Sign Up

Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.

Have an account? Sign In

Have an account? Sign In Now

Sign In

Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.

Sign Up Here

Forgot Password?

Don't have account, Sign Up Here

Forgot Password

Lost your password? Please enter your email address. You will receive a link and will create a new password via email.

Have an account? Sign In Now

You must login to ask a question.

Forgot Password?

Need An Account, Sign Up Here

Please briefly explain why you feel this question should be reported.

Please briefly explain why you feel this answer should be reported.

Please briefly explain why you feel this user should be reported.

Sign InSign Up

The Archive Base

The Archive Base Logo The Archive Base Logo

The Archive Base Navigation

  • SEARCH
  • Home
  • About Us
  • Blog
  • Contact Us
Search
Ask A Question

Mobile menu

Close
Ask a Question
  • Home
  • Add group
  • Groups page
  • Feed
  • User Profile
  • Communities
  • Questions
    • New Questions
    • Trending Questions
    • Must read Questions
    • Hot Questions
  • Polls
  • Tags
  • Badges
  • Buy Points
  • Users
  • Help
  • Buy Theme
  • SEARCH
Home/ Questions/Q 738889
In Process

The Archive Base Latest Questions

Editorial Team
  • 0
Editorial Team
Asked: May 14, 20262026-05-14T08:20:15+00:00 2026-05-14T08:20:15+00:00

I’ve something like this Object[] myObjects = …(initialized in some way)… int[] elemToRemove =

  • 0

I’ve something like this

Object[] myObjects = ...(initialized in some way)...
int[] elemToRemove = new int[]{3,4,6,8,...}

What’s the most efficient way of removing the elements of index position 3,4,6,8… from myObjects ?

I’d like to implement an efficient Utility method with a signature like

public Object[] removeElements(Object[] object, int[] elementsToRemove) {...}

The Object[] that is returned should be a new Object of size myObjects.length – elemToRemove.length

  • 1 1 Answer
  • 0 Views
  • 0 Followers
  • 0
Share
  • Facebook
  • Report

Leave an answer
Cancel reply

You must login to add an answer.

Forgot Password?

Need An Account, Sign Up Here

1 Answer

  • Voted
  • Oldest
  • Recent
  • Random
  1. Editorial Team
    Editorial Team
    2026-05-14T08:20:16+00:00Added an answer on May 14, 2026 at 8:20 am

    You will need to create a new array, since arrays do not change in size (you cannot “remove” elements while keeping the same array, it would create holes somewhere). I suggest this:

    Object[] nobjs = Arrays.copyOf(myObjects, myObjects.length - elemToRemove.length);
    for (int i = 0, j = 0, k = 0; i < myObjects.length; i ++) {
        if (j < elemToRemove.length && i == elemToRemove[j]) {
            j ++;
        } else {
            nobjs[k ++] = myObjects[i];
        }
    }
    

    This assumes that elemToRemove is already sorted, with no duplicates, and contains only indices valid within the source array. The Arrays.copyOf() call is used only to make sure that the runtime type of the new array is identical to that of the source array.

    If you often have to “remove” data from arrays, thus requiring the creation of new arrays, then Java arrays might not be the most efficient data structure for you. LinkedList or ArrayList may be more appropriate.

    • 0
    • Reply
    • Share
      Share
      • Share on Facebook
      • Share on Twitter
      • Share on LinkedIn
      • Share on WhatsApp
      • Report

Sidebar

Ask A Question

Stats

  • Questions 404k
  • Answers 404k
  • Best Answers 0
  • User 1
  • Popular
  • Answers
  • Editorial Team

    How to approach applying for a job at a company ...

    • 7 Answers
  • Editorial Team

    What is a programmer’s life like?

    • 5 Answers
  • Editorial Team

    How to handle personal stress caused by utterly incompetent and ...

    • 5 Answers
  • Editorial Team
    Editorial Team added an answer A problem I've often had with layouts such as that… May 15, 2026 at 5:36 am
  • Editorial Team
    Editorial Team added an answer This is undoubtedly an error. The comma here is the… May 15, 2026 at 5:36 am
  • Editorial Team
    Editorial Team added an answer Filter out the pieces with the correct category, then count… May 15, 2026 at 5:36 am

Trending Tags

analytics british company computer developers django employee employer english facebook french google interview javascript language life php programmer programs salary

Top Members

Explore

  • Home
  • Add group
  • Groups page
  • Communities
  • Questions
    • New Questions
    • Trending Questions
    • Must read Questions
    • Hot Questions
  • Polls
  • Tags
  • Badges
  • Users
  • Help
  • SEARCH

Footer

© 2021 The Archive Base. All Rights Reserved
With Love by The Archive Base

Insert/edit link

Enter the destination URL

Or link to existing content

    No search term specified. Showing recent items. Search or use up and down arrow keys to select an item.