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Home/ Questions/Q 6703455
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Editorial Team
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Editorial Team
Asked: May 26, 20262026-05-26T07:10:23+00:00 2026-05-26T07:10:23+00:00

With an array like this: [google, google, yahoo, yahoo, yahoo, msn, msn, msn, google]

  • 0

With an array like this:

[google, google, yahoo, yahoo, yahoo, msn, msn, msn, google]

what’s the best way to arrange it with as few adjacent values as possible?
The goal is to get something like this, for instance:

[google, msn, yahoo, google, yahoo, msn, yahoo, msn, google]

Even better would be an algoritnm that ensures that any similar values are as far as possible from each other.

And even better would be to do it directly in SQL.

I’m doing that for a script I’m working on that sends newsletters, and I’d like to avoid sending too many emails to the same domain at once.

[edit] I’m using MySQL

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-26T07:10:24+00:00Added an answer on May 26, 2026 at 7:10 am

    Using SQL Server 2005 and up, you might take advantage of the ROW_NUMBER functionality

    • Add a row number to each item, partitioning on sitename.
    • select from this subquery
    • order the results on the rownumber and name.

    Your requirement that they should be as far away from each other as possible might not be met but all in all, this would suffice.

    SQL Statement

    SELECT  Site
    FROM    (
                SELECT  Site, o = ROW_NUMBER() OVER (PARTITION BY Site ORDER BY Site)
                FROM    q
            ) q
    ORDER BY                
            o, Site
    

    Results

    google msn yahoo google msn yahoo google msn yahoo

    Test script

    ;WITH q ([site])AS (
        SELECT 'google'
        UNION ALL SELECT 'google'
        UNION ALL SELECT 'yahoo'
        UNION ALL SELECT 'yahoo'
        UNION ALL SELECT 'yahoo'
        UNION ALL SELECT 'msn'
        UNION ALL SELECT 'msn'
        UNION ALL SELECT 'msn'
        UNION ALL SELECT 'google'   
    )
    SELECT  Site
    FROM    (
                SELECT  Site, o = ROW_NUMBER() OVER (PARTITION BY Site ORDER BY Site)
                FROM    q
            ) q
    ORDER BY                
            o, Site
    
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