Pascal's triangle

//To print the full pascal traingle TC. O(n^2)
class Solution {
public:
    vector<vector<int>> generate(int n) {
        vector <vector <int>> arr(n);
        for(int i=0;i<n;i++){
            arr[i].resize(i+1);  //making the row of required size
            arr[i][0]=arr[i][i]=1;  //making first and last element always 1
            for(int j=1;j<i;j++){   //taking care of middle elements
                arr[i][j]=arr[i-1][j-1]+arr[i-1][j];
            }
        }
        return arr;

    }
};

//To print a row of pascal's traingle
/*
Logic:
If you observe then you will find that the nth row elements are :
nc0,nc1,nc2,nc3,nc4,nc5....ncn(where n is 1 based indexing)
so we can calculate nc0,nc1,nc2...one by one.

Improvisation:
NCr = (NC(r-1) * (N - r + 1)) / r where 1 ≤ r ≤ N
so if NC0=1 keep it in prev variable
so NC1=1*(N-1+1) and so on.

#include <bits/stdc++.h>
using namespace std;
int main()
{
    int N;cin>>N;
    int prev = 1;   // nC0 = 1
    cout << prev;

    for (int i = 1; i <= N; i++) {
        // nCr = (nCr-1 * (n - r + 1))/r
        int curr = (prev * (N - i + 1)) / i;
        cout << ", " << curr;  //printing ith element of row
        prev = curr;
    return 0;
}


*/

/*
Same above approach if row needed considering large number as the modulo of 10^9+7
*/
ll val=pow(10,9)+7;
ll gcdExtended(ll a, ll b, ll *x, ll *y)
{
    // Base Case
    if (a == 0)
    {
        *x = 0, *y = 1;
        return b;
    }

    ll x1, y1; // To store results of recursive call
    ll gcd = gcdExtended(b%a, a, &x1, &y1);

    // Update x and y using results of recursive
    // call
    *x = y1 - (b/a) * x1;
    *y = x1;

    return gcd;
}
ll modInverse(ll b, ll m)
{
    ll x, y; // used in extended GCD algorithm
    ll g = gcdExtended(b, m, &x, &y);

    // Return -1 if b and m are not co-prime
    if (g != 1)
        return -1;

    // m is added to handle negative x
    return (x%m + m) % m;
}
ll modDivide(ll a, ll b, ll m){
    a = a % m;
    ll inv = modInverse(b, m);
    return ((inv * a) % m);
}
class Solution{
public:
    vector<ll> nthRowOfPascalTriangle(int n) {
        n--;
        ll prev = 1;
        vector <ll> res;
        res.push_back(1);
    for (int i = 1; i <= n; i++) {
        // nCr = (nCr-1 * (n - r + 1))/r
        ll curr = (prev * (n - i + 1))%val;  //This is NC(r-1)*(n-i+1)
        curr=modDivide(curr,i,val);  //This is curr/i
        res.push_back(curr);
        prev = curr;
    }
    return res;
    }
};