The ultimate guide to recursion interviews in JavaScript. Developed by FAANG engineers. Practice with real-world interview questions and get interview-ready in just a few hours.

If you’ve ever struggled with solving problems using recursion, or if you have to brush up your skills for an interview, this course is for you.

We'll start with the basics of what recursion is and why it’s important before diving into practicing solving actual questions. You’ll have access to detailed explanations and visualizations for each problem to help you along the way. 

By the time you have completed this course, you’ll be able to use all the concepts to solve complex real-world problems. We hope that through practicing recursion, you will be able to ace all recursion related questions in interviews at top tech companies.

Recursion for Coding Interviews in JavaScript

function average(testVariable, currentIndex = 0) {
	// Base Case
	if (currentIndex == testVariable.length - 1) { 
		return testVariable[currentIndex]
	}
	
  	// Recursive case1
	// When currentIndex is 0
	if (currentIndex == 0) { 
		return ((testVariable[currentIndex] + average(testVariable, currentIndex + 1)) / testVariable.length)
	}
	
  	// Recursive case2
	// Compute sum 
	return (testVariable[currentIndex] + average(testVariable, currentIndex + 1)) 
}

// Driver code 
array = [10, 2, 3, 4, 8, 0] 
console.log(average(array)) 

### Explanation

> The average of an array containing only $1$ number, is the number itself. This condition can be our **base case**. 
$$
\displaystyle average \: of \: array \: of \: length \: 1={\frac {a_{0}}{1}} = a_{0}
$$

For the **recursive case** we have two conditions:
1. If the `currentIndex == 0`.
   * This is our initial condition. The element `testVariable[currentIndex]` is saved for later use. It will be added in the final result. The `currentIndex` is moved one step forward and another instance of `average()` function is called. 
   * When this child function will return, the *sum* of all the numbers of the array would have been accumulated. Now, we will *divide* that total sum with the length of the array.

2. If the `currentIndex != 0`.
   * We continue accumulating the sum of the numbers of the array.

In each recursive call, we move the `currentIndex` one step ahead. When `currentIndex` reaches `testVariable.length - 1`, we add all the numbers in the reverse order. When we have added all the numbers, in the last step, i.e., when `currentIndex == 0`, we divide the sum by the length of the array.

Let's take a look at the sequence of function calls for this example:

## Explanation

> The average of an array containing only $1$ number, is the number itself. This condition can be our **base case**. 
$$
\displaystyle average \: of \: array \: of \: length \: 1={\frac {a_{0}}{1}} = a_{0}
$$

For the **recursive case** we have two conditions:
1. If the `currentIndex == 0`.
   * This is our initial condition. The element `testVariable[currentIndex]` is saved for later use. It will be added in the final result. The `currentIndex` is moved one step forward and another instance of `average()` function is called. 
   * When this child function will return, the *sum* of all the numbers of the array would have been accumulated. Now, we will *divide* that total sum with the length of the array.

2. If the `currentIndex != 0`.
   * We continue accumulating the sum of the numbers of the array.

In each recursive call, we move the `currentIndex` one step ahead. When `currentIndex` reaches `testVariable.length - 1`, we add all the numbers in the reverse order. When we have added all the numbers, in the last step, i.e., when `currentIndex == 0`, we divide the sum by the length of the array.

Let's take a look at the sequence of function calls for this example:

This review provides a detailed analysis of the solution to find the average of numbers.

Recursion Fundamentals

Iteration Vs Recursion

Recursion with Numbers

Recursion with Strings

Recursion with Arrays

Recursion with Data Structures

Conclusion

Solution Review: Average of Numbers

Solution: Using Recursion

Explanation