leetcode-844-backspace-string-compare.swift

/**

Copyright (c) 2020 David Seek, LLC

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OR OTHER DEALINGS IN THE SOFTWARE.





https://leetcode.com/problems/backspace-string-compare/


Given two strings S and T, return if they are equal when both are typed into empty text editors. # means a backspace character.

Note that after backspacing an empty text, the text will continue empty.

Example 1:

Input: S = "ab#c", T = "ad#c"
Output: true
Explanation: Both S and T become "ac".
Example 2:

Input: S = "ab##", T = "c#d#"
Output: true
Explanation: Both S and T become "".
Example 3:

Input: S = "a##c", T = "#a#c"
Output: true
Explanation: Both S and T become "c".
Example 4:

Input: S = "a#c", T = "b"
Output: false
Explanation: S becomes "c" while T becomes "b".
Note:

1 <= S.length <= 200
1 <= T.length <= 200
S and T only contain lowercase letters and '#' characters.
Follow up:

Can you solve it in O(N) time and O(1) space?

*/

/**
Big O Annotation
Time complexity O(n) where n is the combined character length of S and T.
Space complexity O(n) where n is the combined character length of S and T.
*/
func backspaceCompare(_ S: String, _ T: String) -> Bool {

	/**
	Stacks to hold the characters of each S and T
	*/
	var sStack = Stack<Character>()
	var tStack = Stack<Character>()
		
	// Iterate through S
	for character in S {
			
		// And for every # pop the last..
		if character == "#" {
				
			// ... character in the stack
			sStack.pop()
				
		} else {
				
			// Otherwise push the characters into the stack
			sStack.push(character)
		}
	}
		
	// Same as for S
	for character in T {
			
		if character == "#" {
				
			tStack.pop()
				
		} else {
				
			tStack.push(character)
		}
	}

	/**
	Check if the amount of elements in the stack is equal
	if not then there is no chance for us to create the same strings.
	*/	
	guard sStack.count == tStack.count else {
		return false
	}
		
	/**
	We know the length of the stacks is equal
	so we can iterate through either of them...
	*/
	for _ in 0..<sStack.count {
			
		// To get the last elements of both stacks
		let s = sStack.pop()
		let t = tStack.pop()
		
		// If they differ return false
		guard s == t else {
			return false
		}
	}
	
	/**
	If we fell through the iteration we 
	can return true as we've checked for proper equality.
	*/
	return true
}

struct Stack<Element: Equatable> {
	var array: [Element]
	
	init() {
		array = []
	}
	
	public var isEmpty: Bool {
		return array.isEmpty
	}
	
	public var peek: Element? {
		return array.last
	}
	
	public var count: Int {
		return array.count
	}
	
	mutating public func push(_ element: Element) {
		array.append(element)
	}
	
	@discardableResult
	mutating public func pop() -> Element? {
		return array.popLast()
	}
}


print(backspaceCompare("ab#c", "ad#c"))