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Daily Challenge #6 - Grandma and her friends

Another day, another challenge. Today, your task is to help your grandmother visit her friends.

This challenge comes from user @g964 on CodeWars:

Your granny, who lives in town X0, has friends. These friends are given in an array, for example: array of friends is

[ "A1", "A2", "A3", "A4", "A5" ].

Note: the order of friends is this array and it must not be changed since this order gives the order in which they will be visited.

These friends inhabit towns and you have an array with friends and towns, for example:

[ ["A1", "X1"], ["A2", "X2"], ["A3", "X3"], ["A4", "X4"] ]
[ ("A1", "X1"), ("A2", "X2"), ("A3", "X3"), ("A4", "X4") ]
{"A1", "X1", "A2", "X2", "A3", "X3", "A4", "X4"}

which means A1 is in town X1, A2 in town X2... It can happen that we don't know the town of one of the friends.

Your granny wants to visit her friends and to know how many miles she will have to travel. Can you help her figure it out?

Thank you to CodeWars, who has licensed redistribution of this challenge under the 2-Clause BSD License!

Want to propose a challenge for a future post? Email with your suggestions!

Top comments (16)

shayaulman profile image
Shaya Ulman • Edited

I think that some of the details are missing here...
What are the distances?

ryansmith profile image
Ryan Smith

I was able to find which has additional information.

coreyja profile image
Corey Alexander

Ya agreed...
Feels like we are missing some kind of map, or distances between different towns

spaciecat profile image
Spacie • Edited

I'm making the following assumptions, but I think I have a solution:

  • Town X1 is 1 mile from X0, X2 is 2 miles from X0, etc... (otherwise I don't believe there is enough information to come up with a solution)
  • friendTownList uses format C (flattened list)
  • the return trip from the last friend's house back to X0 is included in the total

Could be one (ugly) line if not counting error handling in the case where we don't know where a friend lives, could also be smaller if friendTownList was a hash to start with~

def getDistance(visitOrder, friendTownList)
  townHash = Hash[*friendTownList]
  visitOrder.each {|f| raise "IDK where #{f} lives!" if !townHash.key? f }
  [0, * {|f| townHash[f][1 .. -1].to_i }), 0]
    .map(&proc {|a, b| (a - b).abs })

getDistance(%w(A2 A1 A3), %w(A1 X1 A2 X2 A3 X3))
# -> 8
# Because (0, 2[2], 1[1], 3[2], 0[3]) -> 2 + 1 + 2 + 3 = 8

I'll update / post another solution if the question is updated with more information!

coreyja profile image
Corey Alexander

Wow! Love that we are all over here just complaining about not having enough data and you just went for it!

Love it! Keep up the awesome work on these challenges!

stevemoon profile image
Steve Moon • Edited

note: I followed the link to the codewars page and am using their list of distances. I also assume Granny is coming home after visiting her friends. A different list of distances can be passed in, without limit on length. But if the correlation between friend1 is in town1 breaks down then I'd need to add a map/hashtable for that lookup.


find_distance([B, C | Rest]) ->
    TotalDist = B,
    A = math:sqrt(C * C - B * B),
    find_distance(TotalDist + A, C, Rest).

find_distance(TotalDist, B, [C | Rest]) ->
    A = math:sqrt(C * C - B * B),
    find_distance(TotalDist + A, C, Rest);
find_distance(TotalDist, C, []) ->
    TotalDist + C.

devto6:find_distance([100, 200, 250, 300]).

valerionarcisi profile image
Valerio Narcisi • Edited


enum Map {
    X1 = 100.0,
    X2 = 200.0,
    X3 = 250.0,
    X4 = 300.0,

const grannyFriends = [
    ['A1', 'X1'],
    ['A2', 'X2'],
    ['A3', 'X3'],
    ['A4', 'X4'],

export default (friends: string[]): number => {
    return friends
        .map(f => {
            for (const gf of grannyFriends) {
                if (gf.includes(f)) {
                    return Map[gf[1]];
            return 0;
        .reduce((acc, curr) => (curr > 0 ? acc + (curr - acc) : acc), 0);


import grannyTrip from './help-your-granny';

describe('tdd granny trip', () => {
    test('should calculate distances', () => {
        expect(grannyTrip(['A1', 'A2', 'A3', 'A4', 'A5'])).toEqual(300);
        expect(grannyTrip(['A1', 'A2', 'A3'])).toEqual(250);
        expect(grannyTrip(['A1', 'A2', 'A4'])).toEqual(300);
celyes profile image
Ilyes Chouia • Edited

this is my solution in PHP, although the description is quite ambiguous...

$friends = [["A1", "X1"], ["A2", "X2"], ["A3", "X3"], ["A4", "X4"]];
$distances = [["X1", 100], ["X2", 200], ["X3", 350], ["X4", 420]];

$distances = array_reduce($distances, function($x, $y){
    $x[$y[0]] = $y[1]; 
    return $x;

function getTotalDistance($friends, $distances){
    $total = 0;
    for($x = 0; $x < count($friends); $x++){
        if(count($friends[$x]) === 2){
            $total += ($distances[$friends[$x][1]] * 2);
    return $total;
echo getTotalDistance($friends, $distances);
margo1993 profile image
package utils

import (

type CityDistance struct {
    City string
    Distance float64

func VisitFriend(friendsToVisit []string, friendsCities [][]string, cityDistances []CityDistance) (int, error) {
    totalDistance := 0.0

    var lastCityName string
    for _, friend := range friendsToVisit {
        friendCity, e := findFriendCity(friend, friendsCities)

        if e != nil {
            return 0, e

        if lastCityName == "" {
            distanceToCity, e := findDistanceToCity(friendCity, cityDistances)

            if e != nil {
                return 0, e

            totalDistance += distanceToCity
        } else {
            distanceBetweenCities, e := findDistanceBetweenCities(lastCityName, friendCity, cityDistances)

            if e != nil {
                return 0, e

            totalDistance += distanceBetweenCities

        lastCityName = friendCity
    distanceBackToHome, _ := findDistanceToCity(lastCityName, cityDistances)

    totalDistance += distanceBackToHome
    return int(math.Floor(totalDistance)), nil

func findDistanceToCity(city string, cityDistances []CityDistance) (float64, error) {
    for _, cityDistance := range cityDistances {
        cityName := cityDistance.City
        distance := cityDistance.Distance

        if cityName == city {
            return distance, nil

    return 0, errors.New("Didn't find city")

func findDistanceBetweenCities(fromCity string, toCity string, cityDistances []CityDistance) (float64, error) {
    totalDistance := 0.0

    lastDistance := 0.0;
    for _, cityDistance := range cityDistances {
        city := cityDistance.City
        distance := cityDistance.Distance

        if lastDistance > 0 {
            totalDistance += math.Sqrt((lastDistance * lastDistance) + (distance * distance))

        if (city == fromCity || city == toCity) && lastDistance > 0 {
            return totalDistance, nil

        if (city == fromCity || city == toCity) || lastDistance > 0 {
            lastDistance = distance

    return 0, errors.New("Didn't find city")

func findFriendCity(friend string, friendCities [][]string) (string, error) {
    for _, friendCity := range friendCities {
        if friend == friendCity[0] {
            return friendCity[1], nil

    return "", errors.New("Didn't found city for friend!")
kesprit profile image
kesprit • Edited

My solution in Swift, first time I use nested functions :

let distanceReferences = [ "X1": 100.0, "X2": 200.0, "X3": 250.0, "X4": 300.0 ]
let friendsReferences = [ "A1": "X1", "A2": "X2", "A3": "X3", "A4": "X4" ]
let friendsOrder = [ "A1", "A2", "A3", "A4", "A5"]

func grandmaTravel(friendsToVisit: [String]) -> Int {
    func distanceBetween(firstPoint: String, to secondPoint: String) -> Int {
        let first = distanceReferences.first {
            $0.key == firstPoint

        let second = distanceReferences.first {
            $0.key == secondPoint

        guard let distanceToFirst = first?.value, let distanceToSecond = second?.value
            else { return 0 }
        let result = (pow(distanceToSecond, 2) - pow(distanceToFirst, 2)).squareRoot()

        return Int(result.rounded())

    var distance = 0
    var points = friendsToVisit.compactMap {
    var currentPoint = points.removeFirst()

    // X0 to fist town
    distance += Int(distanceReferences[currentPoint] ?? 0)

    points.forEach {
        distance += distanceBetween(firstPoint: currentPoint, to: $0)
        currentPoint = $0

    // last town to X0
    distance += Int(distanceReferences[currentPoint] ?? 0)

    return distance

grandmaTravel(friendsToVisit: friendsOrder)
alvaromontoro profile image
Alvaro Montoro


let friends = [["A1", "X1"], ["A2", "X2"], ["A3", "X3"], ["A4", "X4"]];
let distances = [["X1", 100.0], ["X2", 200.0], ["X3", 250.0], ["X4", 300.0]];

// transform the array into an object to make it easier to process later
// solution from
distances = distances.reduce(function(p, c) { p[c[0]]=c[1]; return p; }, {});

function calculateDistance(friends, distances) {
  let total = 0;
  for (let x = 0; x < friends.length; x++) 
    if (friends[x].length === 2) 
      total += (distances[friends[x][1]] * 2);
  return total;

It doesn't use a smart algorithm to calculate the distance: grandma will visit a friend and then go back home, and continue with the next friend. If she doesn't know the city where a friend lives, she'll stay at home and meet the next friend.

Another version using array methods (and kind of unreadable):

let distances = { "X1": 100.0, "X2": 200.0, "X3": 250.0, "X4": 300.0 };
let friends = [ ["A1", "X1"], ["A2", "X2"], ["A3", "X3"], ["A4", "X4"] ];

const calculateDistance = (friends, distances) => friends.reduce((acc, curr) => curr.length === 2 ? acc + (distances[curr[1]] * 2) : acc, 0);

Here is a demo on CodePen.

alvaromontoro profile image
Alvaro Montoro

This is a bit confusing. If you don't know the next city to visit, how will you calculate the distance between the two cities?

dylanesque profile image
Michael Caveney

The problem says that the cities must be visited in the order in which they appear in the array, but yes, details are missing, we don't know the distance between the cities, which is necessary information to solve the problem.

alvaromontoro profile image
Alvaro Montoro

It's even worse: "It can happen that we don't know the town of one of the friends." Even if we had the distances, we may not even know which is our next destination.

coreyja profile image
Corey Alexander

Take this comment as a commitment to do this challenge sometime over the holiday weekend, pinky promise!

coreyja profile image
Corey Alexander

Here I am to fulfill my promise!

I followed the directions at:

Buttt... Even then they left something to be desired lol

But I made do! I made a few assumptions, changes to the rules.

The one rule I ignored (because honestly I didn't read it till later) is:

If a town Xi is not visited you will suppose that the triangle
X0Xi-1Xi+1 is still a right triangle.

I'm gonna justify it by saying that this rule means that the map changes depending on where Granny goes. I didn't think granny's plan should alter the map, but also I kinda just didn't read that rule!

I also created the rule that there are only roads between adjacent towns, and there is a road from each down to your grannies Home town. This makes the map a lot like the wheel and spoke map that is drawn in the linked problem.

These two rules combine to mean that not every map is valid, so I return an Error when we encounter an invalid map.

This change of rules, actually led to some more complexity lol. Now it wasn't straight forward what the best path from any arbitrary town to another was. Maybe you should travel "town to town" and maybe it was better to go home in the middle.
I take whichever of these two paths turns out to be shorter!

It's close to 200 lines, so instead of posting it here I'm just gonna post a link to the Github!