Examples usages of recursion

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
}