There are n stairs, a person standing at the bottom wants to reach the top. The person can climb either 1 stair or 2 stairs at a time. Count the number of ways, the person can reach the top.
// C# program to count the // number of ways to reach // n'th stair using System;
class GFG {
// A simple recursive
// program to find n'th
// fibonacci number
static int fib( int n)
{
if (n <= 1)
return n;
return fib(n - 1) + fib(n - 2);
}
// Returns number of ways
// to reach s'th stair
static int countWays( int s)
{
return fib(s + 1);
}
// Driver Code
static public void Main()
{
int s = 4;
Console.WriteLine( "Number of ways = " + countWays(s));
}
} // This code is contributed // by akt_mit |
Number of ways = 5
The time complexity of the above implementation is exponential (golden ratio raised to power n). It can be optimized to work in O(Logn) time using the previously discussed Fibonacci function optimizations.
// C# program to Count ways to reach // the n’th stair using System;
class GFG {
// A recursive function used by
// countWays
static int countWaysUtil( int n, int m)
{
if (n <= 1)
return n;
int res = 0;
for ( int i = 1; i <= m && i <= n; i++)
res += countWaysUtil(n - i, m);
return res;
}
// Returns number of ways to reach
// s'th stair
static int countWays( int s, int m)
{
return countWaysUtil(s + 1, m);
}
/* Driver program to test above function */
public static void Main()
{
int s = 4, m = 2;
Console.Write( "Number of ways = "
+ countWays(s, m));
}
} // This code is contributed by nitin mittal. |
Number of ways = 5
Time complexity: O(2n)
Auxiliary Space: O(n)
// C# program to count number // of ways to reach n't stair when // a person can climb 1, 2, ..m // stairs at a time using System;
class GFG {
// A recursive function
// used by countWays
static int countWaysUtil( int n, int m)
{
int [] res = new int [n];
res[0] = 1;
res[1] = 1;
for ( int i = 2; i < n; i++) {
res[i] = 0;
for ( int j = 1; j <= m && j <= i; j++)
res[i] += res[i - j];
}
return res[n - 1];
}
// Returns number of ways
// to reach s'th stair
static int countWays( int s, int m)
{
return countWaysUtil(s + 1, m);
}
// Driver Code
public static void Main()
{
int s = 4, m = 2;
Console.WriteLine( "Number of ways = " + countWays(s, m));
}
} // This code is contributed by anuj_67. |
Number of ways = 5
Time Complexity: O(m*n)
Auxiliary Space: O(n)
Please refer complete article on Count ways to reach the n’th stair for more details!