Reverse words in a sentence - Swift

Question

How to perform reversing of all words in a sentence.

Example

let str = \"Hello playground\"

Result should be like \"olleH dnuorg

Answer 1

Building on the OPs answer, if you wanted to follow English capitalization rules on the output:

var str = "Hello playground"

let result = str.split(separator: " ").enumerated().map {
    let reversed = String($0.1.reversed())
    return $0.0 == 0 ? reversed.capitalized : reversed.lowercased()
    }
    .joined(separator: " ")
print("Rerversing letters in each word = \"" + result + "\"") // Olleh dnuorgyalp

Also note that multiple spaces would mess this up, as would commas, periods, and other word/sentence delimiters.

Answer 2

Working code!

 import UIKit

 var str = "Hello playground"

 let result = str.split(separator: " ").map { String($0.reversed())}.joined(separator: " ")
 print(result) // olleH dnuorgyalp

Answer 3

let result = String(str.reversed())
    .components(separatedBy: .whitespaces)
    .reversed()
    .joined(separator: " ")

Answer 4

You can use String method enumerateSubstrings using .byWords options and replace the subrange of each word with the substring reversed. Note that this way the punctuation will remain in place:

---

import Foundation

Mutating approach:

var str = "Hello, playground!!!"

str.enumerateSubstrings(in: str.startIndex..., options: .byWords) { _, range, _, _ in
    str.replaceSubrange(range, with: str[range].reversed())
}
print(str)  // "olleH, dnuorgyalp!!!"

---

Non mutating:

let str = "Hello, playground!!!"
var result = ""
str.enumerateSubstrings(in: str.startIndex..., options: .byWords) { string, range, enclosingRange, _ in
    result.append(contentsOf: string!.reversed())
    result.append(contentsOf: str[range.upperBound..<enclosingRange.upperBound])
}
print(result)  // "olleH, dnuorgyalp!!!"

Answer 5

Here's the typical functional programming approach (also posted by @Saranjith)

let result = str
    .components(separatedBy: " ")
    .map { $0.reversed() }
    .joined()

Complexity

First of all lets define

• n: the number of characters in the input string
• k: the number of words in the input string (of course k<=n)

⏳ Time complexity

Now lets look at the time complexity of each piece of our code

.components(separatedBy: " ")

This instructions need to go through the entire string so O(n)

.map { $0.reversed() }

Here each word is reversed. So if we have k words we have a time complexity of

O(m0) + O(m1) + ... + O(mk-1)

where mi is the length of the i-th word. However the sum of the length of all the words is <= n so we can say that

O(m0) + O(m1) + ... + O(mk-1) <= O(n)

.joined()

Finally we have k words which need to be joined togheter. This can be done in O(k) that, again, is <= O(n).

Wrap up

let result = str
    .components(separatedBy: " ") // O(n)
    .map { $0.reversed() } // O(m0) + O(m1) + ... + O(mk-1) <= O(n)
    .joined() // O(k) <= O(n)

Time complexity = O(n) + O(n) + O(n) = O(n)