Solution for kth row of Pascal's triangle for a job interview

Question:


Given an index \$k\$, return the \$k^{th}\$ row of the Pascal's triangle.


For example, given \$k = 3\$, return \$[1,3,3,1]\$. Bonus points for using \$O(k)\$ space.

Can it be further optimized using this way or another?

class Solution {
   public:
    long long C(int n, int r) {
       if(r > n / 2) r = n - r; // because C(n, r) == C(n, n - r)
       long long ans = 1;
       int i;

       for(i = 1; i  getRow(int rowIndex) {
           vector v;
           long long sol;
           for(int i=0;i<=rowIndex;i++)
           {sol=C(rowIndex,i);
           v.push_back(sol);}

          return v; 
       }
   };

For one thing, the inconsistency with whitespace use and curly brace placing may demonstrate a lack of attention to detail. Before you attack the actual problem, make sure your code is written cleanly.

This:

for(int i=0;i<=rowIndex;i++)

should use some whitespace:

for (int i = 0; i <= rowIndex; i++)

You already do this in other places, and it should be done everywhere.

Also, this:

for(int i=0;i<=rowIndex;i++)
{sol=C(rowIndex,i);
v.push_back(sol);}

looks a bit sloppy and is harder to read. Again, use more whitespace, but also separate the loop body from the curly braces.

for (int i = 0; i <= rowIndex; i++) {
    sol = C(rowIndex, i);
    v.push_back(sol);
}

StackExchange Code Review Q#52028, answer score: 32