# How many possible ways the child can run up the ladder

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)

### One Thought on “How many possible ways the child can run up the ladder”

1. 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);
}