### Example 1: Recursively count the number of elements in an array

This is a solution to this kata.

``````function deepCount(arr) {
let count = 0

// Loop through each item
for (let item of arr) {
// If element is an array, call deepCount again
if (Array.isArray(item)) {
count += deepCount(item)
}

// Otherwise, increment count
count += 1
}

return count
}

// deepCount([]) => 0
// deepCount([1, 2, 3]) => 3
// deepCount(["x", "y", ["z"]]) => 4
// deepCount([1, 2, [3, 4, [5]]]) => 7``````

### Example 2: Manipulate function argument until some condition is true

This is a solution to this kata:

So, your task is to sum up all the digits in string, possibly multiple times, till you get a one digit result. You should then return it as a string. The input will be always valid.

``````const str = '1234'
digitSum(str) // returns 1

/* 1 + 2 + 3 + 4 = 10 => this result doesn't have one digit => 1 + 0 = 1 */``````
``````function digitSum(numStr) {
// Get digits sum (e.g., 10)
const digitsSum = numStr.split('').reduce((sum, num) => sum + Number(num), 0).toString()

// If sum.length === 1, return sum
if (digitsSum.length === 1)  {
return digitsSum
}

// Else, call function again
return digitSum(digitsSum)
}``````

### Example 3

Solution to this kata. I need to elaborate on how this works.

``````function getElement(array, indices) {
if (indices.length === 1) {
return array[indices]
}

return getElement(array[indices[0]], indices. Slice(1))
}``````

### Example 4

Solution to this kata:

``````function removeParentheses(str) {
const result = str.replace(/\([A-Za-z0-9\s]*\)/g, '')

if (result.includes('(') && result.includes(')')) {
return removeParentheses(result)
}

return result
}``````
Tagged: Snippets