Balance-Parentheses.swift

// Run in an Xcode Playground

let input = ")(()("
let input2 = "))()))("
let input3 = "))))(((("
let input4 = ")))()(()()(("
let input5 = "(()())"
let input6 = ""
let input7 = "()"

func solution(parentheses: String) -> String {
    var solution = parentheses
    
    // the strategy is to handle the outward facing leading & trailing parentheses first
    // once those are calculated, just find the difference in the inner parentheses & add as many as are needed to even them out
    
    let leadingLeftFacingParenthesesCount = parentheses.prefix { $0 == ")" }.count
    let trailingRightFacingParenthesesCount = parentheses.reversed().prefix { $0 == "(" }.count
    let stringWithLeadingLeftFacingParenthesesRemoved = parentheses.suffix(parentheses.count - leadingLeftFacingParenthesesCount)
    let stringWithLeadingAndTrailingOpenParenthesesRemoved = stringWithLeadingLeftFacingParenthesesRemoved.prefix(stringWithLeadingLeftFacingParenthesesRemoved.count - trailingRightFacingParenthesesCount)
    
    let innerRightFacingParenthesisCount = stringWithLeadingAndTrailingOpenParenthesesRemoved.replacingOccurrences(of: ")", with: "").count
    let innerLeftFacingParenthesisCount = stringWithLeadingAndTrailingOpenParenthesesRemoved.replacingOccurrences(of: "(", with: "").count
    let innerParenthesesDifference = innerLeftFacingParenthesisCount - innerRightFacingParenthesisCount
    
    if leadingLeftFacingParenthesesCount > 0 {
        (1...leadingLeftFacingParenthesesCount).forEach { _ in
            solution = "(" + solution
        }
    }
    if trailingRightFacingParenthesesCount > 0 {
        (1...trailingRightFacingParenthesesCount).forEach { _ in
            solution = solution + ")"
        }
    }
    
    if innerParenthesesDifference < 0 {
        (innerParenthesesDifference ... -1).forEach { _ in
            solution = solution + ")"
        }
    } else if innerParenthesesDifference > 0 {
        (1 ... innerParenthesesDifference).forEach { _ in
            solution = "(" + solution
        }
    }
    
    return solution
}

solution(parentheses: input) == "()(()())"
solution(parentheses: input2) == "(((())()))()"
solution(parentheses: input3) == "(((())))(((())))"
solution(parentheses: input4) == "((()))()(()()(()))"
solution(parentheses: input5) == "(()())"
solution(parentheses: input6) == ""
solution(parentheses: input7) == "()"