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Posted on Aug 27, 2019

Daily Challenge #50 - Number Neighbor

#challenge

Given a phone number of integer length N (1 ≤ N ≤ 10) as a number string (e.g. 555-555-5555 would be passed as "5555555555"), return an array of all phone numbers as number strings that would be considered neighbors of that phone number (which is any numbers ±1).

For example: someone with the phone number 555-555-5555 has neighbors 555-555-5554 and 555-555-5556.

Want to propose a challenge idea for a future post? Email yo+challenge@dev.to with your suggestions!

This challenge comes from SaladFork on CodeWars. Thank you to CodeWars, who has licensed redistribution of this challenge under the 2-Clause BSD License!

Top comments (20)

Chris Achard • Aug 27 '19

Huh, this one was harder than I expected it to be... that probably means I missed something 🤣

Also, TIL that setting a character of a string in JS with square brackets doesn't throw an error - but also doesn't do anything! example:

let myStr = "abc123"
myStr[0] = 'X'
// myStr is unchanged here!

Solution in javascript:


const numberNeighbor = (number) => {
  return [...number].reduce((arr, num, i) => {
    num = Number(num)

    if(num - 1 >= 0) {
      arr.push([ 
        number.slice(0, i), 
        num - 1, 
        number.slice(i+1) 
      ].join(""))
    }

    if(num + 1 <= 9) {
      arr.push([ 
        number.slice(0, i), 
        num + 1, 
        number.slice(i+1) 
      ].join(""))
    }

    return arr
  }, [])
}

Alfredo Salzillo • Aug 27 '19

Strings in javascript are immutable, so even if you change a character using the [] notation, it always returns a new string.

This instead work as you expected

// The string is transformed into an Array
const arrString = [..."abc123"]
arrString[0] = "X"
// arrString is now ["X", "b", "c", "1", "2", "3"]

Chris Achard • Aug 27 '19

Makes sense - thanks! (forgot (or never really knew?) that strings in javascript were immutable)

Jarod Smith • Aug 27 '19

// areaCode :: String -> String
const areaCode = str => str.substring(0,3);
// threeCode :: String -> String
const threeCode = str => str.substring(3,6);
// fourCode :: String -> String
const fourCode = str => str.substring(6);
// formatPhone :: String -> String
const formatPhone = str => `${areaCode(str)}-${threeCode(str)}-${fourCode(str)}`;

// getNeighbors :: String -> [Number] -> [String]
const getNeighbors = number => [parseInt(number) + 1, parseInt(number) - 1]
  .map(neighbor => formatPhone(`${neighbor}`));

Attempting to learn and understand more functional paradigms. Is this right way to solve this problem functionally?

Amin • Aug 27 '19

Way to go buddy!

From my perspective, to format the phone number in 3 segments, I would have used this instead of the 3 calls you made.

"use strict"

const phoneNumber = "123456789"

const formatPhoneNumber = (string, glue) => string.match(/.{1,3}/g).join(glue)

console.log(formatPhoneNumber(phoneNumber, "-")) // 123-456-789
console.log(formatPhoneNumber(phoneNumber, " ")) // 123 456 789
console.log(formatPhoneNumber(phoneNumber, ".")) // 123.456.789

This leverage the regular expressions to match three characters at most. I know regular expressions is not the easiest thing to use but in this case it can be quite handy. As it returns an array, joining the items will produce a string back so the signature of the function can be written as follow in a purely functional programming language like PureScript.

formatPhoneNumber :: String -> String -> String

Unfortunately, currying (partially applying parameters) is not built-in in JavaScript. You could implement your own curry function and use it on all your future function definitions. But this would be overkill for this challenge. But if you would ask me how I would implement a currying function in JavaScript, here is my take.

"use strict"

function curry(callable, ...initialParameters) {
  if (typeof callable !== "function") {
    throw new TypeError("Function expected as first argument")
  }

  return function(...additionalParameters) {
    const parameters = [...initialParameters, ...additionalParameters]

    if (parameters.length >= callable.length) {
      return callable(...parameters)
    }

    return curry(callable, ...parameters)
  }
}

// format :: String -> Number -> String -> String
const format = curry(function(glue, segments, phone) {
  return phone.match(new RegExp(`.{1,${segments}}`, "g")).join(glue)
})

// usFormat :: String -> String
const usFormat = format("-", 3)

// frenchFormat :: String -> String
const frenchFormat = format(".", 2)

console.log(usFormat("123456789")) // 123-456-789
console.log(frenchFormat("1234567890")) // 12.34.56.78.90

Last tip, when creating a function, try to put the data structure at the end, that way you can construct partial application of your function without pain, even in JavaScript. In this case, this means puting the phone number to format at the very end of the parameter's list in your function definition. Hope that helps you in your functional programing journey pal.

Jarod Smith • Aug 28 '19

Hey thanks for the response I will look at regexes next time for this kind of problem. Also I thought currying was supported in ES6?
via

const f = a => b=> a + b

Amin • Aug 28 '19

What you did is indeed currying but it is not transparent in the sense that if you want to pass all three arguments at once you will have to make three calls. The curry implementation allows you to either pass all arguments at once like a regular function call or partially calling the function. I kinda dislike calling my function multiple times and the only two reasons are aesthetic and the fact that it is less natural to define functions like that but arrow functions syntax makes it a little bit cleaner.

La blatte • Aug 27 '19

Here goes some ugly code!

f=a=>{p=parseInt,l=(''+a).length,i=0,r=[a];while(i<l){r.push(...[('0'+(p(a)-(10**i))).slice(-l),''+(p(a)+(10**i))]);++i}return r}

It works like that:

Take the input's length
Add and subtract 10ⁿ (where n goes from 0 to the input's length - 1) to the input number
Add a trailing 0 if necessary
Return the built array!

f('5555555555').join('\n');
"5555555555
5555555554
5555555556
5555555545
5555555565
5555555455
5555555655
5555554555
5555556555
5555545555
5555565555
5555455555
5555655555
5554555555
5556555555
5545555555
5565555555
5455555555
5655555555
4555555555
6555555555"

Alfredo Salzillo • Aug 27 '19

One line JS generate all possible neighbor number for any position.

const allNumberNeighbor = number => [...number]
  .map(Number)
  .flatMap((n, i) => Number(n) && [
    n + 1 <= 9 && Object.values({ ...number, [i]: n + 1 }),
    n - 1 >= 0 && Object.values({ ...number, [i]: n - 1 }),
  ])
  .filter(Boolean)
  .map(n => n.join(''));

allNumberNeighbor('555-555-555');
// return
/*
 [
  '655-555-555',
  '455-555-555',
  '565-555-555',
  '545-555-555',
  '556-555-555',
  '554-555-555',
  '555-655-555',
  '555-455-555',
  '555-565-555',
  '555-545-555',
  '555-556-555',
  '555-554-555',
  '555-555-655',
  '555-555-455',
  '555-555-565',
  '555-555-545',
  '555-555-556',
  '555-555-554',
 ];
*/

André Adriano • Aug 27 '19

My take on this challenge with JS

const neighbors = number => {
    const format = n => {
        n = `${n}`.padStart(10, '0');
        return `${`${n}`.substr(0,3)}-${`${n}`.substr(3, 3)}-${`${n}`.substr(6, 4)}`;
    };

    const nArray = [...number].reduce((t, d, i) => {
        const n1 = Number(number) + Math.pow(10, i);
        const n2 = number - Math.pow(10, i);

        t.push(format(n1));
        t.push(format(n2));

        return t;
    }, [])

    return nArray;
}

/**

neighbors("5555555555");

[
  "555-555-5556",
  "555-555-5554",
  "555-555-5565",
  "555-555-5545",
  "555-555-5655",
  "555-555-5455",
  "555-555-6555",
  "555-555-4555",
  "555-556-5555",
  "555-554-5555",
  "555-565-5555",
  "555-545-5555",
  "555-655-5555",
  "555-455-5555",
  "556-555-5555",
  "554-555-5555",
  "565-555-5555",
  "545-555-5555",
  "655-555-5555",
  "455-555-5555"
]

**/

K.V.Harish • Oct 14 '19

My solution in js

const numberNeighbour = (num) => {
  const neighbours = [];
  num.split('').forEach((value, index) => {
    if(parseInt(value) - 1 > 0) {
      let prevNum = num.split('');
      prevNum[index] = parseInt(value) - 1;
      prevNum = prevNum.join('');
      neighbours.push(prevNum.slice(0, 3) + "-" + prevNum.slice(3, 6) + "-" + prevNum.slice(6));
    }
    if(parseInt(value) + 1 < 10) {
      let nextNum = num.split('');
      nextNum[index] = parseInt(value) + 1;
      nextNum = nextNum.join('');
      neighbours.push(nextNum.slice(0, 3) + "-" + nextNum.slice(3, 6) + "-" + nextNum.slice(6));
    }
  });
  return neighbours;
};

numberNeighbour('5555555555');
/*[
"455-555-5555",
"655-555-5555",
"545-555-5555",
"565-555-5555",
"554-555-5555",
"556-555-5555",
"555-455-5555",
"555-655-5555",
"555-545-5555",
"555-565-5555",
"555-554-5555",
"555-556-5555",
"555-555-4555",
"555-555-6555",
"555-555-5455",
"555-555-5655",
"555-555-5545",
"555-555-5565",
"555-555-5554",
"555-555-5556"
]*/
numberNeighbour('1111111111');
/*[
"211-111-1111",
"121-111-1111",
"112-111-1111",
"111-211-1111",
"111-121-1111",
"111-112-1111",
"111-111-2111",
"111-111-1211",
"111-111-1121",
"111-111-1112"
]*/
numberNeighbour('9999999999');
/*[
"899-999-9999",
"989-999-9999",
"998-999-9999",
"999-899-9999",
"999-989-9999",
"999-998-9999",
"999-999-8999",
"999-999-9899",
"999-999-9989",
"999-999-9998"
]*/

Yaroslav Barkovskiy • Aug 28 '19

Simple ruby solution(with edge cases):

def neighbors str
  num = str.to_i
  res = case num
  when 0
    [num + 1]
  when 9999999999
    [num - 1]
  when -Float::INFINITY..0, 9999999999..Float::INFINITY
    raise 'Not a phone number'
  else
    [num - 1, num + 1]
  end

  res.map(&:to_s)
end

Oleksii Filonenko • Aug 28 '19

I liked the "any number can differ by 1" approach more - it's a but more work :)

Rust:

pub fn number_neighbors(number: u32) -> Vec<u32> {
    let len = number.to_string().len() as u32;
    (0..len)
        .map(|delta| 10u32.pow(delta))
        .flat_map(|delta| vec![number - delta, number + delta])
        .collect()
}

And a test (I usually omit them from my submissions, but I think they have a place here):

#[test]                                                                   
fn test_number_neighbors() {                                              
    assert_eq!(vec![221, 223, 212, 232, 122, 322], number_neighbors(222));
}

Amin • Aug 27 '19

My take at the challenge written in Elm.

computeNeighbors : String -> Maybe (List Int)
computeNeighbors phoneNumber = 
  case String.toInt phoneNumber of
    Just integerPhoneNumber ->
      Just [integerPhoneNumber + 1, integerPhoneNumber - 1]
    Nothing ->
      Nothing

Try it online.

Aadi Bajpai • Aug 27 '19

Python one-liner to the rescue again. (Horribly formatted this time tho)

f = lambda x: print(*[x[:len(x)-1]+str(int(x[len(x)-1])+1), x[:len(x)-1]+str(int(x[len(x)-1])-1)])

>>> f('555-555-5555')
555-555-5556 555-555-5554

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Daily Challenge #50 - Number Neighbor

Top comments (20)

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