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.)
Related
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...])
I'm working through the Swift Guided Playground and came across this example.
The example code takes an array of dictionaries (kinds) and searches for the largest number. The exercise is to find both the largest number, and return the number's kind.
let interestingNumbers = [
"Prime": [2, 3, 5, 7, 11, 13],
"Fibonacci": [1, 1, 2, 3, 5, 8],
"Square": [1, 4, 9, 16, 25],
]
var largest = 0
for (kind, numbers) in interestingNumbers {
for number in numbers {
if number > largest {
largest = number
}
}
}
print(largest)
In this example, 25 is the largest number and is the "Square" kind.
My immediate instinct was to add a new string variable called "largestKind" and declare it after largest(Int).
let interestingNumbers = [
"Prime": [2, 3, 5, 7, 11, 13],
"Fibonacci": [1, 1, 2, 3, 5, 8],
"Square": [1, 4, 9, 16, 25],
]
var largest = 0
var largestKind: String
for (kind, numbers) in interestingNumbers {
for number in numbers {
if number > largest {
largest = number
largestKind = kind
}
}
}
print(largest)
print(largestKind) // error: Variable 'largestKind' used before being initialized
However, Swift throws this error stating that I have not initialized largestKind before using it. I suppose setting largestKind equal to kind does not count as initialization?
I found two work arounds to this error:
Initializing largestKind with a placeholder string (i.e. var largestKind = "")
Treat largestKind as an optional and then unwrap with print(largestKind!)
Example 1:
let interestingNumbers = [
"Prime": [2, 3, 5, 7, 11, 13],
"Fibonacci": [1, 1, 2, 3, 5, 8],
"Square": [1, 4, 9, 16, 25],
]
var largest = 0
var largestKind = ""
for (kind, numbers) in interestingNumbers {
for number in numbers {
if number > largest {
largest = number
largestKind = kind
}
}
}
print(largest)
print(largestKind)
Example 2:
let interestingNumbers = [
"Prime": [2, 3, 5, 7, 11, 13],
"Fibonacci": [1, 1, 2, 3, 5, 8],
"Square": [1, 4, 9, 16, 25],
]
var largest = 0
var largestKind: String?
for (kind, numbers) in interestingNumbers {
for number in numbers {
if number > largest {
largest = number
largestKind = kind
}
}
}
print(largest)
print(largestKind!)
What would be the best practice for fixing this error? Also, why did I get the initialization error when I was setting the value equal to another variable?
I can look at your data and your code and I can see that largestKind will always be initialised, because there are numbers > 0.
However, according to the rules that the compiler uses, there is no guarantee that largestKind is ever set. That's why you get an error message at compile time: Because according to the compiler, largestKind might not be initialised when it is used.
You can fix this in two ways: One, you initialise var largestKind = "". Now largestKind is always initialised. Two, you declare var largestKind: String? Now it is an optional string, and optionals are always automatically initialised to nil.
In the second case, if you call print(largestKind!) you tell the compiler "please crash my program at runtime if largestKind is nil". That's an accident waiting to happen. Forced unwrapping using ! should only be used if you are absolutely sure that it cannot be nil. You could a check around the print like if let largestKind = largestKind { } or you could write print (largestKind ?? "") which will print largestKind if not nil, and an empty string otherwise.
Based on your example I would say var largestKind = "" because you will always have a largest kind value. However, in another situation, where this is not guaranteed, you should use var largestKind: String?
Not sure what you mean with your 2nd question. The error results from var largestKind: String
You got the error because there was no guarantee that largestKind would be initialized. Btw in your example 2, don't force unwrap your optional. If it is not initialised your app would crash.
The best practice is to unwrap your optional using if let or guard let.
if let largestKind = largestKind {
// your code
}
or provide a default value using the nil coalescing operator
largestKind ?? "none"
Note that if your intent is to find the maximum value there are higher functions exactly for this purpose. You can compactMap your dictionary values to find their maximum value and them get the maximum key value pair:
if let (kind, largest) = interestingNumbers.compactMapValues({$0.max()}).max(by: {$0.value < $1.value}) {
print("kind:", kind, "largest:", largest) // kind: Square largest: 25
}
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 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
I've got a list of some integers, e.g. [1, 2, 3, 4, 5, 10]
And I've another integer (N). For example, N = 19.
I want to check if my integer can be represented as a sum of any amount of numbers in my list:
19 = 10 + 5 + 4
or
19 = 10 + 4 + 3 + 2
Every number from the list can be used only once. N can raise up to 2 thousand or more. Size of the list can reach 200 integers.
Is there a good way to solve this problem?
4 years and a half later, this question is answered by Jonathan.
I want to post two implementations (bruteforce and Jonathan's) in Python and their performance comparison.
def check_sum_bruteforce(numbers, n):
# This bruteforce approach can be improved (for some cases) by
# returning True as soon as the needed sum is found;
sums = []
for number in numbers:
for sum_ in sums[:]:
sums.append(sum_ + number)
sums.append(number)
return n in sums
def check_sum_optimized(numbers, n):
sums1, sums2 = [], []
numbers1 = numbers[:len(numbers) // 2]
numbers2 = numbers[len(numbers) // 2:]
for sums, numbers_ in ((sums1, numbers1), (sums2, numbers2)):
for number in numbers_:
for sum_ in sums[:]:
sums.append(sum_ + number)
sums.append(number)
for sum_ in sums1:
if n - sum_ in sums2:
return True
return False
assert check_sum_bruteforce([1, 2, 3, 4, 5, 10], 19)
assert check_sum_optimized([1, 2, 3, 4, 5, 10], 19)
import timeit
print(
"Bruteforce approach (10000 times):",
timeit.timeit(
'check_sum_bruteforce([1, 2, 3, 4, 5, 6, 7, 8, 9, 10], 200)',
number=10000,
globals=globals()
)
)
print(
"Optimized approach by Jonathan (10000 times):",
timeit.timeit(
'check_sum_optimized([1, 2, 3, 4, 5, 6, 7, 8, 9, 10], 200)',
number=10000,
globals=globals()
)
)
Output (the float numbers are seconds):
Bruteforce approach (10000 times): 1.830944365834205
Optimized approach by Jonathan (10000 times): 0.34162875449254027
The brute force approach requires generating 2^(array_size)-1 subsets to be summed and compared against target N.
The run time can be dramatically improved by simply splitting the problem in two. Store, in sets, all of the possible sums for one half of the array and the other half separately. It can now be determined by checking for every number n in one set if the complementN-n exists in the other set.
This optimization brings the complexity down to approximately: 2^(array_size/2)-1+2^(array_size/2)-1=2^(array_size/2 + 1)-2
Half of the original.
Here is a c++ implementation using this idea.
#include <bits/stdc++.h>
using namespace std;
bool sum_search(vector<int> myarray, int N) {
//values for splitting the array in two
int right=myarray.size()-1,middle=(myarray.size()-1)/2;
set<int> all_possible_sums1,all_possible_sums2;
//iterate over the first half of the array
for(int i=0;i<middle;i++) {
//buffer set that will hold new possible sums
set<int> buffer_set;
//every value currently in the set is used to make new possible sums
for(set<int>::iterator set_iterator=all_possible_sums1.begin();set_iterator!=all_possible_sums1.end();set_iterator++)
buffer_set.insert(myarray[i]+*set_iterator);
all_possible_sums1.insert(myarray[i]);
//transfer buffer into the main set
for(set<int>::iterator set_iterator=buffer_set.begin();set_iterator!=buffer_set.end();set_iterator++)
all_possible_sums1.insert(*set_iterator);
}
//iterator over the second half of the array
for(int i=middle;i<right+1;i++) {
set<int> buffer_set;
for(set<int>::iterator set_iterator=all_possible_sums2.begin();set_iterator!=all_possible_sums2.end();set_iterator++)
buffer_set.insert(myarray[i]+*set_iterator);
all_possible_sums2.insert(myarray[i]);
for(set<int>::iterator set_iterator=buffer_set.begin();set_iterator!=buffer_set.end();set_iterator++)
all_possible_sums2.insert(*set_iterator);
}
//for every element in the first set, check if the the second set has the complemenent to make N
for(set<int>::iterator set_iterator=all_possible_sums1.begin();set_iterator!=all_possible_sums1.end();set_iterator++)
if(all_possible_sums2.find(N-*set_iterator)!=all_possible_sums2.end())
return true;
return false;
}
Ugly and brute force approach:
a = [1, 2, 3, 4, 5, 10]
b = []
a.size.times do |c|
b << a.combination(c).select{|d| d.reduce(&:+) == 19 }
end
puts b.flatten(1).inspect