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The iterative approach builds each row of Pascal's Triangle from the topmost row down to the target row, updating values in-place. The key observation is that each element at position j in a row is the sum of the element at position j and j-1 from the previous row. By starting from the end of the row, you can use the same array for updating values without overwriting the necessary data.
Time Complexity: O(n^2) where n is rowIndex as it involves nested iteration to build each row.
Space Complexity: O(n) for storing the result row.
1def getRow(rowIndex):
2 row = [0] * (rowIndex + 1)
3 row[0] = 1
4 for i in range(1, rowIndex + 1):
5 for j in range(i, 0, -1):
6 row[j] += row[j - 1]
7 return row
8
9print(getRow(3))This Python implementation uses a list initialized with zeroes. It applies in-place updates for calculating the values based on Pascal's Triangle properties.
This approach utilizes the combinatorial formula for a specific row in Pascal's Triangle. Specifically, the k-th element in the n-th row is given by the binomial coefficient: C(n, k) = n! / (k! * (n-k)!). Using this fact, we can derive all the values of the row without constructing the triangle iteratively.
Time Complexity: O(n).
Space Complexity: O(n) for storing the row.
1using System.Collections.Generic;
public class Solution {
public IList<int> GetRow(int rowIndex) {
IList<int> row = new List<int> { 1 };
long C = 1;
for (int i = 1; i <= rowIndex; i++) {
C = C * (rowIndex - i + 1) / i;
row.Add((int)C);
}
return row;
}
public static void Main() {
Solution sol = new Solution();
var result = sol.GetRow(3);
Console.WriteLine(string.Join(", ", result));
}
}This C# program calculates the row of Pascal's Triangle using the binomial coefficient with iteratively computed products and divisions.