Exam of Oursky

answer1.swift

func isSubset(_ a: [Character], _ b: [Character]) -> Bool {
    return b.filter({ (item) -> Bool in
        return a.index(of: item) != nil
    }).count == b.count
}

isSubset(["A", "B", "C", "D"], ["A", "C"])
isSubset(["A","B","C","D","E"], ["A","D","Z"])
isSubset(["A","D","E"], ["A","A","D","E"])

answer2.md

Answer: O(n^2)

In swift, filter an array means a loop to determine each item available, so the time complexity is O(n) And find an index in an array is the same, time complexity is O(n) .count and equal operation is all constant, so they are O(1)

So the total time complexity is O(n^2)

answer3.swift

func nextFibonacci(numbers: [Int]) -> [Int] {
    let sortedNumbers = numbers.sorted()
    guard let min = sortedNumbers.first,
        let max = sortedNumbers.last else {
            return []
    }
    var prevFibonacci = 1
    var fibonacci = 1
    var fibonacciNumbers = [1,1]
    while fibonacci < max {
        let nextFibonacci = prevFibonacci + fibonacci
        prevFibonacci = fibonacci
        fibonacci = nextFibonacci
        fibonacciNumbers.append(nextFibonacci)
    }
    return numbers.map { (n) -> Int in
        for f in fibonacciNumbers {
            if f > n {
                return f
            }
        }
        preconditionFailure()
    }
}

nextFibonacci(numbers: [1, 9, 22])

answer4.js

function createArrayOfFunctions(y) {
var arr = [];
for(var i = 0; i<y; i++) {
arr[i] = function(x) { return x + i; }(i); // The bug is here
}
return arr;
}

// in the for loop, the value of elements are functions not numbers
// So I add (i) after it to get the return value and assign into arr