Sponsored
Sponsored
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.
1function getRow(rowIndex) {
2 let row = new Array(rowIndex + 1).fill(0);
3 row[0] = 1;
4 for (let i = 1; i <= rowIndex; i++) {
5 for (let j = i; j > 0; j--) {
6 row[j] += row[j - 1];
7 }
8 }
9 return row;
10}
11
12console.log(getRow(3));This JavaScript program uses an array filled with zeroes, updating its values iteratively to achieve the desired row of Pascal's Triangle.
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.
1#include <vector>
class Solution {
public:
std::vector<int> getRow(int rowIndex) {
std::vector<int> row(rowIndex + 1, 1);
long long C = 1;
for (int i = 1; i < rowIndex; ++i) {
C = C * (rowIndex - i + 1) / i;
row[i] = C;
}
return row;
}
};
int main() {
Solution sol;
std::vector<int> result = sol.getRow(3);
for (int num : result) {
std::cout << num << " ";
}
return 0;
}This C++ solution calculates each element of the row using the binomial coefficient, effectively leveraging integer math skills to minimize algebraic errors and inaccuracies.