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  1. Prepare
  2. Algorithms
  3. Warmup
  4. Staircase
  5. Discussions

Staircase

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5177 Discussions

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  • anarod_val25
     + 0 comments

    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;
    }
}

  • abdullah2000_da1
     + 0 comments

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

  • sakthisakthivel4
     + 0 comments

    def staircase(n):
    for i in range(1,n+1):
    print (("#"*i).rjust(n," "))

  • 24f2006425
     + 0 comments
    Here is my solution in java, time complexity O(n^2)`

public static void staircase(int n) {
    // Write your code here
    for(int i=0;i<n;i++)
    {
        for(int j=0;j<n;j++)
        {   
            if(j<n-(i+1))
            System.out.print(" ");
            else
            System.out.print("#");
        }
        System.out.println();
    }
    }

    `

  • oguzhn_dgd
     + 0 comments

    Here is my solution for javascript: 

    function staircase(n) {
    
    let star = ""
    for (let i = 0; i < n ; i++) {
        
        for (let j = n - 1; j > i; j--) {
                star += " ";
        }

        for (let k = i; k >= 0; k--) {
            star += "#";
        }
        
        star += "\n";  
    }
    console.log(star)

}