639. Decode Ways II

func numDecodings(_ s: String) -> Int {
        if s.isEmpty {
            return 0
        }
        let nums = Array(s.utf16).map { Int($0) - 48}
        let star = Int(Array("*".utf16)[0]) - 48

        var dp = [Int](repeating: 0, count: nums.count + 1)

        if nums[0] == 0 {
            // 012 return 0 result
            return 0
        }

        // initialize
        dp[0] = 1
        dp[1] = nums[0] == star ? 9 : 1

        // corner case "*" or "4"
        if nums.count == 1 {
            return dp[1]
        }

        for i in 2...nums.count {
            // first consider single digit at nums[i - 1]
            if nums[i - 1] > 0 && nums[i - 1] < 10 {
                // case normal number: "123"
                dp[i] += dp[i - 1]
            } else if nums[i - 1] == star {
                // * case "12*"
                dp[i] += dp[i - 1] * 9
            }

            // second consider two digits at nums[i - 2], nums[i - 1]
            if nums[i - 1] != star && nums[i - 2] != star {
                //122 check num 10-26 or not
                let num = nums[i - 2] * 10 + nums[i - 1]
                if num >= 10 && num <= 26 {
                    dp[i] += dp[i - 2]
                }
            } else if nums[i - 1] == star && nums[i - 2] == star {
                // 1**:  10-26 so 15
                dp[i] += dp[i - 2] * (9 + 6)
            } else if nums[i - 2] != star {
                // 1* and 2* two cases
                if nums[i - 2] == 2 {
                    dp[i] += dp[i - 2] * 6
                } else if nums[i - 2] == 1 {
                    dp[i] += dp[i - 2] * 9
                }
            } else {
                // *X two case X == 0-6 (10, 20, 11, 21, ..., 16, 26) and X == 7-9 (17, 18, 19)
                if nums[i - 1] < 7 {
                    dp[i] += dp[i - 2] * 2
                } else {
                    dp[i] += dp[i - 2]
                }
            }

            // I think as swift is dynamic, we don't need to worry about this, but the testcase answers are %1000000007
            // case like "*************" would overflow Int32 if we dont mod 1000000007 in order to pass the testcase result
            dp[i] %= 1000000007
        }
        return dp[nums.count]
    }