Organize array elements based on their values - using javascript

Question

Organize array elements based on their values - using javascript

I am currently exploring the most effective approach to divide an array into multiple arrays based on specific operations performed on the values within the array.

Imagine I have an array like this:

var arr = [100, 200, 300, 500, 600, 700, 1000, 1100, 1200]

I aim to separate this array into different arrays whenever the difference between elements exceeds 100.

The desired output would be:

var result = [[100, 200, 300], [500, 600, 700], [1000, 1100, 1200]]

What is the most efficient method to achieve this?

I have attempted using a traditional for loop with if conditions that check the difference between consecutive elements. Whenever the difference surpasses 100, a new array is created. However, I sense there could be a more optimized solution.

EDIT:-

This is the code snippet I have implemented. Although it is close, it is not yet complete. I did not invest further time as I was aiming to find a simpler version utilizing reduce or underscore -

var result = [], temp = [], difference;
for (var i = 0; i < array.length; i += 1) {
 if (difference !== (array[i + 1] - array[i])) {
    if (difference !== undefined) {
        result.push(temp);
        temp = [];
    }
    difference = array[i + 1] - array[i];
}
temp.push(array[i]);
}

if (temp.length) {
  result.push(temp);
}

javascript arrays javascript-objects

Answer 1

Answer №1

To achieve this, one can utilize the reduce method.

var numbers = [100, 200, 300, 500, 600, 700, 1000, 1100, 1200];

var groups = numbers.reduce(function (accumulator, currentValue) {
  if (accumulator.length === 0) { // First Iteration
    accumulator.push([currentValue]);
    return accumulator;
  }

  var lastGroupIndex = accumulator.length - 1;
  var lastGroup = accumulator[lastGroupIndex];

  var lastElementIndex = lastGroup.length - 1;
  var lastElement = lastGroup[lastElementIndex];

  if (currentValue - lastElement <= 100) { // Add to current group
    lastGroup.push(currentValue);
    return accumulator;
  }

  accumulator.push([currentValue]); // Create new group
  return accumulator;
}, [])

Note: This approach avoids using ES6 features as it is based on the assumption that the codebase does not currently support ES6, particularly noted through the use of var.

Answer 2

To achieve this, one can utilize the reduce method.

var numbers = [100, 200, 300, 500, 600, 700, 1000, 1100, 1200];

var groups = numbers.reduce(function (accumulator, currentValue) {
  if (accumulator.length === 0) { // First Iteration
    accumulator.push([currentValue]);
    return accumulator;
  }

  var lastGroupIndex = accumulator.length - 1;
  var lastGroup = accumulator[lastGroupIndex];

  var lastElementIndex = lastGroup.length - 1;
  var lastElement = lastGroup[lastElementIndex];

  if (currentValue - lastElement <= 100) { // Add to current group
    lastGroup.push(currentValue);
    return accumulator;
  }

  accumulator.push([currentValue]); // Create new group
  return accumulator;
}, [])

Note: This approach avoids using ES6 features as it is based on the assumption that the codebase does not currently support ES6, particularly noted through the use of var.

Answer 3

Answer №2

This question really piqued my interest.

Update: Upon taking another look, it's not explicitly stated whether the input array is sorted or not. If it happens to be sorted, then utilizing the reduce function would make things much simpler.

Initially, I pondered if there might be a more efficient solution than O(nlogn). However, even with item grouping, the ultimate combined array remains completely sorted. Thus, if there exists a way to solve your problem faster than nlogn, you could also improve sorting time beyond nlogn.

Considering this, the most direct approach involves sorting the array and then partitioning it based on the last element encountered during iteration. Reduce proves to be an ideal tool in this scenario:

const input = [100, 200, 300, 500, 600, 700, 1000, 1100, 1200];

const sorted = input.slice().sort((a, b) => a - b);

const buckets = sorted.reduce((buckets, item, i) => {
  const lastBucket = buckets[buckets.length - 1];
  if (i === 0) {
    return [[item]];
  }
  const prev = sorted[i - 1];
  if (item - prev > 100) {
    buckets.push([item]);
  } else {
    buckets[buckets.length - 1].push(item);
  }
  return buckets;
}, [])
console.log(buckets);

Answer 4

This question really piqued my interest.

Update: Upon taking another look, it's not explicitly stated whether the input array is sorted or not. If it happens to be sorted, then utilizing the reduce function would make things much simpler.

Initially, I pondered if there might be a more efficient solution than O(nlogn). However, even with item grouping, the ultimate combined array remains completely sorted. Thus, if there exists a way to solve your problem faster than nlogn, you could also improve sorting time beyond nlogn.

Considering this, the most direct approach involves sorting the array and then partitioning it based on the last element encountered during iteration. Reduce proves to be an ideal tool in this scenario:

const input = [100, 200, 300, 500, 600, 700, 1000, 1100, 1200];

const sorted = input.slice().sort((a, b) => a - b);

const buckets = sorted.reduce((buckets, item, i) => {
  const lastBucket = buckets[buckets.length - 1];
  if (i === 0) {
    return [[item]];
  }
  const prev = sorted[i - 1];
  if (item - prev > 100) {
    buckets.push([item]);
  } else {
    buckets[buckets.length - 1].push(item);
  }
  return buckets;
}, [])
console.log(buckets);

Answer 5

Answer №3

In this particular solution, we're utilizing the reduce method to manipulate an array based on a specific condition without sorting.

var arrOrg = [100, 200, 300, 500, 600, 700, 1000, 1100, 1200]
arrOrg.reduce(function (acc, item, idx, arr) {
  if (idx) {
    (idx < arr.length && arr[idx + 1] > item + 100) // the spec condition
      ? acc[acc.length] = [item]
      : acc[acc.length - 1].push(item)
  } else acc[0][0] = item
  return acc
}, [[]])

It's important to note that this function results in an empty nested array when called with no input value.

Furthermore, it disregards any need for sorting within the array elements.

Answer 6

In this particular solution, we're utilizing the reduce method to manipulate an array based on a specific condition without sorting.

var arrOrg = [100, 200, 300, 500, 600, 700, 1000, 1100, 1200]
arrOrg.reduce(function (acc, item, idx, arr) {
  if (idx) {
    (idx < arr.length && arr[idx + 1] > item + 100) // the spec condition
      ? acc[acc.length] = [item]
      : acc[acc.length - 1].push(item)
  } else acc[0][0] = item
  return acc
}, [[]])

It's important to note that this function results in an empty nested array when called with no input value.

Furthermore, it disregards any need for sorting within the array elements.

Organize array elements based on their values - using javascript

Answer №1

Answer №2

Answer №3

Similar questions

What is the best way to extract data from individual columns in a PHP multidimensional array?

Grabbing a collection of GeoPoints from a Firebase backend to incorporate into a map's polygon

Adjust the size of the font in your Bootstrap Table

Having trouble showing an image with jQuery after it has been uploaded

Assign a class to an element depending on the date

Unable to instantiate a class using a module in React

Electron Builder reminder: Make sure to include the author's email in the application package.json file

What makes CVE-2021-33623 susceptible to ReDoS attacks?

What is the best way to retrieve JavaScript Values through Selenium?

Using a Promise to signal the completion of certain tasks

The $scope.$watch function does not activate with each individual change

Conceal the second click action within the anchor tag

Angular 4 - The Promising Outcome: How to Retrieve the Value upon Completion

Utilizing objects from a JSON file within an HTML document

Vue's computed property failing to compute

Angular.js: The Best Way to Attach a "Change" Event to a Service

Is there a method to extract the y value for a specific x value that is not part of the current data set in a Chart.js graph

Convert epoch time to HHMM format using JavaScript

Babel Compile disrupts the flow of commands

Exploring the world of jQuery: Toggling radio buttons with variables on click