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Sort list of Array Linked Object by keys and values

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

0

$begingroup$

Lets say I have this 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!

Any help would be great.

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

share|improve this question

asked 34 mins ago

Shivam BhallaShivam Bhalla

1725

$endgroup$

migrated from stackoverflow.com 13 mins 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
  30 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is from unique for all elements?
  $endgroup$
  – Taplar
  27 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How does the input indicate what the origin is of the trip?
  $endgroup$
  – trincot
  3 mins ago
  

  
  
  
  

add a comment | 

0

$begingroup$

Lets say I have this 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!

Any help would be great.

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

share|improve this question

asked 34 mins ago

Shivam BhallaShivam Bhalla

1725

$endgroup$

migrated from stackoverflow.com 13 mins 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
  30 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is from unique for all elements?
  $endgroup$
  – Taplar
  27 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How does the input indicate what the origin is of the trip?
  $endgroup$
  – trincot
  3 mins ago
  

  
  
  
  

add a comment | 

$begingroup$

Lets say I have this 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!

Any help would be great.

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

share|improve this question

asked 34 mins ago

Shivam BhallaShivam Bhalla

1725

$endgroup$

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

asked 34 mins ago

Shivam BhallaShivam Bhalla

1725

share|improve this question

asked 34 mins ago

Shivam BhallaShivam Bhalla

1725

asked 34 mins ago

Shivam BhallaShivam Bhalla

1725

asked 34 mins ago

Shivam BhallaShivam Bhalla

1725

1 Answer
1

active

oldest

votes

0

$begingroup$

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 map or filter. Here's one way to sort it in place and can handle any number of trips greater 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

answered 6 mins ago

matthewlam.jsmatthewlam.js

1

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.

$endgroup$

add a comment | 

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0

$begingroup$

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 map or filter. Here's one way to sort it in place and can handle any number of trips greater 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

answered 6 mins ago

matthewlam.jsmatthewlam.js

1

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.

$endgroup$

add a comment | 

0

$begingroup$

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 map or filter. Here's one way to sort it in place and can handle any number of trips greater 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

answered 6 mins ago

matthewlam.jsmatthewlam.js

1

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.

$endgroup$

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 map or filter. Here's one way to sort it in place and can handle any number of trips greater 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

answered 6 mins ago

matthewlam.jsmatthewlam.js

1

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

answered 6 mins ago

matthewlam.jsmatthewlam.js

1

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 6 mins ago

matthewlam.jsmatthewlam.js

1

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 6 mins ago

matthewlam.jsmatthewlam.js

1