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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.
1#include <stdio.h>
2#include <stdlib.h>
3
4int* getRow(int rowIndex, int* returnSize) {
5 *returnSize = rowIndex + 1;
6 int* row = (int*)malloc(sizeof(int) * (*returnSize));
7 row[0] = 1;
8 for (int i = 1; i <= rowIndex; ++i) {
9 for (int j = i; j > 0; --j) {
10 row[j] = row[j] + row[j - 1];
11 }
12 }
13 return row;
14}
15
16int main() {
17 int returnSize;
18 int* result = getRow(3, &returnSize);
19 for (int i = 0; i < returnSize; ++i) {
20 printf("%d ", result[i]);
21 }
22 free(result);
23 return 0;
24}This C program achieves the solution using in-place updates for constructing Pascal's Triangle. It manages memory dynamically and prints the resulting row for a given rowIndex.
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.