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Home/ Questions/Q 8820249
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Editorial Team
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Editorial Team
Asked: June 14, 20262026-06-14T05:33:45+00:00 2026-06-14T05:33:45+00:00

I have a 2D vector containing 96 blocks of 600 values, which is what

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I have a 2D vector containing 96 blocks of 600 values, which is what I want.

I need to remove (blocks) that do not contain sufficient energy. I have managed to calculate the energy but do not know which way would be better in removing the (blocks) that do not contain enough energy.

In your opinions would it be better to create a temporary 2D vector, that pushed back the blocks that do contain enough energy and then delete the original vector from memory or…

Should I remove the blocks from the vector at that particular position?

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  1. Editorial Team
    Editorial Team
    2026-06-14T05:33:47+00:00Added an answer on June 14, 2026 at 5:33 am

    I’m assuming you have this:

    typedef std::vector<value> Block;
    typedef std::vector< Block > my2dVector;
    

    and you have a function like this:

    bool BlockHasInsufficientEnergy( Block const& vec );
    

    and you want to remove the Blocks that do not have sufficient energy.

    By remove, do you mean you want there to be fewer than 96 Blocks afterwards? I will assume so.

    Then the right way to do this is:

    void RemoveLowEnergyBlocks( my2dVector& vec )
    {
      my2dVector::iterator erase_after = std::remove_if( vec.begin(), vec.end(), BlockHasInsufficientEnergy );
      vec.erase( erase_after, vec.end() );
    }
    

    the above can be done in one line, but by doing it in two what is going on should be more clear.

    remove_if finds everything that passes the 3rd argument condition, and filters it out of the range. It returns the point where the “trash” at the end of the vector lives. We then erase the trash. This is called the remove-erase idiom.

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