Know Your Neighbor

The challenge:

Create a function that takes a string as an argument and returns true if each letter in the string is surrounded by a plus sign. Return false otherwise.

Examples

plusSign(“+f+d+c+#+f+”) ➞ true

plusSign(“+d+=3=+s+”) ➞ true

plusSign(“f++d+g+8+”) ➞ false

plusSign(“+s+7+fg+r+8+”) ➞ false

Notes

For clarity, each letter must have a plus sign on both sides.

My Solution

My solution was a bit more verbose than some.

function plusSign(str) {
	// Put all the string's characters onto an array.
	const arr = [...str]
	//Initiate an array to hold all letters that are flanked by '+'
  let letters = []
  for (let i=0; i< arr.length; i++) {
    letters.push(arr[i].match(/[a-z]/g) && arr[i-1] === '+' && arr[i+1] === '+' ? arr[i] : '')
  }
	// return a if the length of the letters array is the same as the length of the array `arr` with the letters filtered out.  
  return letters.filter(x=> x !== '').length === arr.filter(x=> /[a-z]/.test(x) ? x : '').filter(x=> x !== '').length
}

plusSign("+f+d+c+#+f+") // true
plusSign("+d+=3=+s+") // true
plusSign("f++d+g+8+") // false
plusSign("+s+7+fg+r+8+") // false

Other Solutions

This solution uses purely regex and the test method to determine if the string meets the criteria.

const plusSign = str =>
!/(^|[^+])[a-z]|[a-z]([^+]|$)/i.test(str);
//DreamArdor

This solution uses the .every() method to test if the array of items from the string has each letter flanked by a ’+‘.

function plusSign(str) {
	return [...str].every((c,i)=>(/[a-z]/gi.test(c))?(str[i-1]=='+'&&str[i+1]=='+'):true)
}