Last Updated: 15 Oct, 2020

# Count Ways To Reach The N-th Stairs

Moderate

## Problem statement

#### You are supposed to return the number of distinct ways you can climb from the 0th step to the Nth step.

##### Example :
``````N=3
``````

``````We can climb one step at a time i.e. {(0, 1) ,(1, 2),(2,3)} or we can climb the first two-step and then one step i.e. {(0,2),(1, 3)} or we can climb first one step and then two step i.e. {(0,1), (1,3)}.
``````

#### Input format :

``````The first line contains an integer 'T', which denotes the number of test cases or queries to be run. Then the test cases follow.

The first and the only argument of each test case contains an integer 'N', representing the number of stairs.
``````
##### Output format :
``````For each test case/query, print the number of distinct ways to reach the top of stairs. Since the number can be huge, so return output modulo 10^9+7.

Output for every test case will be printed in a separate line.
``````
##### Note :
``````You do not need to print anything. It has already been taken care of.
``````
##### Constraints :
``````1 <= 'T' <= 10
0 <= 'N' <= 10^5

Where 'T' is the number of test cases, and 'N' is the number of stairs.

It is guaranteed that sum of 'N' over all test cases is <= 10^5.
``````

## Approaches

### 01 Approach

One basic approach is to explore all possible steps which can be climbed with either taking one step or two steps. So at every step, we have two options to climb the stairs either we can climb with one step, or we can climb with two steps. So the number of ways  can be recursively defined as :

``countDistinctWayToClimbStair ( currStep, N ) = countDistinctWayToClimbStair ( currStep+1, N ) + countDistinctWayToClimbStair ( currStep + 2, N )``

Where “currStep” denotes the current step on which the person is standing, and N denotes the destination step.