Skip to content
Related Articles

Related Articles

Improve Article
Total number of different staircase that can made from N boxes
  • Difficulty Level : Hard
  • Last Updated : 21 May, 2021

Given N boxes 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: 

  • The staircase must be in strictly descending order.
  • Each staircase contains at least two steps. ( The total steps are equal to the breadth of the staircase.)

Examples

Input : N = 5 
Output : 2 
The two staircases are the following : 

Input : N = 6 
Output : 3 
The three staircases are the following :  



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

Now, if the total steps are 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 the staircase and put another staircase on it of steps size K and K – 1 have 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++




// 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];
 
    // Initialize 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;
}

Java




// 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];
         
            // Initialize 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

Python 3




# 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)]
 
    # Initialize 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

C#




// 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];
 
    // Initialize 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

PHP




<?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)
{
     
    // Initialize 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
?>

Javascript




<script>
        // Javascript 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)
        {
            // DP table, there are two states.
            // First describes the number of boxes
            // and second describes the step
            let memo=new Array(N + 5);
             for(let i=0;i<N+5;i++)
            {
                memo[i]=new Array(N+5);
                for(let j=0;j<N+5;j++)
                {
                    memo[i][j]=0;
                }
            }
            // Initialize all the elements of
            // the table to zero
            for (let i = 0; i <= N; i++) {
                for (let j = 0; j <= N; j++) {
                    memo[i][j] = 0;
                }
            }
          
            // Base case
            memo[3][2] = memo[4][2] = 1;
          
            for (let i = 5; i <= N; i++) {
                for (let 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
            let answer = 0;
            for (let i = 1; i <= N; i++)
                answer = answer + memo[N][i];
          
            return answer;
           }
     
        // Driver Code
        let N = 7;
        document.write(countStaircases(N));
     
 
// This code is contributed by rag2127
</script>
Output: 
4

 

Time Complexity: O(n^2   ).
 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with industry experts, please refer Geeks Classes Live




My Personal Notes arrow_drop_up
Recommended Articles
Page :