A different and more efficient way to address the issue at hand

Question

A different and more efficient way to address the issue at hand

During a recent interview, I was asked the following question and despite being able to solve it, the interviewer pointed out that my code was not optimized. Is there anyone who can help me with an optimized solution?

Question : Given an array array1, rearrange the array in the format:

All negative numbers and their absolute values should be inserted first in the array (sorted).
The remaining items should be concatenated to the new array (sorted).

//input 
var array = [8,7,0,6,4,-9,-7,2,-1,3,5,1,10];
//output
var array2 = [-7,7,-1,1,-9,0,2,3,4,5,6,8,10];

My Code :

function sort(arr){  
  var arrange = arr.sort((a, b)=>{ return a-b});
  var array1=[];
  for(let i=0; i<arrange.length; i++){
    let firstItem = Math.abs(arrange[i]);
    for(let j=i+1; j<arrange.length; j++){
       if(firstItem === Math.abs(arrange[j])){
         array1.push(arrange[i], arrange[j])       
       }
   }
  }
  arrange = arrange.filter((item, i)=>{
     return array1.indexOf(item) === -1
  })
 return [...array1, ...arrange]
} 


console.log(sort(array));

JSBIN

Please give some hints also, what is wrong with my approach.

Note : The code I wrote has a time complexity of o(n^2), but the interviewer wanted a O(log n) solution without nested for loops. Can we achieve this?

javascript arrays optimization logic

Answer 1

Answer №1

This particular approach may not be the most efficient as it consumes more processing time due to complexities like sorting and filtering, resulting in a complexity of O(n^2). This means that as the number of elements increases, the time taken also increases.

When faced with such questions, it is advisable to avoid using nested loops if feasible.

Answer 2

This particular approach may not be the most efficient as it consumes more processing time due to complexities like sorting and filtering, resulting in a complexity of O(n^2). This means that as the number of elements increases, the time taken also increases.

When faced with such questions, it is advisable to avoid using nested loops if feasible.

Answer 3

Answer №2

It appears that there are several issues with your algorithm.

The nested for loops have a time complexity of O(n2).

for(let i=0; i<arrange.length; i++){ // ----------O(n)
  let firstItem = Math.abs(arrange[i]);
  for(let j=i+1; j<arrange.length; j++){ // ---------x O(n(n+1)/2)==O(n^2)
       if(firstItem === Math.abs(arrange[j])){
         array1.push(arrange[i], arrange[j])       
       }
   }
}

The use of filter + indexOf both result in O(n2) complexity as well.

arrange = arrange.filter((item, i)=>{ // O(n)
   return array1.indexOf(item) === -1 // -- x O(n*n) // because indexOf is O(n)
})

A potential solution could involve optimizing the algorithm to run in O(nlog(n)).

One approach could be:

Identify negative numbers along with their corresponding positive counterparts using a map
Separate out the positive numbers into another container
Sort the negative map based on keys
Output the negative map and remove any associated positives from the positive container
Sort the positive container and output it

... (continue with the rest of the text) What I want to highlight is that while incremental optimizations can improve performance, focusing on algorithm efficiency and data structure choices play a more significant role in achieving optimal results.

Answer 4

It appears that there are several issues with your algorithm.

The nested for loops have a time complexity of O(n2).

for(let i=0; i<arrange.length; i++){ // ----------O(n)
  let firstItem = Math.abs(arrange[i]);
  for(let j=i+1; j<arrange.length; j++){ // ---------x O(n(n+1)/2)==O(n^2)
       if(firstItem === Math.abs(arrange[j])){
         array1.push(arrange[i], arrange[j])       
       }
   }
}

The use of filter + indexOf both result in O(n2) complexity as well.

arrange = arrange.filter((item, i)=>{ // O(n)
   return array1.indexOf(item) === -1 // -- x O(n*n) // because indexOf is O(n)
})

A potential solution could involve optimizing the algorithm to run in O(nlog(n)).

One approach could be:

Identify negative numbers along with their corresponding positive counterparts using a map
Separate out the positive numbers into another container
Sort the negative map based on keys
Output the negative map and remove any associated positives from the positive container
Sort the positive container and output it

... (continue with the rest of the text) What I want to highlight is that while incremental optimizations can improve performance, focusing on algorithm efficiency and data structure choices play a more significant role in achieving optimal results.

Answer 5

Answer №3

Presented below is an alternative approach

//input 
var array = [8,7,0,6,4,-9,-7,2,-1,3,5,1,10];
//output
var array2 = [-7,7,-1,1,-9,0,2,3,4,5,6,8,10];


console.clear();

function sort(arr){
  const singular=[];
  const arrange = arr.sort((a, b)=>{ return a-b});

  const output = arrange.flatMap((item, i)=>{
     if(item < 0 && arrange.includes(-item)){
         return [item, -item];       
     }else{
       singular.push(item);
     }     
  }).filter((item)=>{ return item; });

  const result = [...new Set(output), ...singular];

  console.log('result : ', result);
} 


console.log(sort(array));

Answer 6

Presented below is an alternative approach

//input 
var array = [8,7,0,6,4,-9,-7,2,-1,3,5,1,10];
//output
var array2 = [-7,7,-1,1,-9,0,2,3,4,5,6,8,10];


console.clear();

function sort(arr){
  const singular=[];
  const arrange = arr.sort((a, b)=>{ return a-b});

  const output = arrange.flatMap((item, i)=>{
     if(item < 0 && arrange.includes(-item)){
         return [item, -item];       
     }else{
       singular.push(item);
     }     
  }).filter((item)=>{ return item; });

  const result = [...new Set(output), ...singular];

  console.log('result : ', result);
} 


console.log(sort(array));

Answer 7

Answer №4

Your approach includes 2 sorts and a nested for loop, which may not be the most efficient solution from a business standpoint.

The complexity of sorting is O(nlog(n)), while a nested for loop has a complexity of O(n^2).

When tackling problems with arrays, it's worth considering linear solutions first before resorting to sorting. Sorting is necessary in this case, but using nested loops should be avoided unless absolutely required - it comes with added responsibilities!

Take a look at the following solution, which utilizes one sort and iterates through the array twice:

function modifyArray(array) {
    var sorted = array.sort((a, b) => { return a - b });
    let map = {};
    for (var i = 0; i < sorted.length; i++) {
        map[sorted[i]] = true;
    }

    let negativesAndAbsolutes = [];
    let uniqueNegatives = [];
    let uniquePositives = [];
    let currNumber = null;

    for (var i = 0; i < sorted.length; i++) {
        currNumber = sorted[i];
        if (!map[currNumber]) {
            // already sorted.
            continue;
        }
        if (currNumber < 0) {
            if (map[-1 * currNumber]) {
                // negative number with its absolute value present.
                negativesAndAbsolutes.push(currNumber, -1 * currNumber);
                map[currNumber] = false;
                map[-1 * currNumber] = false;       // don't process it again.
                continue;
            } else {
                // unique negative.
                uniqueNegatives.push(currNumber);
            }
            continue;
        }
        uniquePositives.push(currNumber);
    }

    return [...negativesAndAbsolutes, ...uniqueNegatives, ...uniquePositives];
}

modifyArray([8,7,0,6,4,-9,-7,2,-1,3,5,1,10]);

Answer 8

Your approach includes 2 sorts and a nested for loop, which may not be the most efficient solution from a business standpoint.

The complexity of sorting is O(nlog(n)), while a nested for loop has a complexity of O(n^2).

When tackling problems with arrays, it's worth considering linear solutions first before resorting to sorting. Sorting is necessary in this case, but using nested loops should be avoided unless absolutely required - it comes with added responsibilities!

Take a look at the following solution, which utilizes one sort and iterates through the array twice:

function modifyArray(array) {
    var sorted = array.sort((a, b) => { return a - b });
    let map = {};
    for (var i = 0; i < sorted.length; i++) {
        map[sorted[i]] = true;
    }

    let negativesAndAbsolutes = [];
    let uniqueNegatives = [];
    let uniquePositives = [];
    let currNumber = null;

    for (var i = 0; i < sorted.length; i++) {
        currNumber = sorted[i];
        if (!map[currNumber]) {
            // already sorted.
            continue;
        }
        if (currNumber < 0) {
            if (map[-1 * currNumber]) {
                // negative number with its absolute value present.
                negativesAndAbsolutes.push(currNumber, -1 * currNumber);
                map[currNumber] = false;
                map[-1 * currNumber] = false;       // don't process it again.
                continue;
            } else {
                // unique negative.
                uniqueNegatives.push(currNumber);
            }
            continue;
        }
        uniquePositives.push(currNumber);
    }

    return [...negativesAndAbsolutes, ...uniqueNegatives, ...uniquePositives];
}

modifyArray([8,7,0,6,4,-9,-7,2,-1,3,5,1,10]);

A different and more efficient way to address the issue at hand

Answer №1

Answer №2

Answer №3

Answer №4

Similar questions

Can we separate city, state, and country data into individual input fields using Autocomplete and Geonames?

Two interactive dropdown menus featuring custom data attributes, each influencing the options available in the other

Troubleshooting: Unable to load template file content in AngularJS ng-view

What could be causing my cmd to report an error stating that it is unable to locate the node-modules

What is the best way to monitor a document variable that is set after a component has been

Displaying a div after a successful form submission using Jquery

Enhance the material-table component by incorporating dynamic information into the lookup property

Angular's Conditional ng-if Error

Compatibility Issue between Jquery UI CheckBox Button and Click Function in IE8

What's the best way to save data from an HTML input field into a JavaScript array?

The tooltip in a scrollable container of Highcharts moves along with the content as it scrolls

A guide on updating an item in a basic JavaScript file with Node.js

Discover the method for displaying a user's "last seen at" timestamp by utilizing the seconds provided by the server

Incorporating JavaScript to dynamically load an HTML page into an iframe

Determining if a user is already logged in from a different device using express-session

The transfer of character data from PHP to jQuery is not successful

Is it possible to define a 2-D array using the ref( ) function in VueJS?

Whenever I try to include something within the `componentWillUnmount` function,

Populating a textbox with a portion of my PHP array

What are some techniques for preventing Null Pointer Exception errors?