A child is running up a staircase with n steps and can hop either 1 step, 2 steps, or 3 steps at a time. Implement a method to count how many possible ways the child can run up the stairs.
Examples:
Input : 4 Output : 7 Explantion: Below are the four ways 1 step + 1 step + 1 step + 1 step 1 step + 2 step + 1 step 2 step + 1 step + 1 step 1 step + 1 step + 2 step 2 step + 2 step 3 step + 1 step 1 step + 3 step Input : 3 Output : 4 Explantion: Below are the four ways 1 step + 1 step + 1 step 1 step + 2 step 2 step + 1 step 3 step
There are two methods to solve this problem:
- Recursive Method
- Dynamic Programming
Method 1: Recursive.
There are n stairs, and a person is allowed to jump next stair, skip one stair or skip two stairs. So there are n stairs. So if a person is standing at i-th stair, the person can move to i+1, i+2, i+3-th stair. A recursive function can be formed where at current index i the function is recursively called for i+1, i+2 and i+3 th stair.
There is another way of forming the recursive function. To reach a stair i, a person has to jump either from i-1, i-2 or i-3 th stair or i is the starting stair.
Algorithm:
- Create a recursive function (count(int n)) which takes only one parameter.
- Check the base cases. If the value of n is less than 0 then return 0, and if the value of n is equal to zero then return 1 as it is the starting stair.
- Call the function recursively with values n-1, n-2 and n-3 and sum up the values that are returned, i.e. sum = count(n-1) + count(n-2) + count(n-3)
- Return the value of the sum.
C++
// C++ Program to find n-th stair using step size // 1 or 2 or 3. #include <iostream> using namespace std; class GFG { // Returns count of ways to reach n-th stair // using 1 or 2 or 3 steps. public : int findStep( int n) { if (n == 1 || n == 0) return 1; else if (n == 2) return 2; else return findStep(n - 3) + findStep(n - 2) + findStep(n - 1); } }; // Driver code int main() { GFG g; int n = 4; cout << g.findStep(n); return 0; } // This code is contributed by SoM15242 |
C
// Program to find n-th stair using step size // 1 or 2 or 3. #include <stdio.h> // Returns count of ways to reach n-th stair // using 1 or 2 or 3 steps. int findStep( int n) { if (n == 1 || n == 0) return 1; else if (n == 2) return 2; else return findStep(n - 3) + findStep(n - 2) + findStep(n - 1); } // Driver code int main() { int n = 4; printf ( "%d\n" , findStep(n)); return 0; } |
Java
// Program to find n-th stair // using step size 1 or 2 or 3. import java.util.*; import java.lang.*; public class GfG { // Returns count of ways to reach // n-th stair using 1 or 2 or 3 steps. public static int findStep( int n) { if (n == 1 || n == 0 ) return 1 ; else if (n == 2 ) return 2 ; else return findStep(n - 3 ) + findStep(n - 2 ) + findStep(n - 1 ); } // Driver function public static void main(String argc[]) { int n = 4 ; System.out.println(findStep(n)); } } /* This code is contributed by Sagar Shukla */ |
Python
# Python program to find n-th stair # using step size 1 or 2 or 3. # Returns count of ways to reach n-th # stair using 1 or 2 or 3 steps. def findStep( n) : if (n = = 1 or n = = 0 ) : return 1 elif (n = = 2 ) : return 2 else : return findStep(n - 3 ) + findStep(n - 2 ) + findStep(n - 1 ) # Driver code n = 4 print (findStep(n)) # This code is contributed by Nikita Tiwari. |
C#
// Program to find n-th stair // using step size 1 or 2 or 3. using System; public class GfG { // Returns count of ways to reach // n-th stair using 1 or 2 or 3 steps. public static int findStep( int n) { if (n == 1 || n == 0) return 1; else if (n == 2) return 2; else return findStep(n - 3) + findStep(n - 2) + findStep(n - 1); } // Driver function public static void Main() { int n = 4; Console.WriteLine(findStep(n)); } } /* This code is contributed by vt_m */ |
PHP
<?php // PHP Program to find n-th stair // using step size 1 or 2 or 3. // Returns count of ways to // reach n-th stair using // 1 or 2 or 3 steps. function findStep( $n ) { if ( $n == 1 || $n == 0) return 1; else if ( $n == 2) return 2; else return findStep( $n - 3) + findStep( $n - 2) + findStep( $n - 1); } // Driver code $n = 4; echo findStep( $n ); // This code is contributed by m_kit ?> |
Output :
7
Complexity Analysis:
- Time Complexity: O(3n).
The time complexity of the above solution is exponential, a close upper bound will be O(3n). From each state, 3 recursive function are called. So the upperbound for n states is O(3n). - Space Complexity:O(1).
As no extra space is required.
Note: The Time Complexity of the program can be optimized using Dynamic Programming.
Method 2: Dynamic Programming.
The idea is similar, but it can be observed that there are n states but the recursive function is called 3 ^ n times. That means that some states are called repeatedly. So the idea is to store the value of states. This can be done in two ways.
- Top-Down Approach: The first way is to keep the recursive structure intact and just store the value in a HashMap and whenever the function is called again return the value store without computing ().
- Bottom-Up Approach: The second way is to take an extra space of size n and start computing values of states from 1, 2 .. to n, i.e. compute values of i, i+1, i+2 and then use them to calculate the value of i+3.
Algorithm:
- Create an array of size n + 1 and initialize the first 3 variables with 1, 1, 2. The base cases.
- Run a loop from 3 to n.
- For each index i, computer value of ith position as dp[i] = dp[i-1] + dp[i-2] + dp[i-3].
- Print the value of dp[n], as the Count of the number of ways to reach n th step.
C++
// A C++ program to count number of ways // to reach n't stair when #include <iostream> using namespace std; // A recursive function used by countWays int countWays( int n) { int res[n + 1]; res[0] = 1; res[1] = 1; res[2] = 2; for ( int i = 3; i <= n; i++) res[i] = res[i - 1] + res[i - 2] + res[i - 3]; return res[n]; } // Driver program to test above functions int main() { int n = 4; cout << countWays(n); return 0; } //This code is contributed by shubhamsingh10 |
C
// A C program to count number of ways // to reach n't stair when #include <stdio.h> // A recursive function used by countWays int countWays( int n) { int res[n + 1]; res[0] = 1; res[1] = 1; res[2] = 2; for ( int i = 3; i <= n; i++) res[i] = res[i - 1] + res[i - 2] + res[i - 3]; return res[n]; } // Driver program to test above functions int main() { int n = 4; printf ( "%d" , countWays(n)); return 0; } |
Java
// Program to find n-th stair // using step size 1 or 2 or 3. import java.util.*; import java.lang.*; public class GfG { // A recursive function used by countWays public static int countWays( int n) { int [] res = new int [n + 1 ]; res[ 0 ] = 1 ; res[ 1 ] = 1 ; res[ 2 ] = 2 ; for ( int i = 3 ; i <= n; i++) res[i] = res[i - 1 ] + res[i - 2 ] + res[i - 3 ]; return res[n]; } // Driver function public static void main(String argc[]) { int n = 4 ; System.out.println(countWays(n)); } } /* This code is contributed by Sagar Shukla */ |
Python
# Python program to find n-th stair # using step size 1 or 2 or 3. # A recursive function used by countWays def countWays(n) : res = [ 0 ] * (n + 2 ) res[ 0 ] = 1 res[ 1 ] = 1 res[ 2 ] = 2 for i in range ( 3 , n + 1 ) : res[i] = res[i - 1 ] + res[i - 2 ] + res[i - 3 ] return res[n] # Driver code n = 4 print (countWays(n)) # This code is contributed by Nikita Tiwari. |
C#
// Program to find n-th stair // using step size 1 or 2 or 3. using System; public class GfG { // A recursive function used by countWays public static int countWays( int n) { int [] res = new int [n + 2]; res[0] = 1; res[1] = 1; res[2] = 2; for ( int i = 3; i <= n; i++) res[i] = res[i - 1] + res[i - 2] + res[i - 3]; return res[n]; } // Driver function public static void Main() { int n = 4; Console.WriteLine(countWays(n)); } } /* This code is contributed by vt_m */ |
PHP
<?php // A PHP program to count // number of ways to reach // n'th stair when // A recursive function // used by countWays function countWays( $n ) { $res [0] = 1; $res [1] = 1; $res [2] = 2; for ( $i = 3; $i <= $n ; $i ++) $res [ $i ] = $res [ $i - 1] + $res [ $i - 2] + $res [ $i - 3]; return $res [ $n ]; } // Driver Code $n = 4; echo countWays( $n ); // This code is contributed by ajit ?> |
Output :
7
1 -> 1 -> 1 -> 1 1 -> 1 -> 2 1 -> 2 -> 1 1 -> 3 2 -> 1 -> 1 2 -> 2 3 -> 1 So Total ways: 7
Complexity Analysis:
- Time Complexity: O(n).
Only one traversal of the array is needed. So Time Complexity is O(n). - Space Complexity: O(n).
To store the values in a DP, n extra space is needed.
Other Related Articles
http://www.geeksforgeeks.org/count-ways-reach-nth-stair/
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