# Total number of different staircase that can made from N boxes

Given N boxes of unit dimension, i.e. (1×1 ). 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 ` `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

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## Javascript

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Output:

`4`

Time Complexity: O().

Auxiliary Space: O(n ^ 2)

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