vector Solution::getRow(int A) {

vector arr={1};

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

arr.push_back((arr[i]*(A-i))/(i+1));

return arr;

}

# Very Easy and Simple Code ... Whole code in no more than 5 line

**stone**#3

@harshit0001

it is equivalent to

vector arr;

arr.push_back(1);

the first element is initialised to 1

elegent solution by the way

please and thank you

concept of nCr and nCr+1 is used here as elements of the kth row would be [1,nC1,nC2,nC3,…nCr,nCr+1,…1]. and nCr+1=nCr*(n-r)/(r+1). therefore every next element of the row can easily be found by the previous element with the help of this formula.I hope its clear