I have an ordered sequence of numbers, let's say something like
0, 1, 2, 3, 5, 6, 11, 12, 15, 20
Given a number N, how could I get a sequence that starts from the last number that is smaller than N? For example, if N = 7, I'd like to get back
6, 11, 12, 15, 20
Please note that this sequence will get very big and new numbers will be appended.
drop(while:) seemed like a good candidate, but in the example above it would also drop 6 so I can't use it.
For huge sorted arrays the most efficient way is binary search. It cuts the array in half until the index was found.
extension RandomAccessCollection where Element : Comparable {
func lastIndex(before value: Element) -> Index {
var slice : SubSequence = self[...]
while !slice.isEmpty {
let middle = slice.index(slice.startIndex, offsetBy: slice.count / 2)
if value < slice[middle] {
slice = slice[..<middle]
} else {
slice = slice[index(after: middle)...]
}
}
return slice.startIndex == self.startIndex ? startIndex : index(before: slice.startIndex)
}
}
let array = [0, 1, 2, 3, 5, 6, 11, 12, 15, 20]
let index = array.lastIndex(before: 7)
print(array[index...])
Related
Given an array of integers for example let array = [1, 3, 4, 7, 8, 9, 12, 14, 15]
What is the best way of finding the largest amount of consecutive integers preferably without using a for-in loop. If we would pass this array into a function it would return 3 as '7, 8, 9' is the largest amount of consecutive integers.
let array = [1, 3, 4, 7, 8, 9, 12, 14, 15]
func getMaxConsecutives(from array: [Int]) -> Int {
var maxCount = 1
var tempMaxCount = 1
var currentNumber = array[0]
for number in array {
if currentNumber == number - 1 {
tempMaxCount += 1
maxCount = tempMaxCount > maxCount ? tempMaxCount : maxCount
currentNumber = number
} else {
tempMaxCount = 1
currentNumber = number
}
}
return maxCount
}
getMaxConsecutives(from: array)
This works as intended but I would like a more efficient solution something that is not O(n).
I appreciate any creative answers.
You can do it like this:
let array = [1, 3, 4, 7, 8, 9, 12, 14, 15]
if let maxCount = IndexSet(array).rangeView.max(by: {$0.count < $1.count})?.count {
print("The largest amount of consecutive integers: \(maxCount)")
//prints 3
}
I think I can write it more tightly (basically as a one-liner):
let array = [1, 3, 4, 7, 8, 9, 12, 14, 15]
let (_,_,result) = array.reduce((-1000,1,0)) {
$1 == $0.0+1 ? ($1,$0.1+1,max($0.2,$0.1+1)) : ($1,1,$0.2)
}
print(result) // 3
But we are still looping through the entire array — so that we are O(n) — and there is no way to avoid that. After all, think about what your eye does as it scans the array looking for the answer: it scans the whole array.
(One way to achieve some savings: You could short-circuit the loop when we are not in the middle of a run and the maximum run so far is longer than what remains of the array! But the gain might not be significant.)
I have a string that consists of few numbers. First number is the number of the row, remaining are numbers in this row.(Array but string, kind of). The problem is that remaining numbers are unsorted, and I want to find the clearest way of sorting them without creating new List and sorting everything there.
String unsorted = '9, 12, 14, 11, 2, 10';
print(unsorted.sort()); // '9, 2, 10, 11, 12, 14'
You cant really avoid converting to a list of numbers if you want to sort it.
void main() {
print(sortNumString('9, 12, 14, 11, 2, 10')); // 2, 9, 10, 11, 12, 14
}
String sortNumString(String numString, [String separator = ', ']) =>
(numString.split(separator).map(int.parse).toList()..sort())
.join(separator);
The .. means to return the previous thing, the list, since sort returns void.
I'm not exactly experienced when it comes to dart programming language but this is what i came up with.
void main() {
var unsorted = "9, 12, 14, 11, 2, 10";
var nums_int = unsorted.split(", ").map(int.parse).toList();
nums_int.sort();
for (var n in nums_int) {
stdout.write(n.toString() + ", ");
}
}
That should give the expected output: "2, 9, 10, 11, 12, 14"
Hope this helps.
I have an array with integers [1, 2, 3, 7, 13, 11, 4] and an integer value 12. I need to return an array of a closest combination of sum of the integer values in the new array.
For example: [1, 2, 3, 7, 13, 11, 4] with value 12. The array that need to returned is [1, 2, 3, 4] because the sum of elements 1 + 2 + 3 + 4 <= 12. This is the longest array and prefered over [1, 7, 4] 1 + 7 + 4 = 12.
this solution works if you dont have repeated numbers in array.
var array = [1, 2, 3, 7, 13, 11, 4, -12, 22, 100]
print(array.sorted())
let value = 19
func close(array: [Int], value: Int) -> (Int , Int) {
let sortedArray = array.sorted()
var lastNumberBeforValue = sortedArray.first!
for (index,number) in sortedArray.enumerated() {
let sub = value - number
if sub > 0 {
lastNumberBeforValue = number
}
if sortedArray.contains(sub) && sub != number {
return (sub, number)
} else if index == sortedArray.count - 1 {
if sub < 0 {
let near = close(array: array, value: value - lastNumberBeforValue)
return (near.0, lastNumberBeforValue)
}
let near = close(array: array, value: sub)
return (near.0, number)
}
}
return (-1,-1)
}
let numbers = close(array: array, value: value)
print(numbers) //prints (4, 13)
This solution also finds the two Integers in an array which sum is closest to a given value:
let array = [1, 2, 3, 7, 13, 11, 4]
let value = 5
var difference = Int.max
var result = [Int(), Int()]
for firstInt in array {
for secondInt in array {
let tempDifference = abs((firstInt + secondInt) - value)
if tempDifference < difference {
difference = tempDifference
result[0] = firstInt
result[1] = secondInt
}
}
}
Or a solution that doesn't allow multiple use of the same value:
for (firstIndex, firstInt) in array.enumerated() {
for (secondIndex, secondInt) in array.enumerated() {
guard firstInt != secondInt else { break }
let tempDifference = abs((firstInt + secondInt) - value)
if tempDifference < difference {
difference = tempDifference
result[0] = firstInt
result[1] = secondInt
}
}
}
You can keep on nesting for-loops and guarding for multiple use of same index, if you want longer result arrays. Finally you can take the valid array with the largest result.count. This is not a vary nice solution though, since it computationally heavy, and requires the array to have a static length.
I'm not necessary good with algorithms but I would do it like that :
Since your rule is <= 5 you can sort your array : [1, 2, 3, 7, 13, 11, 4] => [1, 2, 3, 4, 7, 11, 13]
Then, as you don't need integer > 5 you can split your array in 2 parts and take only the first one : [1, 2, 3, 4]
From there you have to add 4 + 3 and match with your value. If it doesn't work do 4 + 2, then 4 + 1. If it's still > 5, loop with 3,2 ; 3,1 ; etc. I did something like this in a Swift way :
// Init
let array = [1, 2, 3, 7, 13, 11, 4]
let value = 5
let sortedArray = array.sorted()
let usefulArray = sortedArray.filter() {$0 < value}
var hasCombination = false
var currentIndex = usefulArray.count - 1
var indexForSum = currentIndex - 1
// Process
if currentIndex > 0 {
hasCombination = true
while usefulArray[currentIndex] + usefulArray[indexForSum] > value {
indexForSum -= 1
if indexForSum < 0 {
currentIndex -= 1
indexForSum = currentIndex - 1
}
if currentIndex == 0 {
hasCombination = false
break
}
}
}
// Result
if hasCombination {
let combination = [usefulArray[indexForSum], usefulArray[currentIndex]]
} else {
print("No combination")
}
I guess it works, tell me if it doesn't !
I understand how to write a for loop using swift syntax (using .enumerate() and .revers()), however how would I re-write (in swift) the following javascript version of for loop, considering I have multiple conditions to adhere to:
for(var j = 10; j >= 0 && array[j] > value; j--) {
array[j + 1] = array[j];
}
What about this?
for j in (0...10).reversed() {
guard array[j] > value else { break }
array[j + 1] = array[j]
}
I'm not sure that produces exactly the same result, but this is one of the approaches in Swift 3
for j in stride(from:10, through:0, by: -1) {
if array[j] <= value { break }
array[j + 1] = array[j]
}
For the sake of completion (I would personally prefer to use a for loop with a check for an early break, as others have already suggested) – in Swift 3.1 you could use Sequence's prefix(while:) method in order to get the suffix of the array's indices where the elements meet a given predicate.
var array = [2, 3, 6, 19, 20, 45, 100, 125, 7, 9, 21, 22]
let value = 6
for i in array.indices // the array's indices.
.dropLast() // exclude the last index as we'll be accessing index + 1 in the loop.
.reversed() // reversed so we can get the suffix that meets the predicate.
.prefix(while: {array[$0] > value}) // the subsequence from the start of the
{ // collection where elements meet the predicate.
array[i + 1] = array[i]
}
print(array) // [2, 3, 6, 19, 19, 20, 45, 100, 125, 7, 9, 21]
This is assuming that you're looking to begin iterating at the penultimate index of the array. If you want to start at a particular index, you can say:
for i in (0...10).reversed().prefix(while: {array[$0] > value}) {
array[i + 1] = array[i]
}
This will start at the index 10 and iterate down to 0, giving you the same behaviour to the code in your question.
It's worth noting that both of the above variants will first iterate through the reversed indices (until the predicate isn't met), and then through the array's elements. Although, in Swift 3.1, there is a version of prefix(while:) which operates on a lazy sequence – which would allow for just a single iteration through the elements until the predicate isn't met.
Until Swift 3.1, you can use the below extension to get prefix(while:) on a Collection:
extension Collection {
func prefix(while predicate: (Self.Iterator.Element) throws -> Bool) rethrows -> Self.SubSequence {
var index = startIndex
while try index < endIndex && predicate(self[index]) {
formIndex(after: &index)
}
return self[startIndex..<index]
}
}
I am trying to solve the second problem on Project Euler. The problem is as follows:
Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 terms will be:
1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...
By considering the terms in the Fibonacci sequence whose values do not exceed four million, find the sum of the even-valued terms.
I think I've written a solution, but when I try to run my code it crashes my Swift playground and gives me this error message:
Playground execution aborted: Execution was interrupted, reason: EXC_BAD_INSTRUCTION (code=EXC_I386_INVOP, subcode=0x0)
var prev = 0
var next = 1
var num = 0
var sum = 0
for var i = 1; i < 400; i++ {
num = prev + next
if next % 2 == 0 {
sum += next
}
prev = next
next = num
}
print(sum)
The weird thing is, if I set the counter on my loop to less than 93, it works fine. Explicitly setting the variable names to Double does not help. Anyone know what's going on here?
There is nothing weird about this at all. Do you know how large the 400 fibonacci number is?
176023680645013966468226945392411250770384383304492191886725992896575345044216019675
Swift Int64 or UInt64 simply cannot handle that large of a number. The later can go up to 18446744073709551615 at max - not even close.
If you change your variables to be doubles it works but will be inaccurate:
var prev : Double = 0
var next : Double = 1
var num : Double = 0
var sum : Double = 0
will yield
2.84812298108489e+83
which is kind of close to the actual value of
1.76e+83
Luckily you do not need to get values that big. I would recommend not writing a for loop but a while loop that calculates the next fibonacci number until the break condition is met whose values do not exceed four million.
The Fibonacci numbers become very large quickly. To compute large Fibonacci numbers, you need to implement some kind of BigNum. Here is a version the makes a BigNum that is implemented internally as an array of digits. For example, 12345 is implemented internally as [1, 2, 3, 4, 5]. This makes it easy to represent arbitrarily large numbers.
Addition is implemented by making the two arrays the same size, then map is used to add the elements, finally the carryAll function restores the array to single digits.
For example 12345 + 67:
[1, 2, 3, 4, 5] + [6, 7] // numbers represented as arrays
[1, 2, 3, 4, 5] + [0, 0, 0, 6, 7] // pad the shorter array with 0's
[1, 2, 3, 10, 12] // add the arrays element-wise
[1, 2, 4, 1, 2] // perform carry operation
Here is the implementation of BigNum. It is also CustomStringConvertible which makes it possible to print the result as a String.
struct BigNum: CustomStringConvertible {
var arr = [Int]()
// Return BigNum value as a String so it can be printed
var description: String { return arr.map(String.init).joined() }
init(_ arr: [Int]) {
self.arr = carryAll(arr)
}
// Allow BigNum to be initialized with an `Int`
init(_ i: Int = 0) {
self.init([i])
}
// Perform the carry operation to restore the array to single
// digits
func carryAll(_ arr: [Int]) -> [Int] {
var result = [Int]()
var carry = 0
for val in arr.reversed() {
let total = val + carry
let digit = total % 10
carry = total / 10
result.append(digit)
}
while carry > 0 {
let digit = carry % 10
carry = carry / 10
result.append(digit)
}
return result.reversed()
}
// Enable two BigNums to be added with +
static func +(_ lhs: BigNum, _ rhs: BigNum) -> BigNum {
var arr1 = lhs.arr
var arr2 = rhs.arr
let diff = arr1.count - arr2.count
// Pad the arrays to the same length
if diff < 0 {
arr1 = Array(repeating: 0, count: -diff) + arr1
} else if diff > 0 {
arr2 = Array(repeating: 0, count: diff) + arr2
}
return BigNum(zip(arr1, arr2).map { $0 + $1 })
}
}
// This function is based upon this question:
// https://stackoverflow.com/q/52975875/1630618
func fibonacci(to n: Int) {
guard n >= 2 else { return }
var array = [BigNum(0), BigNum(1)]
for i in 2...n {
array.append(BigNum())
array[i] = array[i - 1] + array[i - 2]
print(array[i])
}
}
fibonacci(to: 400)
Output:
1
2
3
5
8
...
67235063181538321178464953103361505925388677826679492786974790147181418684399715449
108788617463475645289761992289049744844995705477812699099751202749393926359816304226
176023680645013966468226945392411250770384383304492191886725992896575345044216019675