A child is running up a ladder with n steps, and can hop either 1 step or 2 steps at a time. Calculate how many possible ways the child can run up the ladder.

This problem is similar to Fibonacci Problem. Each step n can be reached from either two steps below (n-2) or one step below (n-1), thus the number of possibilities to reach that step is the sum of the possibilities to reach those other two steps. Base cases would be:

1. If there is only 1 step then child can run up the ladder only one way by taking 1 step.

2. If there are 2 steps then child can run up the ladder 2 ways by taking a single 2 step or taking two 1 steps.

int countWays(int n) { if(n==0) return 0; else if(n==1||n==2) return n; else return countWays(n-1)+countWays(n-2); }

Like the Fibonacci problem,the runtime of this problem is exponential O(2^n)

My Solution

//starts from n = 0 and has to reach n = N. Everytime it is able to reach there,

//one unique path is found.

int countWays(int n, final int N){

if (n > N) //over shoots

{ return 0; }

else if(n == N)// reaches the top, or climbs exactly N steps

{ return 1; } //one unique way

else{ //for current n, find no of unique ways from n+1 and n+2 to reach N

return countWays(n+1,N) + countWays(n+2,N);

}