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Editorial Team
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Editorial Team
Asked: May 24, 20262026-05-24T04:06:43+00:00 2026-05-24T04:06:43+00:00

I am just getting started with MongoDB and trying to understand how indexes work.

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I am just getting started with MongoDB and trying to understand how indexes work. I have a list of items in a collection. Each item has a version that gets incremented. Then, all previous versions (less than current version) get removed (record is not updated so that both versions are available for a while). There is a compound index on item ID and version. For removing, does it make a difference (in terms of performance) whether you use $ne versus $lt?

I would assume no, but I just want to confirm.

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  1. Editorial Team
    Editorial Team
    2026-05-24T04:06:44+00:00Added an answer on May 24, 2026 at 4:06 am

    Without knowing the details of the implementation $lt can be more efficient than $ne. On a B-tree index $ne would be two range scans ($lt and $gt), whereas $lt is just one.

    But in your case $lt seems to be what you want anyway (to find the older versions). If you used $ne, you could accidentally also remove newer versions that you just assume do not exist, but might actually have been created in the mean-time. Remember that MongoDB does not support transactions or consistent views across documents. Concurrent updates might bite you here.

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