Count ways to reach the n'th stair - GeeksforGeeks


Count ways to reach the n'th stair - GeeksforGeeks
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.


    ways(n) = ways(n-1) + ways(n-2)
there is one thing to notice, the value of ways(n) is equal to fibonacci(n+1).
ways(1) = fib(2) = 1
ways(2) = fib(3) = 2
ways(3) = fib(4) = 3

Generalization of the above problem
How to count number of ways if the person can climb up to m stairs for a given value m? For example if m is 4, the person can climb 1 stair or 2 stairs or 3 stairs or 4 stairs at a time.
We can write the recurrence as following.
   ways(n, m) = ways(n-1, m) + ways(n-2, m) + ... ways(n-m, m) 
int countWaysUtil(int n, int m)
{
    int res[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
int countWays(int s, int m)
{
    return countWaysUtil(s+1, m);
}

// A recursive function used by countWays
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;
}
Program for Fibonacci numbers
http://www.geeksforgeeks.org/program-for-nth-fibonacci-number/
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 141, ……..
DP:
  for (i = 2; i <= n; i++)
  {
     c = a + b;
     a = b;
     b = c;
  }
Method 4 ( Using power of the matrix {{1,1},{1,0}} ) Log(N)
The matrix representation gives the following closed expression for the Fibonacci numbers:
fibonaccimatrix

int fib(int n)
{
  int F[2][2] = {{1,1},{1,0}};
  if (n == 0)
      return 0;
  power(F, n-1);
 
  return F[0][0];
}
 
void multiply(int F[2][2], int M[2][2])
{
  int x =  F[0][0]*M[0][0] + F[0][1]*M[1][0];
  int y =  F[0][0]*M[0][1] + F[0][1]*M[1][1];
  int z =  F[1][0]*M[0][0] + F[1][1]*M[1][0];
  int w =  F[1][0]*M[0][1] + F[1][1]*M[1][1];
 
  F[0][0] = x;
  F[0][1] = y;
  F[1][0] = z;
  F[1][1] = w;
}
 
void power(int F[2][2], int n)
{
  int i;
  int M[2][2] = {{1,1},{1,0}};
 
  // n - 1 times multiply the matrix to {{1,0},{0,1}}
  for (i = 2; i <= n; i++)
      multiply(F, M);
}

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