My solution in C# builds the staircase row by row. For each row i (from 1 to n):

I print (n - i) spaces to align it to the right

Then I print i hash symbols (#)

This way the output grows step by step until the full staircase is printed.

    public static void staircase(int n)
{
    for(int i =1; i <= n; i++){
        int spaces = n - i;
        int hashtags = i;
        string staircase = new string(' ', spaces) + new string('#', hashtags);
        Console.WriteLine(staircase);
    }
}

Here is my solution in C (I know could be better) time complexity: O(n^2)

void staircase(int n) {

int m = 0;

for(int i = 1 ; i <= n ; i++)
{
    m++;
    for(int j = 0 ; j < n - i ; j++)
    {
        printf(" ");
    }

    for(int k = 0 ; k < m ; k++)
    {
        printf("#");
    }

    printf("\n");
}

}

var sb=new StringBuilder();

    for(var i=1;i<=n;i++)
    {
        sb.Append(new String(' ',n-i));        
        sb.Append(new String('#',i)+"\n");
    }

    Console.Write(sb.ToString());

Ussing operator overloading for fun, we all know it is not necessary.

string operator*(const std::string& str, int count){
    std::string result = "";
    for (int i = 0; i < count; ++i) {
        result += str;
    }
    return result;
}

string operator*(int count, const string& str) {
    return str * count;
}

void staircase(int n) {
    int steps = n;
    while (n-->0) {
        cout<<" "s*n + "#"s*(steps-n)<<endl;
    }
}

for (int i = 1; i <= n; i++) { string bosluklar = new string(' ', n - i); string kareler = new string('#', i); Console.WriteLine(bosluklar + kareler); }