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.
There are two methods to solve this problem
1. Recursive Method
2. Dynamic Programming
Examples:
Input : 4 Output : 7 Input : 3 Output : 4
1. Recursive Method
How Code is Working :
Suppose you have n stairs then you can hop either 1 step, 2 step, 3 step.
1. If you hop 1 step then remaining stairs = n-1
2. If you hop 2 step then remaining stairs = n-2
3. If you hop 3 step then remaining stairs = n-3
If you hop 1 step then again you can hop 1 step, 2 step, 3 step until n equals 0.
Repeat this process and count total number of ways to reach at nth stair using step 1, 2, 3.
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 */
Output :
7
How Code is Working….
The Time Complexity of the program is Optimized using Dynamic Programming.
2. Using Dynamic Programming
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 + 1)
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 + 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()
{
int n = 4;
Console.WriteLine(countWays(n));
}
}
/* This code is contributed by vt_m */
Output :
7
Output Explanation:
1 -> 1 -> 1 -> 1
1 -> 1 -> 2
1 -> 2 -> 1
1 -> 3
2 -> 1 -> 1
2 -> 2
3 -> 1
So Total ways: 7
Other Related Articles
http://www.geeksforgeeks.org/count-ways-reach-nth-stair/
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