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.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0;
}

2

$begingroup$

I have this sample data:

let trips = [

  {

    from: "DEN",

    to: "JFK"

  },

  {

    from: "SEA",

    to: "DEN"

  },

  {

    from: 'JFK',

    to: 'SEA'

  },

];

and my origin is 'JFK', I want to sort the list by how I have traveled. So for instance, this should be the end result:

let trips = [

  {

    from: 'JFK',

    to: 'SEA'

  },

  {

    from: "SEA",

    to: "DEN"

  },

  {

    from: "DEN",

    to: "JFK"

  },

];

My solution works but it's not very well written, but I tried!

function sortByLinked(trips, origin = 'JFK') {

  let sortedArray = [];



  let first = trips.filter(trip => trip.from === origin)[0];

  sortedArray.push(first);



  for(var i = 0; i < trips.length; i++) {

    if(sortedArray[0].to === trips[i].from) {

      sortedArray.push(trips[i]);

    }

  }



  for(var i = 0; i < trips.length; i++) {

    if(sortedArray[1].to === trips[i].from) {

      sortedArray.push(trips[i]);

    }

  }



  return sortedArray;

}



sortByLinked(trips) 

javascript algorithm sorting graph

share|improve this question

edited 9 mins ago

200_success

131k17157422

asked 2 hours ago

Shivam BhallaShivam Bhalla

1825

$endgroup$

migrated from stackoverflow.com 1 hour ago

This question came from our site for professional and enthusiast programmers.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  is it like the origin and final destination always the same? and how many intermediate trips are going to be present?
  $endgroup$
  – karthick
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is from unique for all elements?
  $endgroup$
  – Taplar
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @karthick yes the origin and final should be the same.
  $endgroup$
  – Shivam Bhalla
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Taplar yes from is unique
  $endgroup$
  – Shivam Bhalla
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The task you want to perform is a simple kind of topological sorting.
  $endgroup$
  – 200_success
  7 mins ago
  

  
  
  
  

add a comment | 

2

$begingroup$

I have this sample data:

let trips = [

  {

    from: "DEN",

    to: "JFK"

  },

  {

    from: "SEA",

    to: "DEN"

  },

  {

    from: 'JFK',

    to: 'SEA'

  },

];

and my origin is 'JFK', I want to sort the list by how I have traveled. So for instance, this should be the end result:

let trips = [

  {

    from: 'JFK',

    to: 'SEA'

  },

  {

    from: "SEA",

    to: "DEN"

  },

  {

    from: "DEN",

    to: "JFK"

  },

];

My solution works but it's not very well written, but I tried!

function sortByLinked(trips, origin = 'JFK') {

  let sortedArray = [];



  let first = trips.filter(trip => trip.from === origin)[0];

  sortedArray.push(first);



  for(var i = 0; i < trips.length; i++) {

    if(sortedArray[0].to === trips[i].from) {

      sortedArray.push(trips[i]);

    }

  }



  for(var i = 0; i < trips.length; i++) {

    if(sortedArray[1].to === trips[i].from) {

      sortedArray.push(trips[i]);

    }

  }



  return sortedArray;

}



sortByLinked(trips) 

javascript algorithm sorting graph

share|improve this question

edited 9 mins ago

200_success

131k17157422

asked 2 hours ago

Shivam BhallaShivam Bhalla

1825

$endgroup$

migrated from stackoverflow.com 1 hour ago

This question came from our site for professional and enthusiast programmers.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  is it like the origin and final destination always the same? and how many intermediate trips are going to be present?
  $endgroup$
  – karthick
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is from unique for all elements?
  $endgroup$
  – Taplar
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @karthick yes the origin and final should be the same.
  $endgroup$
  – Shivam Bhalla
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Taplar yes from is unique
  $endgroup$
  – Shivam Bhalla
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The task you want to perform is a simple kind of topological sorting.
  $endgroup$
  – 200_success
  7 mins ago
  

  
  
  
  

add a comment | 

$begingroup$

I have this sample data:

let trips = [

  {

    from: "DEN",

    to: "JFK"

  },

  {

    from: "SEA",

    to: "DEN"

  },

  {

    from: 'JFK',

    to: 'SEA'

  },

];

and my origin is 'JFK', I want to sort the list by how I have traveled. So for instance, this should be the end result:

let trips = [

  {

    from: 'JFK',

    to: 'SEA'

  },

  {

    from: "SEA",

    to: "DEN"

  },

  {

    from: "DEN",

    to: "JFK"

  },

];

My solution works but it's not very well written, but I tried!

function sortByLinked(trips, origin = 'JFK') {

  let sortedArray = [];



  let first = trips.filter(trip => trip.from === origin)[0];

  sortedArray.push(first);



  for(var i = 0; i < trips.length; i++) {

    if(sortedArray[0].to === trips[i].from) {

      sortedArray.push(trips[i]);

    }

  }



  for(var i = 0; i < trips.length; i++) {

    if(sortedArray[1].to === trips[i].from) {

      sortedArray.push(trips[i]);

    }

  }



  return sortedArray;

}



sortByLinked(trips) 

javascript algorithm sorting graph

share|improve this question

edited 9 mins ago

200_success

131k17157422

asked 2 hours ago

Shivam BhallaShivam Bhalla

1825

$endgroup$

let trips = [

  {

    from: "DEN",

    to: "JFK"

  },

  {

    from: "SEA",

    to: "DEN"

  },

  {

    from: 'JFK',

    to: 'SEA'

  },

];

let trips = [

  {

    from: 'JFK',

    to: 'SEA'

  },

  {

    from: "SEA",

    to: "DEN"

  },

  {

    from: "DEN",

    to: "JFK"

  },

];

function sortByLinked(trips, origin = 'JFK') {

  let sortedArray = [];



  let first = trips.filter(trip => trip.from === origin)[0];

  sortedArray.push(first);



  for(var i = 0; i < trips.length; i++) {

    if(sortedArray[0].to === trips[i].from) {

      sortedArray.push(trips[i]);

    }

  }



  for(var i = 0; i < trips.length; i++) {

    if(sortedArray[1].to === trips[i].from) {

      sortedArray.push(trips[i]);

    }

  }



  return sortedArray;

}



sortByLinked(trips) 

share|improve this question

edited 9 mins ago

200_success

131k17157422

asked 2 hours ago

Shivam BhallaShivam Bhalla

1825

share|improve this question

edited 9 mins ago

200_success

131k17157422

asked 2 hours ago

Shivam BhallaShivam Bhalla

1825

edited 9 mins ago

200_success

131k17157422

edited 9 mins ago

200_success

131k17157422

asked 2 hours ago

Shivam BhallaShivam Bhalla

1825

asked 2 hours ago

Shivam BhallaShivam Bhalla

1825

2 Answers
2

active

oldest

votes

1

$begingroup$

Your solution only works if there's exactly 3 trips. After you find the first trip, put it at the front of the array and then find each subsequent one but instead of using a for loop like you're doing there are other ways to find the next trip like using Array.map() or Array.filter().

Here's one way to sort it in place and can handle any number of trips greater than 1.

function sortByLinked(trips, origin = 'JFK') {



  // this will be useful

  function swap(array, index1, index2){

    let temp = array[index1];

    array[index1] = array[index2];

    array[index2] = temp;

  }



  // find first one

  let first = trips.filter(trip => trip.from === origin)[0];



  // put him in the front of the list

  swap(trips, trips.map(trip => trip.from).indexOf(first.from), 0);



  // sort it in place

  for(let i=1; i<trips.length; i++){

    swap(trips, i, trips.map(trip => trip.from).indexOf(trips[i-1].to));

  }

}

share|improve this answer

edited 1 hour ago

Sᴀᴍ Onᴇᴌᴀ

10.3k62168

answered 1 hour ago

matthewlam.jsmatthewlam.js

111

New contributor

matthewlam.js is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Because ecmascript-6 is used here: "Two variables values can be swapped in one destructuring expression."
  $endgroup$
  – Sᴀᴍ Onᴇᴌᴀ
  1 hour ago
  
  
  

  
  
  
  

add a comment | 

0

$begingroup$

It would be good if your function could work for more than 3 trips.

And for larger chains of trips it becomes important to make it efficient. It is not efficient to search for the next trip by scanning the whole array. This will make the solution have O(n²) time complexity. So I would suggest creating a Map first, so that you can access a trip by its from property in constant time:

function sortByLinked(trips, origin = "JFK") {

    const map = new Map(trips.map(trip => [trip.from, trip]));

    const result = [];

    for (let trip; trip = map.get(origin); origin = trip.to) {

        result.push(trip);

        map.delete(origin);

    }

    return result;

}



const trips = [{from: "DEN",to: "JFK"},{from: "SEA",to: "DEN"},{from: 'JFK', to: 'SEA'}];

const sorted = sortByLinked(trips, "JFK");

console.log(sorted);

share|improve this answer

answered 1 hour ago

trincottrincot

42937

$endgroup$

add a comment | 

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1

$begingroup$

Your solution only works if there's exactly 3 trips. After you find the first trip, put it at the front of the array and then find each subsequent one but instead of using a for loop like you're doing there are other ways to find the next trip like using Array.map() or Array.filter().

Here's one way to sort it in place and can handle any number of trips greater than 1.

function sortByLinked(trips, origin = 'JFK') {



  // this will be useful

  function swap(array, index1, index2){

    let temp = array[index1];

    array[index1] = array[index2];

    array[index2] = temp;

  }



  // find first one

  let first = trips.filter(trip => trip.from === origin)[0];



  // put him in the front of the list

  swap(trips, trips.map(trip => trip.from).indexOf(first.from), 0);



  // sort it in place

  for(let i=1; i<trips.length; i++){

    swap(trips, i, trips.map(trip => trip.from).indexOf(trips[i-1].to));

  }

}

share|improve this answer

edited 1 hour ago

Sᴀᴍ Onᴇᴌᴀ

10.3k62168

answered 1 hour ago

matthewlam.jsmatthewlam.js

111

New contributor

matthewlam.js is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Because ecmascript-6 is used here: "Two variables values can be swapped in one destructuring expression."
  $endgroup$
  – Sᴀᴍ Onᴇᴌᴀ
  1 hour ago
  
  
  

  
  
  
  

add a comment | 

1

$begingroup$

Your solution only works if there's exactly 3 trips. After you find the first trip, put it at the front of the array and then find each subsequent one but instead of using a for loop like you're doing there are other ways to find the next trip like using Array.map() or Array.filter().

Here's one way to sort it in place and can handle any number of trips greater than 1.

function sortByLinked(trips, origin = 'JFK') {



  // this will be useful

  function swap(array, index1, index2){

    let temp = array[index1];

    array[index1] = array[index2];

    array[index2] = temp;

  }



  // find first one

  let first = trips.filter(trip => trip.from === origin)[0];



  // put him in the front of the list

  swap(trips, trips.map(trip => trip.from).indexOf(first.from), 0);



  // sort it in place

  for(let i=1; i<trips.length; i++){

    swap(trips, i, trips.map(trip => trip.from).indexOf(trips[i-1].to));

  }

}

share|improve this answer

edited 1 hour ago

Sᴀᴍ Onᴇᴌᴀ

10.3k62168

answered 1 hour ago

matthewlam.jsmatthewlam.js

111

New contributor

matthewlam.js is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Because ecmascript-6 is used here: "Two variables values can be swapped in one destructuring expression."
  $endgroup$
  – Sᴀᴍ Onᴇᴌᴀ
  1 hour ago
  
  
  

  
  
  
  

add a comment | 

$begingroup$

Your solution only works if there's exactly 3 trips. After you find the first trip, put it at the front of the array and then find each subsequent one but instead of using a for loop like you're doing there are other ways to find the next trip like using Array.map() or Array.filter().

Here's one way to sort it in place and can handle any number of trips greater than 1.

function sortByLinked(trips, origin = 'JFK') {



  // this will be useful

  function swap(array, index1, index2){

    let temp = array[index1];

    array[index1] = array[index2];

    array[index2] = temp;

  }



  // find first one

  let first = trips.filter(trip => trip.from === origin)[0];



  // put him in the front of the list

  swap(trips, trips.map(trip => trip.from).indexOf(first.from), 0);



  // sort it in place

  for(let i=1; i<trips.length; i++){

    swap(trips, i, trips.map(trip => trip.from).indexOf(trips[i-1].to));

  }

}

share|improve this answer

edited 1 hour ago

Sᴀᴍ Onᴇᴌᴀ

10.3k62168

answered 1 hour ago

matthewlam.jsmatthewlam.js

111

New contributor

matthewlam.js is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

function sortByLinked(trips, origin = 'JFK') {



  // this will be useful

  function swap(array, index1, index2){

    let temp = array[index1];

    array[index1] = array[index2];

    array[index2] = temp;

  }



  // find first one

  let first = trips.filter(trip => trip.from === origin)[0];



  // put him in the front of the list

  swap(trips, trips.map(trip => trip.from).indexOf(first.from), 0);



  // sort it in place

  for(let i=1; i<trips.length; i++){

    swap(trips, i, trips.map(trip => trip.from).indexOf(trips[i-1].to));

  }

}

function sortByLinked(trips, origin = 'JFK') {



  // this will be useful

  function swap(array, index1, index2){

    let temp = array[index1];

    array[index1] = array[index2];

    array[index2] = temp;

  }



  // find first one

  let first = trips.filter(trip => trip.from === origin)[0];



  // put him in the front of the list

  swap(trips, trips.map(trip => trip.from).indexOf(first.from), 0);



  // sort it in place

  for(let i=1; i<trips.length; i++){

    swap(trips, i, trips.map(trip => trip.from).indexOf(trips[i-1].to));

  }

}

function sortByLinked(trips, origin = 'JFK') {



  // this will be useful

  function swap(array, index1, index2){

    let temp = array[index1];

    array[index1] = array[index2];

    array[index2] = temp;

  }



  // find first one

  let first = trips.filter(trip => trip.from === origin)[0];



  // put him in the front of the list

  swap(trips, trips.map(trip => trip.from).indexOf(first.from), 0);



  // sort it in place

  for(let i=1; i<trips.length; i++){

    swap(trips, i, trips.map(trip => trip.from).indexOf(trips[i-1].to));

  }

}

share|improve this answer

edited 1 hour ago

Sᴀᴍ Onᴇᴌᴀ

10.3k62168

answered 1 hour ago

matthewlam.jsmatthewlam.js

111

New contributor

matthewlam.js is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited 1 hour ago

Sᴀᴍ Onᴇᴌᴀ

10.3k62168

edited 1 hour ago

Sᴀᴍ Onᴇᴌᴀ

10.3k62168

answered 1 hour ago

matthewlam.jsmatthewlam.js

111

New contributor

matthewlam.js is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered 1 hour ago

matthewlam.jsmatthewlam.js

111

0

$begingroup$

It would be good if your function could work for more than 3 trips.

And for larger chains of trips it becomes important to make it efficient. It is not efficient to search for the next trip by scanning the whole array. This will make the solution have O(n²) time complexity. So I would suggest creating a Map first, so that you can access a trip by its from property in constant time:

function sortByLinked(trips, origin = "JFK") {

    const map = new Map(trips.map(trip => [trip.from, trip]));

    const result = [];

    for (let trip; trip = map.get(origin); origin = trip.to) {

        result.push(trip);

        map.delete(origin);

    }

    return result;

}



const trips = [{from: "DEN",to: "JFK"},{from: "SEA",to: "DEN"},{from: 'JFK', to: 'SEA'}];

const sorted = sortByLinked(trips, "JFK");

console.log(sorted);

share|improve this answer

answered 1 hour ago

trincottrincot

42937

$endgroup$

add a comment | 

0

$begingroup$

It would be good if your function could work for more than 3 trips.

And for larger chains of trips it becomes important to make it efficient. It is not efficient to search for the next trip by scanning the whole array. This will make the solution have O(n²) time complexity. So I would suggest creating a Map first, so that you can access a trip by its from property in constant time:

function sortByLinked(trips, origin = "JFK") {

    const map = new Map(trips.map(trip => [trip.from, trip]));

    const result = [];

    for (let trip; trip = map.get(origin); origin = trip.to) {

        result.push(trip);

        map.delete(origin);

    }

    return result;

}



const trips = [{from: "DEN",to: "JFK"},{from: "SEA",to: "DEN"},{from: 'JFK', to: 'SEA'}];

const sorted = sortByLinked(trips, "JFK");

console.log(sorted);

share|improve this answer

answered 1 hour ago

trincottrincot

42937

$endgroup$

add a comment | 

$begingroup$

It would be good if your function could work for more than 3 trips.

And for larger chains of trips it becomes important to make it efficient. It is not efficient to search for the next trip by scanning the whole array. This will make the solution have O(n²) time complexity. So I would suggest creating a Map first, so that you can access a trip by its from property in constant time:

function sortByLinked(trips, origin = "JFK") {

    const map = new Map(trips.map(trip => [trip.from, trip]));

    const result = [];

    for (let trip; trip = map.get(origin); origin = trip.to) {

        result.push(trip);

        map.delete(origin);

    }

    return result;

}



const trips = [{from: "DEN",to: "JFK"},{from: "SEA",to: "DEN"},{from: 'JFK', to: 'SEA'}];

const sorted = sortByLinked(trips, "JFK");

console.log(sorted);

share|improve this answer

answered 1 hour ago

trincottrincot

42937

$endgroup$

function sortByLinked(trips, origin = "JFK") {

    const map = new Map(trips.map(trip => [trip.from, trip]));

    const result = [];

    for (let trip; trip = map.get(origin); origin = trip.to) {

        result.push(trip);

        map.delete(origin);

    }

    return result;

}



const trips = [{from: "DEN",to: "JFK"},{from: "SEA",to: "DEN"},{from: 'JFK', to: 'SEA'}];

const sorted = sortByLinked(trips, "JFK");

console.log(sorted);

function sortByLinked(trips, origin = "JFK") {

    const map = new Map(trips.map(trip => [trip.from, trip]));

    const result = [];

    for (let trip; trip = map.get(origin); origin = trip.to) {

        result.push(trip);

        map.delete(origin);

    }

    return result;

}



const trips = [{from: "DEN",to: "JFK"},{from: "SEA",to: "DEN"},{from: 'JFK', to: 'SEA'}];

const sorted = sortByLinked(trips, "JFK");

console.log(sorted);

function sortByLinked(trips, origin = "JFK") {

    const map = new Map(trips.map(trip => [trip.from, trip]));

    const result = [];

    for (let trip; trip = map.get(origin); origin = trip.to) {

        result.push(trip);

        map.delete(origin);

    }

    return result;

}



const trips = [{from: "DEN",to: "JFK"},{from: "SEA",to: "DEN"},{from: 'JFK', to: 'SEA'}];

const sorted = sortByLinked(trips, "JFK");

console.log(sorted);

share|improve this answer

answered 1 hour ago

trincottrincot

42937

answered 1 hour ago

trincottrincot

42937

answered 1 hour ago

trincottrincot

42937