I'm learning recursive functions in Swift, and I did the following:
func recursive(i: Int) -> Int {
if i == 1 {
return 1
} else if i >= 2 {
return recursive(i: i - 1) + 1
}
return 0
}
I couldn't figure out why the function above is not working. I've tested it by doing the below doing print(recursive(10)), which gives me an output of 10. I expected the output to be 1. Can anyone help me with this? Thank you in advance.
I'm using Playgrounds on XCode 8.3.
When you do this:
recursive(i: i - 1) + 1
… then you are in effect decrementing i and then incrementing it again. That cancels out and you arrive at i again.
Let's write down what calculation would be done for i = 3:
(3 - 1) + 1 = ((2 - 1) + 1) + 1 = (((1) + 1) + 1) = 3
This is a perfect example of printing numbers without using any loop.
The recursive functions are very useful to handle such cases.
func printCount( count : inout Int , limit : Int) {
print(count, terminator: " ")
count += 1
if count > limit {
return
}
printCount(count: &count , limit: limit)
}
var count = 11
let limit = 20
printCount(count: &count , limit: limit)
Output : 11 12 13 14 15 16 17 18 19 20
Related
I am trying to build a Binary to Decimal calculator for the Apple Watch using Swift 4.
The code I am having trouble is this:
var i = 0
var labelInputInt = 0
let labelOutputString = "10010" // Random number in binary
let reverse = String(labelOutputString.reversed()) // Reversing the original string
while i <= reverse.count {
let indexOfString = reverse.index(reverse.startIndex, offsetBy: i)
if reverse[indexOfString] == "1" {
labelInputInt += 2^i * 1
}
i += 1
}
I am using a while loop to get the index indexOfString and check if in the string reverse at the specific index it is equal with "1".
The problem is that I get a runtime error when the if statement is executed.
The error looks like this:
2 libpthread.so.0 0x00007fc22f163390
3 libswiftCore.so 0x00007fc22afa88a0 _T0s18_fatalErrorMessages5NeverOs12Stati
cStringV_A2E4fileSu4lines6UInt32V5flagstFTfq4nnddn_n + 96
4 libswiftCore.so 0x00007fc22afb3323
5 libswiftCore.so 0x00007fc22afdf9a2
6 libswiftCore.so 0x00007fc22aedca19 _T0SS9subscripts9CharacterVSS5IndexVcfg
+ 9
7 libswiftCore.so 0x00007fc22f591294 _T0SS9subscripts9CharacterVSS5IndexVcfg
+ 74139780
8 swift 0x0000000000f2925f
9 swift 0x0000000000f2d402
10 swift 0x00000000004bf516
11 swift 0x00000000004ae461
12 swift 0x00000000004aa411
13 swift 0x0000000000465424
14 libc.so.6 0x00007fc22d88d830 __libc_start_main + 240
15 swift 0x0000000000462ce9
Stack dump:
0. Program arguments: /home/drkameleon/swift4/usr/bin/swift -frontend -inte
rpret tmp/XfwP0oM7FJ.swift -disable-objc-interop -suppress-warnings -module-na
me XfwP0oM7FJ
Illegal instruction (core dumped)
So, how can I get a specific character of a String and compare it with another character without getting this crash?
Your approach to get a specific character from a string is actually correct, there are two other problems in your code:
The index i should run up to and excluding reverse.count.
This is conveniently done with the "half-open range" operator (..<).
^ is the bitwise-xor operator, not exponentiation. Exponentiation is done with the pow() function, in your case
labelInputInt += Int(pow(2.0, Double(i)))
or with the "shift-left" operator << if the base is 2.
So this would be a working variant:
for i in 0 ..< reverse.count {
let indexOfString = reverse.index(reverse.startIndex, offsetBy: i)
if reverse[indexOfString] == "1" {
labelInputInt += 1 << i
}
i += 1
}
But you can simply enumerate the characters of a string in reverse order instead of subscripting (which is also more efficient):
let binaryString = "10010"
var result = 0
for (i, char) in binaryString.reversed().enumerated() {
if char == "1" {
result += 1 << i
}
}
print(result)
Even simpler with forward iteration, no reversed() or << needed:
let binaryString = "10010"
var result = 0
for char in binaryString {
result = 2 * result
if char == "1" {
result += 1
}
}
print(result)
Which suggests to use reduce():
let binaryString = "10010"
let result = binaryString.reduce(0) { 2 * $0 + ($1 == "1" ? 1 : 0) }
print(result)
But why reinvent the wheel? Just use init?(_:radix:) from the Swift standard library (with error-checking for free):
let binaryString = "10010"
if let result = Int(binaryString, radix: 2) {
print(result)
} else {
print("invalid input")
}
To find how many ways we have of making change for the amount 4 given the coins [1,2,3], we can create a DP algorithm that produces the following table:
table[amount][coins.count]
0 1 2 3 4
-----------
(0) 1 | 1 1 1 1 1
(1) 2 | 1 1 2 2 3
(2) 3 | 1 1 2 3 4
The last position being our answer. The answer is 4 because we have the following combinations: [1,1,1,1],[2,1],[2,2],[3,1].
My question is, is it possible to retrieve these combinations from the table I just generated? How?
For completeness, here's my algorithm
func coinChange(coins: [Int], amount: Int) -> Int {
// int[amount+1][coins]
var table = Array<Array<Int>>(repeating: Array<Int>(repeating: 0, count: coins.count), count: amount + 1)
for i in 0..<coins.count {
table[0][i] = 1
}
for i in 1...amount {
for j in 0..<coins.count {
//solutions that include coins[j]
let x = i - coins[j] >= 0 ? table[i - coins[j]][j] : 0
//solutions that don't include coins[j]
let y = j >= 1 ? table[i][j-1] : 0
table[i][j] = x + y
}
}
return table[amount][coins.count - 1];
}
Thanks!
--
Solution
Here's an ugly function that retrieves the combinations, based on #Sayakiss 's explanation:
func getSolution(_ i: Int, _ j: Int) -> [[Int]] {
if j < 0 || i < 0 {
//not a solution
return []
}
if i == 0 && j == 0 {
//valid solution. return an empty array where the coins will be appended
return [[]]
}
return getSolution(i - coins[j], j).map{var a = $0; a.append(coins[j]);return a} + getSolution(i, j - 1)
}
getSolution(amount, coins.count-1)
Output:
[[1, 3], [2, 2], [1, 1, 2], [1, 1, 1, 1]]
Sure you can. We define a new function get_solution(i,j) which means all solution for your table[i][j].
You can think it returns an array of array, for example, the output of get_solution(4,3) is [[1,1,1,1],[2,1],[2,2],[3,1]]. Then:
Case 1. Any solution from get_solution(i - coins[j], j) plus coins[j] is a solution for table[i][j].
Case 2. Any solution from get_solution(i, j - 1) is a solution for table[i][j].
You can prove Case 1 + Case 2 is all possible solution for table[i][j](note you get table[i][j] by this way).
The only problem remains is to implement get_solution(i,j) and I think it's good for you to do it by yourself.
If you still got any question, please don't hesitate to leave a comment here.
I have following line in my code:
for (i = 0, j = count - 1; i < count; j = i++)
Can anyone help to remove the two compiler warnings, that i++ will be removed in Swift 3.0 and C-style for statement is depreciated?
You could use this:
var j = count-1
for i in 0..<count {
defer { j = i } // This will keep the cycle "logic" all together, similarly to "j = i++"
// Cycle body
}
EDIT
As #t0rst noted, be careful using defer, since it will be executed no matter how its enclosing scope is exited, so it isn't a 100% replacement.
So while the standard for ( forInit ; forTest ; forNext ) { … } will not execute forNext in case of a break statement inside the cycle, a return or an exception, the defer will.
Read here for more
Alternatively, lets go crazy to avoid having to declare j as external to the loop scope!
Snippet 1
let count = 10
for (i, j) in [count-1..<count, 0..<count-1].flatten().enumerate() {
print(i, j)
}
/* 0 9
1 0
2 1
3 2
4 3
5 4
6 5
7 6
8 7
9 8 */
Snippet 2
for (i, j) in (-1..<count-1).map({ $0 < 0 ? count-1 : $0 }).enumerate() {
print(i, j)
}
Trying to win the prize for the craziest solution in this thread
Snippet 1
extension Int {
func j(count:Int) -> Int {
return (self + count - 1) % count
}
}
for i in 0..<count {
print(i, i.j(count))
}
Snippet 2
let count = 10
let iList = 0..<count
let jList = iList.map { ($0 + count - 1) % count }
zip(iList, jList).forEach { (i, j) in
print(i, j)
}
You could use a helper function to abstract away the wrapping of j as:
func go(count: Int, block: (Int, Int) -> ()) {
if count < 1 { return }
block(0, count - 1)
for i in 1 ..< count {
block(i, i - 1)
}
}
This is a LeetCode question. I wrote 4 answers in different versions of that question. When I tried to use "Bit manipulation", I got the error. Since no one in LeetCode can answer my question, and I can't find any Swift doc about this. I thought I would try to ask here.
The question is to get the majority element (>n/2) in a given array. The following code works in other languages like Java, so I think it might be a general question in Swift.
func majorityElement(nums: [Int]) -> Int {
var bit = Array(count: 32, repeatedValue: 0)
for num in nums {
for i in 0..<32 {
if (num>>(31-i) & 1) == 1 {
bit[i] += 1
}
}
}
var ret = 0
for i in 0..<32 {
bit[i] = bit[i]>nums.count/2 ? 1 : 0
ret += bit[i] * (1<<(31-i))
}
return ret
}
When the input is [-2147483648], the output is 2147483648. But in Java, it can successfully output the right negative number.
Swift doc says :
Even on 32-bit platforms, Int can store any value between -2,147,483,648 and 2,147,483,647, and is large enough for many integer ranges.
Well, it is 2,147,483,647, the input is 1 larger than that number. When I ran pow(2.0, 31.0) in playground, it shows 2147483648. I got confused. What's wrong with my code or what did I miss about Swift Int?
A Java int is a 32-bit integer. The Swift Int is 32-bit or 64-bit
depending on the platform. In particular, it is 64-bit on all OS X
platforms where Swift is available.
Your code handles only the lower 32 bits of the given integers, so that
-2147483648 = 0xffffffff80000000
becomes
2147483648 = 0x0000000080000000
So solve the problem, you can either change the function to take 32-bit integers as arguments:
func majorityElement(nums: [Int32]) -> Int32 { ... }
or make it work with arbitrary sized integers by computing the
actual size and use that instead of the constant 32:
func majorityElement(nums: [Int]) -> Int {
let numBits = sizeof(Int) * 8
var bit = Array(count: numBits, repeatedValue: 0)
for num in nums {
for i in 0..<numBits {
if (num>>(numBits-1-i) & 1) == 1 {
bit[i] += 1
}
}
}
var ret = 0
for i in 0..<numBits {
bit[i] = bit[i]>nums.count/2 ? 1 : 0
ret += bit[i] * (1<<(numBits-1-i))
}
return ret
}
A more Swifty way would be to use map() and reduce()
func majorityElement(nums: [Int]) -> Int {
let numBits = sizeof(Int) * 8
let bitCounts = (0 ..< numBits).map { i in
nums.reduce(0) { $0 + ($1 >> i) & 1 }
}
let major = (0 ..< numBits).reduce(0) {
$0 | (bitCounts[$1] > nums.count/2 ? 1 << $1 : 0)
}
return major
}
Starting from the Simple C program found here on stackoverflow, I've done it again in Swift and here's the code:
import Foundation
// variables and constants:
var dice1, dice2: UInt32
var score, scoreToWin, diceSum: Int
dice1 = 0
dice2 = 0
diceSum = 0
// functions:
func rollDice() ->Int {
dice1 = arc4random() % 6 + 1
dice2 = arc4random() % 6 + 1
diceSum = Int(dice1 + dice2)
println("\(diceSum)")
return diceSum
}
// main:
score = rollDice()
println("\(dice1) \(dice2) \(score)")
switch score {
case 7, 11:
println("score=\(score)\nYou WIN")
case 2, 3, 12:
println("score=\(score)\nYou LOOSE")
default:
println("You have to roll a \(score) to WIN")
do {
scoreToWin = score
diceSum = rollDice()
if diceSum == 7 { println("You LOOSE") }
else if diceSum == scoreToWin { println("You WIN") }
} while (diceSum != scoreToWin && diceSum != 7)
}
This is a possible output:
6
3 3 6
You have to roll a 6 to WIN
6
You WIN
Program ended with exit 0
I was not expecting the first line of output, because the first line indicate the function rollDice() was run while been defined.
How can I define a function without actually running it?
score = rollDice() would print the 6 you are seeing due to the println("\(diceSum)") you are doing in the rollDice method.
Declaring a method does not run it. Not in swift, nor in any other language I can think of.