Minimum value from range of Array

Question

I have a list of prices and want to find the minimum price EXCLUDING the first element (this is a subset of another problem on HackerRank).

My version is too slow and times out. I suspect this is due to my ArraySlice.

Here is my (working) code:

func calculateMins(prices: [Int]) {
    for j in 1..<prices.count  {
        let lowestPreviousPrice = prices[1...j].min()!
        print (lowestPreviousPrice)
    }
}

calculateMins(prices: [4,8,2,4,3])

Is there a better performing version of this, perhaps one that does not use an ArraySlice?

Needs to be a subset of the array (I've overly reduced the problem) I'll change this example to excluding the first element instead. — stevenpcurtis
– stevenpcurtis, Commented Jan 29, 2019 at 9:47
Let's just take it as the min of an array, EXCLUDING the first element. Have updated the question. — stevenpcurtis
– stevenpcurtis, Commented Jan 29, 2019 at 9:50

Mohmmad S · Accepted Answer · 2019-01-29 09:57:55Z

6

Just use dropFirst() function.

var array = [1,8,2,4,3]
var array2 = array.dropFirst() // Returns a subsequence containing all but the first element of the sequence.
array2.min() // 2

answered Jan 29, 2019 at 9:57

Mohmmad S

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Comments

Joakim Danielson · Accepted Answer · 2019-01-29 09:55:53Z

2

Why not keep it simple

func calculateMins(prices: [Int]) {
    var min = Int.max
    for i in 1..<prices.count {
        if prices[i] < min { min = prices[i] }
    }
    print(min)    
}

answered Jan 29, 2019 at 9:55

Joakim Danielson

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Comments

Oleg Gordiichuk · Accepted Answer · 2019-01-29 09:57:16Z

2

You have few options to solve this issue.

//Default way
prices.dropFirst().min()

//Functional way
prices.dropFirst().reduce(Int.max, { min($0, $1) })

answered Jan 29, 2019 at 9:57

Oleg Gordiichuk

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Comments

Sandeep · Accepted Answer · 2019-01-29 10:26:49Z

1

You could also use suffix, which is quite same as dropFirst that this version could crash if in case array is empty.

array.suffix(from: 1).min()

answered Jan 29, 2019 at 10:26

Sandeep

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