# How to implement backtracking for a climbing stairs practice in JavaScript?

In the given problem statement we are asked to implement backtracking for a climbing stairs practice with the help of javascript functionalities. In the data structures backtracking is popularly known to explore all the possible solutions. We can solve this problem with the help of a backtracking algorithm.

## What is the Backtracking Technique ?

Let's understand the working of a Backtracking technique.

Backtracking is a well known algorithm technique, which is basically used to solve the problem statements that include searching for all possible solutions. It is based on a depthâˆ’first strategy and it is used in combination with a recursive method.

The basic ideology for this technique is to explore all the paths in a search environment by increasingly making a candidate solution and testing if it satisfies the constraints. This process is repeated until all possible paths give a valid solution.

The key components of this technique are:

• Candidate solution is the partial solution that is being constructed incrementally.

• Constraints are the rules that the candidate solution must satisfy to be considered valid.

• Feasible solution is a candidate solution that satisfies all the constraints.

• Backtracking is the process of leaving a candidate solution that violates a constraint and returning to the previous decision point to explore a different path.

It is a widely used technique to solve various kinds of problems like graph problems, constraint satisfaction problems and combinatorial optimization problems. It can also help to significantly reduce the search space and improve the performance of algorithms.

## Logic for The Above Problem

The easiest way to implement backtracking for a climbing stairs practice is by using the helper function.

So let's understand the logic of the problem given. The problem of climbing stairs is a common example that can be solved using backtracking. If we have a staircase of n steps and the possibility to climb either one or two steps at a time, we need to find out the total number of distinct ways to climb to the top of the staircase.

### Algorithm

Step 1: Define a function named climbStairs() takes an integer n as an argument. The n is the number of stairs. This function will return the total number of distinct ways to climb to the top of the staircase.

Step 2: Now define a backtrack function which is a helper function. It is taking a parameter step representing the current step in the climb.

Step 3: Now the above function checks that the current step is greater than the total number of steps n.

Step 4: This step will determine that if the third step is true then the function returns as this is an invalid solution.

Step 5: If the current step is equal to n, we have reached the top of the staircase and we will increment the count variable to indicate that we have found a valid solution.

Step 6: If the fifth step is not valid then we recursively call the backtrack function with step +1 and step +2 to explore all the possible paths up to the staircase.

Step 7: So in the end the climbStairs function will be called with an initial step value to 0 and gives the count variable once all possible paths have been explored.

### Example

//function to calculate climbing stairs
function climbStairs(n) {
let count = 0;
//define backtrack function
function backtrack(step) {
if (step > n) {
return;
}
if (step === n) {
count++;
return;
}
backtrack(step + 1);
backtrack(step + 2);
}

backtrack(0);
return count;
}

const num = 8;
console.log(climbStairs(num));


### Output

34


## Complexity

This implementation is taking O(2^n) time to complete the execution of the function named climbStairs(). Because there are only two choices at each step of the climb. So we can use dynamic programming to optimize this solution and reduce the time complexity to O(n).

## Conclusion

In this code we have implemented a backtracking for climbing stairs practice in javascript. Here we have created all the possible solutions to get the final output. The time complexity for this implementation is O(2^n).

Updated on: 23-Aug-2023

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