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Home/ Questions/Q 421895
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Editorial Team
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Editorial Team
Asked: May 12, 20262026-05-12T19:01:40+00:00 2026-05-12T19:01:40+00:00

I’ve got a sorted array: array = [[4, 13], [1, 12], [3, 8], [2,

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I’ve got a sorted array:

array = [[4, 13], [1, 12], [3, 8], [2, 8], [0, 3]]

Which shows me a position (array[n][0]) and the number of occurrences of that position (array[n][1]).

I need to test to see if more than one item in the array has the same number of occurrences as the last item.

I thought I might be able to do it with this:

array.detect {|i| i[1] == array.last[1] }.length

But it returns 2 for the above array, and seems to also return 2 for the following array:

array = [[4, 13], [1, 12], [3, 8], [2, 3], [0, 3]]

When I run it without length it always returns the first occurrence.

Is there a way to get this to count the occurrences?

EDIT:

Sorry, will ask my follow up question in a new question.

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  1. Editorial Team
    Editorial Team
    2026-05-12T19:01:41+00:00Added an answer on May 12, 2026 at 7:01 pm

    Try using find_all instead of detect. detect returns the first match. In your first example, that’s array[3], which is another array of length 2. That’s why it’s returning 2 (it should always be returning 2 or nil for your arrays). find_all will return an array of the matches (instead of the first match itself), and its length will be the value you want.

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    Editorial Team added an answer That sounds like a reasonable approach to me. Later on,… May 13, 2026 at 12:49 am

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