Decode Ways II Python Solution with Tests

LeetCode 639, Hard. Topics: String, Dynamic Programming. View on LeetCode. Generate this problem as a practice environment: tested reference solution, 30 parametrized pytest cases, and a playground notebook:

lcpy gen -n 639   # by problem number
lcpy gen -s decode_ways_ii   # by problem name

Problem

A message containing letters from A-Z can be encoded into numbers using the following mapping:

'A' -> "1"
'B' -> "2"
...
'Z' -> "26"

To decode an encoded message, all the digits must be grouped then mapped back into letters using the reverse of the mapping above (there may be multiple ways). For example, “11106” can be mapped into:

“AAJF” with the grouping (1 1 10 6)
“KJF” with the grouping (11 10 6)

Note that the grouping (1 11 06) is invalid because “06” cannot be mapped into ‘F’ since “6” is different from “06”. In addition to the mapping above, an encoded message may contain the '' character, which can represent any digit from ‘1’ to ‘9’ (‘0’ is excluded). For example, the encoded message “1” may represent any of the encoded messages “11”, “12”, “13”, “14”, “15”, “16”, “17”, “18”, or “19”. Decoding “1*” is equivalent to decoding any of the encoded messages it can represent. Given a string s consisting of digits and ’*’ characters, return the number of ways to decode it. Since the answer may be very large, return it modulo 10^9 + 7.

Examples

Input: s = "*"
Output: 9
Explanation: The encoded message can represent any of the encoded messages "1" through "9". Each of these can be decoded to the strings "A" through "I" respectively. Hence, there are a total of 9 ways to decode "*".

Input: s = "1*"
Output: 18
Explanation: The encoded message can represent any of the encoded messages "11" through "19". Each of these encoded messages have 2 ways to be decoded (e.g. "11" can be decoded to "AA" or "K"). Hence, there are a total of 9 * 2 = 18 ways to decode "1*".

Input: s = "2*"
Output: 15
Explanation: The encoded message can represent any of the encoded messages "21" through "29". "21" through "26" have 2 ways of being decoded, but "27" through "29" only have 1 way. Hence, there are a total of (6 * 2) + (3 * 1) = 15 ways to decode "2*".

Constraints

1 <= s.length <= 10^5
s[i] is a digit or ’*’.

Solution

Reference implementation from solution.py on GitHub, full suite in test_solution.py:

class Solution:
    # Time: O(n)
    # Space: O(1)
    def num_decodings(self, s: str) -> int:
        mod = 1_000_000_007
        prev = 1
        curr = self._ways_single(s[0])
        for i in range(1, len(s)):
            pair = self._ways_pair(s[i - 1], s[i])
            prev, curr = curr, (curr * self._ways_single(s[i]) + prev * pair) % mod
        return curr

    def _ways_single(self, ch: str) -> int:
        if ch == "*":
            return 9
        return 0 if ch == "0" else 1

    def _ways_pair(self, a: str, b: str) -> int:
        if a == "*":
            if b == "*":
                return 15
            return 2 if b <= "6" else 1
        if b == "*":
            return 9 if a == "1" else (6 if a == "2" else 0)
        if a == "0":
            return 0
        return 1 if int(a + b) <= 26 else 0

Complexity

Time	Space
O(n)	O(1)

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Decode Ways II Python Solution with Tests

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