

Initialize Row :

Start by creating a list named

row

with a length of

rowIndex + 1

, and fill it with

1

s. This lays the foundation, as the edges of Pascal's Triangle are always

1

Update Elements :

Beginning from the 2nd row to
rowIndex

-th row (since the 0th and 1st rows are trivially simple):

For each row
i

, iterate backwards from the second last element to the first element. This ensures that the updating of
row[j]

does not interfere with other updates yet to be performed in the iteration.
Update the current element
row[j]

by adding the value directly above it (
row[j - 1]

from the previous iteration).

Return Result :

By the end of the iterations, the list

row

will contain the values of the

rowIndex

-th row of Pascal's Triangle, which is then returned as the output.

This detailed explanation and pseudocode illustrate how to efficiently generate the specific row of Pascal's Triangle using only linear extra space relative to the input size.