Three Largest Number Interview Question: iOS Interviews (Algorithms + FAANG)

[@iOSAcademy]

Like for a cookie 🍪Also, lets be friends! linkedin.com/in/afrazsiddiqui

[@dandownscolorado]

Did you try duplicate large numbers? [12, 12, 10, 9] How about [1, 2]? I bet it will give you [Int.min, 1, 2] for an output.

[@user-gu1sl1mn2m]

More interview vids please!! I don’t see many iOS interview videos and this was great!!

[@1000ylovers]

You finally posted interview question! Hopefully we will see more of these!!

[@ipaktulane4748]

thanks for keeping doing this

[@maqusss]

Playing with indexes is a little confusing. Following seems simpler to me: func solve(_ arr: [Int]) -> (Int,Int,Int)? { guard arr.count >= 3 else { return nil } // m1 >= m2 >= m3 var m1: Int = arr.min()! var m2: Int = m1 var m3: Int = m1 for x in arr { if x >= m1 { m3 = m2 m2 = m1 m1 = x } else if x >= m2 { m3 = m2 m2 = x } else if x >= m3 { m3 = x } } return (m3,m2,m1) }

[@victorriurean]

welcome back

[@waqarzahoor7201]

The solution that you have provided If we commented out the below lines it also works. func rearrange(number: Int, result: inout [Int]) { var toBeInsertedIdx = -1 if number > result[2] { toBeInsertedIdx = 2 } else if number > result[1] { toBeInsertedIdx = 1 } else if number > result[0] { toBeInsertedIdx = 0 } else { return } var currentIdx = toBeInsertedIdx while currentIdx > 0 { // let temp = result[currentIdx - 1] result[currentIdx - 1] = result[toBeInsertedIdx] currentIdx -= 1 // result[toBeInsertedIdx] = temp } result[toBeInsertedIdx] = number }

[@waqarzahoor7201]

Gotchas: It fails on this input [7, 8, 9, 15] if we commented out

[@MaxTymchii]

What do you think about this? func threeLargestDigits(at inputArray: [Int]) -> [Int] { guard inputArray.count > 2 else { return [] } var result: [Int] = .init(repeating: Int.min, count: 3) inputArray.forEach { newValue in rearrange(newValue, result: &result) } return result.reversed() } func rearrange(_ number: Int, result: inout [Int]) { var candidateNumber = number for index in 0..

[@techwithchavan]

LGTM

[@Денис-ж3ф5р]

Unbelievable that Google asks such easy questions

[@venuvenu2719]

Is native ios development dying due to KMP release?

[@ihorzhukov]

no, as it haven't died yet after React Native release, and after Flutter release.

[@shaikrahim8071]

Never and ever, vision OS Coming, so all orther cross platform will be die very soon.

[@ihorzhukov]

@@shaikrahim8071 why?

[@thinhphung7417]

Never. Because Apple will never let its son die. Cross platform will be depend on native, there are many APIs and functions that cross platform will never be easy to implement, and never has performance good as native code.

[@fitdevelopertv5148]

=====SOLUTION IN SWIFT====== let array = [1,88,2,12,11,23,104] //[88,23,12] var resultArray = Array(Array(repeating: Int.min, count: 3)) for i in 0.. resultArray[0] { resultArray[2] = resultArray[1] resultArray[1] = resultArray[0] resultArray[0] = array[i] } else if array[i] > resultArray[1] { resultArray[2] = resultArray[1] resultArray[1] = array[i] } else { resultArray[2] = array[i] } } print(resultArray)

[@XsiadzBiskup]

func threeLargest(numbers:[int]) -> [int] { guard numbers.count >= 3 else {return []} return numbers.sorted(by: >).prefix(3) } You can do this in 2 lines of code. This is going to have a complexity of O(n log(n)). A more optimal way O(n) would be to do this without sorting and iterate though the numbers finding the 3 largest ones. func findThreeLargestNumbers(in array: [Int]) -> [Int] { guard array.count >= 3 else { print("Array should have at least 3 elements.") return [] } var largest = Int.min var secondLargest = Int.min var thirdLargest = Int.min for number in array { if number > largest { (largest, secondLargest, thirdLargest) = (number, largest, secondLargest) } else if number > secondLargest { (secondLargest, thirdLargest) = (number, secondLargest) } else if number > thirdLargest { thirdLargest = number } } return [largest, secondLargest, thirdLargest] }

[@virajpadsala2572]

let sortedNumbers = numbers.sorted(by: >) // Sort the numbers in descending order return Array(sortedNumbers.prefix(3)) // Select the first three elements This will do the trick instead of repeating and rearranging

[@markaurelius61]

If you can use sorting, it looks like this: let nums = [10, 9, 20, 22, 15, 100, 2] let upper = min(3, nums.count) let a1 = Array(nums.sorted().reversed()[0..

[@timbass9822]

Should have used sort(). Writing code to do what sort() or sorted() does means money wasted managing unnecessary code. In my past, would not hire anyone who rewrites built-in functionality.

[@anupdsouza]

That's not the point. Higher order functions obviously make life easier but companies prefer that candidates come up with solutions without using them to gauge their problem solving & data structure skills.

[@r4lphTv]

func threeLargestNumber(from array: [Int]) -> [Int]{ guard array.count > 2 else { return [] } var result = array.sorted(by: { $0 < $1 })[array.count-3...array.count-1] return Array(result) } let array = [11, 122, 32, 46, 78, 1, 3, 91, 95, 101, 17, 103] print(threeLargestNumber(from: array)) //[101, 103, 122]