Total number of different staircase that can made from N boxes

Given N boxes each of unit dimension i.e. (1×1 meter^2). The task is to find the total number of different staircases that can be made from those boxes with the following rules:

  • Staircase must be in strictly descending order.
  • Each staircase contains at least two steps. (Total steps is equal to the breadth of the staircase.)

Examples:

Input : N = 5
Output : 2
The two staircase are following :

Input : N = 6
Output : 3
The three staircase are following :

If we consider total steps = 2, we can observe the fact that the number of staircase is incremented by 1 if N is incremented by 2. We can illustrate the above thing from the following image :

Now, if total steps is greater than 2 (assume, total steps = K) then we can take this thing as first create a base (base requires boxes equal to the total steps) for staircase and put another staircase on it of steps size K and K – 1 having boxes N – K. (because K boxes already used to create base). Thus, we can solve this problem using bottom-up dynamic programming.

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ program to find the total number of
// different staircase that can made
// from N boxes
#include <iostream>
using namespace std;
  
// Function to find the total number of
// different staircase that can made
// from N boxes
int countStaircases(int N)
{
    // DP table, there are two states.
    // First describes the number of boxes
    // and second describes the step
    int memo[N + 5][N + 5];
  
    // Initilize all the elements of
    // the table to zero
    for (int i = 0; i <= N; i++) {
        for (int j = 0; j <= N; j++) {
            memo[i][j] = 0;
        }
    }
  
    // Base case
    memo[3][2] = memo[4][2] = 1;
  
    for (int i = 5; i <= N; i++) {
        for (int j = 2; j <= i; j++) {
  
            // When step is equal to 2
            if (j == 2) {
                memo[i][j] = memo[i - j][j] + 1;
            }
  
            // When step is greater than 2
            else {
                memo[i][j] = memo[i - j][j] + 
                             memo[i - j][j - 1];
            }
        }
    }
  
    // Count the total staircase
    // from all the steps
    int answer = 0;
    for (int i = 1; i <= N; i++) 
        answer = answer + memo[N][i];    
  
    return answer;
}
  
// Driver Code
int main()
{
    int N = 7;
  
    cout << countStaircases(N);
  
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java program to find the total number of
// different staircase that can made
// from N boxes
  
import java.util.*;
  
class GFG
{
        // Function to find the total number of
        // different staircase that can made
        // from N boxes
        static int countStaircases(int N)
        {
            // DP table, there are two states.
            // First describes the number of boxes
            // and second describes the step
            int [][] memo=new int[N + 5][N + 5];
          
            // Initilize all the elements of
            // the table to zero
            for (int i = 0; i <= N; i++) {
                for (int j = 0; j <= N; j++) {
                    memo[i][j] = 0;
                }
            }
          
            // Base case
            memo[3][2] = memo[4][2] = 1;
          
            for (int i = 5; i <= N; i++) {
                for (int j = 2; j <= i; j++) {
          
                    // When step is equal to 2
                    if (j == 2) {
                        memo[i][j] = memo[i - j][j] + 1;
                    }
          
                    // When step is greater than 2
                    else {
                        memo[i][j] = memo[i - j][j] + 
                                    memo[i - j][j - 1];
                    }
                }
            }
          
            // Count the total staircase
            // from all the steps
            int answer = 0;
            for (int i = 1; i <= N; i++) 
                answer = answer + memo[N][i]; 
          
            return answer;
        }
          
        // Driver Code
        public static void main(String [] args)
        {
            int N = 7;
          
            System.out.println(countStaircases(N));
          
              
        }
  
}
  
// This code is contributed 
// by ihritik

chevron_right


Python 3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python 3 program to find the total 
# number of different staircase that 
# can made from N boxes
  
# Function to find the total number 
# of different staircase that can 
# made from N boxes
def countStaircases(N):
  
    # DP table, there are two states.
    # First describes the number of boxes
    # and second describes the step
    memo = [[0 for x in range(N + 5)]
               for y in range(N + 5)]
  
    # Initilize all the elements of
    # the table to zero
    for i in range(N + 1):
        for j in range (N + 1):
            memo[i][j] = 0
          
    # Base case
    memo[3][2] = memo[4][2] = 1
  
    for i in range (5, N + 1) :
        for j in range (2, i + 1) :
  
            # When step is equal to 2
            if (j == 2) :
                memo[i][j] = memo[i - j][j] + 1
              
            # When step is greater than 2
            else :
                memo[i][j] = (memo[i - j][j] + 
                              memo[i - j][j - 1])
      
    # Count the total staircase
    # from all the steps
    answer = 0
    for i in range (1, N + 1):
        answer = answer + memo[N][i] 
  
    return answer
  
# Driver Code
if __name__ == "__main__":
  
    N = 7
  
    print (countStaircases(N))
  
# This code is contributed
# by ChitraNayal

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# program to find the total number 
// of different staircase that can made
// from N boxes
using System;
  
class GFG
{
      
// Function to find the total number 
// of different staircase that can 
// made from N boxes
static int countStaircases(int N)
{
    // DP table, there are two states.
    // First describes the number of boxes
    // and second describes the step
    int [,] memo = new int[N + 5, N + 5];
  
    // Initilize all the elements 
    // of the table to zero
    for (int i = 0; i <= N; i++) 
    {
        for (int j = 0; j <= N; j++)
        {
            memo[i, j] = 0;
        }
    }
  
    // Base case
    memo[3, 2] = memo[4, 2] = 1;
  
    for (int i = 5; i <= N; i++)
    {
        for (int j = 2; j <= i; j++) 
        {
  
            // When step is equal to 2
            if (j == 2) 
            {
                memo[i, j] = memo[i - j, j] + 1;
            }
  
            // When step is greater than 2
            else 
            {
                memo[i, j] = memo[i - j, j] + 
                             memo[i - j, j - 1];
            }
        }
    }
  
    // Count the total staircase
    // from all the steps
    int answer = 0;
    for (int i = 1; i <= N; i++) 
        answer = answer + memo[N, i]; 
  
    return answer;
}
  
// Driver Code
public static void Main()
{
    int N = 7;
  
    Console.WriteLine(countStaircases(N));
}
}
  
// This code is contributed 
// by Subhadeep

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?php
// PHP program to find the total 
// number of different staircase
// that can made from N boxes
  
// Function to find the total 
// number of different staircase 
// that can made from N boxes
function countStaircases($N)
{
      
    // Initilize all the elements 
    // of the table to zero
    for ($i = 0; $i <= $N; $i++)
    {
        for ($j = 0; $j <= $N; $j++) 
        {
            $memo[$i][$j] = 0;
        }
    }
  
    // Base case
    $memo[3][2] = $memo[4][2] = 1;
  
    for ($i = 5; $i <= $N; $i++) 
    {
        for ($j = 2; $j <= $i; $j++) 
        {
  
            // When step is equal to 2
            if ($j == 2) 
            {
                $memo[$i][$j] = $memo[$i - $j][$j] + 1;
            }
  
            // When step is greater than 2
            else 
            {
                $memo[$i][$j] = $memo[$i - $j][$j] + 
                                $memo[$i - $j][$j - 1];
            }
        }
    }
  
    // Count the total staircase
    // from all the steps
    $answer = 0;
    for ($i = 1; $i <= $N; $i++) 
        $answer = $answer + $memo[$N][$i]; 
  
    return $answer;
}
  
// Driver Code
$N = 7;
  
echo countStaircases($N);
  
// This code is contributed
// by Shivi_Aggarwal
?>

chevron_right


Output:

4

Time Complexity: O(n^2).



My Personal Notes arrow_drop_up

My name is Saurav Chandra I am from Rajasthan I am currently pursuing my B Tech from National Institute of Technology Hamirpur (H P) in Computer Science and Engineering Department My area of interests are Algorithms and Data Structures

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.