leetcode-58-length-of-last-word.swift

/**

Copyright (c) 2020 David Seek, LLC

Permission is hereby granted, free of charge, to any person obtaining a copy
of this software and associated documentation files (the "Software"), to deal
in the Software without restriction, including without limitation the rights
to use, copy, modify, merge, publish, distribute, sublicense, and/or sell
copies of the Software, and to permit persons to whom the Software is
furnished to do so, subject to the following conditions:

The above copyright notice and this permission notice shall be included in all
copies or substantial portions of the Software.

THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND,
EXPRESS OR IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF
MERCHANTABILITY, FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT.
IN NO EVENT SHALL THE AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM,
DAMAGES OR OTHER LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR
OTHERWISE, ARISING FROM, OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE
OR OTHER DEALINGS IN THE SOFTWARE.





https://leetcode.com/problems/length-of-last-word/


Given a string s consists of upper/lower-case alphabets and empty space characters ' ', return the length of last word (last word means the last appearing word if we loop from left to right) in the string.

If the last word does not exist, return 0.

Note: A word is defined as a maximal substring consisting of non-space characters only.

Example:

Input: "Hello World"
Output: 5

*/

/**
Big O Annotation
Time complexity O(n) where n is the amount of words in s.
Space complexity O(n) where n is the amount of words in s.
*/
func lengthOfLastWord(_ s: String) -> Int {
        
    // Without characters no last word length
    guard s.count > 0 else {
        return 0
    }
    
    // Split string by words
    var components: [String] = s.components(separatedBy: " ")

    // Filter out empty words
    components = components.filter { component in 
        return component != ""
    }
    
    // Make sure we have a last word
    guard let last = components.last else {
        return 0
    }
    
    // Return it's length
    return Array(last).count
}