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)
).
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
