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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.
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
This JavaScript implementation uses the properties of binomial coefficients to derive each subsequent value in the row without floating-point inaccuracies due to JavaScript's numeric limits.