CODEWITHSUNDEEP
ruby
nhershy
nhershy

Reputation: 755

Find second smallest element's index in list

I'm going by this algorithm to solve the problem:

  1. Initialize both first and second smallest as INT_MAX.

    first = second = INT_MAX

  2. Loop through all the elements. a. If the current element is smaller than the first, then update the first and the second. b. Else if the current element is smaller than the second, then update the second.

Here's my code:

#finds second smallest element's index in list
def secondSmallest(list)
    first = FIXNUM_MAX
    second = FIXNUM_MAX
    list.each_with_index do |item, index|
        if item < first
            second = first
            first = index
        elsif item < second
            second = index
        end
    end
    return second
end

I do not get the output I expect. I don't know where I'm going wrong.

Upvotes: 0

Views: 181

Answers (2)

Nimish Gupta
Nimish Gupta

Reputation: 3175

Well I think you can do this in a single line with the help of ruby iterators

list.index(list.sort.second)

Upvotes: 1

segfault
segfault

Reputation: 504

You are storing the indexes in first and second. Thus you cannot use them for value comparisons : item < second I think what you would need is 4 variables : maxVal, maxValIndex, minVal, minValIndex if you are going to follow the above mentioned approach.

Upvotes: 3

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