Question

How to find the second largest number in array in Swift

Answer 1

import Foundation

func secondLargestNumber(in array: [Int]) -> Int? {
    let sortedArray = array.sorted(by: >)
    
    return sortedArray.count > 1 ? sortedArray[1] : nil
}

let numbers = [3, 14, 14, 1, 1, 1, 90, 2, 6, 86, 7]

if let secondLargest = secondLargestNumber(in: numbers) {
    print("The second largest number is \(secondLargest)")
} else {
    print("Array does not have enough elements.")
}



/*
run:

The second largest number is 86

*/

 

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Answer 2

import Foundation

func secondLargestNumber(in array: [Int]) -> Int? {
    guard array.count > 1 else { return nil }
    
    var largest = Int.min
    var secondLargest = Int.min
    
    for number in array {
        if number > largest {
            secondLargest = largest
            largest = number
        } else if number > secondLargest && number < largest {
            secondLargest = number
        }
    }
    
    return secondLargest
}

let numbers = [3, 14, 14, 1, 1, 1, 90, 2, 6, 86, 7]

if let secondLargest = secondLargestNumber(in: numbers) {
    print("The second largest number is \(secondLargest)")
} else {
    print("Array does not have enough elements.")
}



/*
run:

The second largest number is 86

*/

 

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Answer 3

import Foundation

func secondLargestNumber(in array: [Int]) -> Int? {
    let uniqueNumbers = Array(Set(array))
    guard uniqueNumbers.count > 1 else { return nil }
    
    let sortedArray = uniqueNumbers.sorted(by: >)
    
    return sortedArray[1]
}

let numbers = [3, 14, 14, 1, 1, 1, 90, 2, 6, 86, 7]

if let secondLargest = secondLargestNumber(in: numbers) {
    print("The second largest number is \(secondLargest)")
} else {
    print("Array does not have enough elements.")
}



/*
run:

The second largest number is 86

*/

 

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