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 <iostream>using namespace std;// Function to find the total number of// different staircase that can made// from N boxesint 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 Codeint 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 boxesimport 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 boxesdef 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 Codeif __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 boxesusing System;class GFG{ // Function to find the total number // of different staircase that can // made from N boxesstatic 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 Codepublic 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 boxesfunction 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(
).
Auxiliary Space: O(n ^ 2)
).
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