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‎Problemset/0639-Decode Ways II/README.md

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# 639. Decode Ways II
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A message containing letters from `A-Z` can be **encoded** into numbers using the following mapping:
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```
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'A' -> "1"
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'B' -> "2"
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...
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'Z' -> "26"
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```
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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:
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* `"AAJF"` with the grouping `(1 1 10 6)`
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* `"KJF"` with the grouping `(11 10 6)`
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Note that the grouping `(1 11 06)` is invalid because `"06"` cannot be mapped into `'F'` since `"6"` is different from `"06"`.
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**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.
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Given a string `s` consisting of digits and `'*'` characters, return *the **number** of ways to **decode** it*.
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Since the answer may be very large, return it **modulo** <code>10<sup>9</sup> + 7</code>.
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#### Example 1:
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<pre>
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<strong>Input:</strong> s = "*"
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<strong>Output:</strong> 9
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<strong>Explanation:</strong> The encoded message can represent any of the encoded messages "1", "2", "3", "4", "5", "6", "7", "8", or "9".
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Each of these can be decoded to the strings "A", "B", "C", "D", "E", "F", "G", "H", and "I" respectively.
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Hence, there are a total of 9 ways to decode "*".
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</pre>
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#### Example 2:
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<pre>
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<strong>Input:</strong> s = "1*"
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<strong>Output:</strong> 18
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<strong>Explanation:</strong> The encoded message can represent any of the encoded messages "11", "12", "13", "14", "15", "16", "17", "18", or "19".
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Each of these encoded messages have 2 ways to be decoded (e.g. "11" can be decoded to "AA" or "K").
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Hence, there are a total of 9 * 2 = 18 ways to decode "1*".
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</pre>
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#### Example 3:
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<pre>
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<strong>Input:</strong> s = "2*"
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<strong>Output:</strong> 15
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<strong>Explanation:</strong> The encoded message can represent any of the encoded messages "21", "22", "23", "24", "25", "26", "27", "28", or "29".
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"21", "22", "23", "24", "25", and "26" have 2 ways of being decoded, but "27", "28", and "29" only have 1 way.
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Hence, there are a total of (6 * 2) + (3 * 1) = 12 + 3 = 15 ways to decode "2*".
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</pre>
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#### Constraints:
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* <code>1 <= s.length <= 10<sup>5</sup></code>
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* `s[i]` is a digit or `'*'`.
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## Solutions (Rust)
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### 1. Solution
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```Rust
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impl Solution {
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pub fn num_decodings(s: String) -> i32 {
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let s = s.as_bytes();
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let mut dp = vec![[0_i64; 2]; s.len()];
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match s[0] {
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b'0' => return 0,
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b'*' => dp[0][0] = 9,
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_ => dp[0][0] = 1,
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}
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if s.len() > 1 {
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dp[1] = match (s[0], s[1]) {
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(b'1', b'0') => [0, 1],
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(b'1', b'*') => [9, 9],
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(b'1', _) => [1, 1],
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(b'2', b'0') => [0, 1],
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(b'2', b'*') => [9, 6],
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(b'2', x) if x < b'7' => [1, 1],
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(b'2', _) => [1, 0],
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(b'*', b'0') => [0, 2],
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(b'*', b'*') => [81, 15],
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(b'*', x) if x < b'7' => [9, 2],
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(b'*', _) => [9, 1],
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(_, b'0') => [0, 0],
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(_, b'*') => [9, 0],
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_ => [1, 0],
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};
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}
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for i in 2..s.len() {
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dp[i][0] = match s[i] {
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b'0' => 0,
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b'*' => 9 * (dp[i - 1][0] + dp[i - 1][1]),
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_ => dp[i - 1][0] + dp[i - 1][1],
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} % 1_000_000_007;
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dp[i][1] = match (s[i - 1], s[i]) {
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(b'1', b'*') => 9 * (dp[i - 2][0] + dp[i - 2][1]),
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(b'1', _) => dp[i - 2][0] + dp[i - 2][1],
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(b'2', b'*') => 6 * (dp[i - 2][0] + dp[i - 2][1]),
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(b'2', x) if x < b'7' => dp[i - 2][0] + dp[i - 2][1],
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(b'*', b'*') => 15 * (dp[i - 2][0] + dp[i - 2][1]),
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(b'*', x) if x < b'7' => 2 * (dp[i - 2][0] + dp[i - 2][1]),
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(b'*', _) => dp[i - 2][0] + dp[i - 2][1],
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_ => 0,
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} % 1_000_000_007;
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}
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((dp[dp.len() - 1][0] + dp[dp.len() - 1][1]) % 1_000_000_007) as i32
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}
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}
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```

‎Problemset/0639-Decode Ways II/README_CN.md

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# 639. 解码方法 II
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一条包含字母 `A-Z` 的消息通过以下的方式进行了 **编码** ：
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```
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'A' -> "1"
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'B' -> "2"
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...
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'Z' -> "26"
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```
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要 **解码** 一条已编码的消息，所有的数字都必须分组，然后按原来的编码方案反向映射回字母（可能存在多种方式）。例如，`"11106"` 可以映射为：
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* `"AAJF"` 对应分组 `(1 1 10 6)`
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* `"KJF"` 对应分组 `(11 10 6)`
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注意，像 `(1 11 06)` 这样的分组是无效的，因为 `"06"` 不可以映射为 `'F'` ，因为 `"6"` 与 `"06"` 不同。
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**除了** 上面描述的数字字母映射方案，编码消息中可能包含 `'*'` 字符，可以表示从 `'1'` 到 `'9'` 的任一数字（不包括 `'0'`）。例如，编码字符串 `"1*"` 可以表示 `"11"`、`"12"`、`"13"`、`"14"`、`"15"`、`"16"`、`"17"`、`"18"` 或 `"19"` 中的任意一条消息。对 `"1*"` 进行解码，相当于解码该字符串可以表示的任何编码消息。
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给你一个字符串 `s` ，由数字和 `'*'` 字符组成，返回 **解码** 该字符串的方法 **数目** 。
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由于答案数目可能非常大，返回 <code>10<sup>9</sup> + 7</code> 的 **模** 。
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#### 示例 1:
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<pre>
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<strong>输入:</strong> s = "*"
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<strong>输出:</strong> 9
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<strong>解释:</strong> 这一条编码消息可以表示 "1"、"2"、"3"、"4"、"5"、"6"、"7"、"8" 或 "9" 中的任意一条。
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可以分别解码成字符串 "A"、"B"、"C"、"D"、"E"、"F"、"G"、"H" 和 "I" 。
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因此，"*" 总共有 9 种解码方法。
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</pre>
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#### 示例 2:
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<pre>
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<strong>输入:</strong> s = "1*"
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<strong>输出:</strong> 18
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<strong>解释:</strong> 这一条编码消息可以表示 "11"、"12"、"13"、"14"、"15"、"16"、"17"、"18" 或 "19" 中的任意一条。
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每种消息都可以由 2 种方法解码（例如，"11" 可以解码成 "AA" 或 "K"）。
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因此，"1*" 共有 9 * 2 = 18 种解码方法。
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</pre>
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#### 示例 3:
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<pre>
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<strong>输入:</strong> s = "2*"
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<strong>输出:</strong> 15
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<strong>解释:</strong> 这一条编码消息可以表示 "21"、"22"、"23"、"24"、"25"、"26"、"27"、"28" 或 "29" 中的任意一条。
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"21"、"22"、"23"、"24"、"25" 和 "26" 由 2 种解码方法，但 "27"、"28" 和 "29" 仅有 1 种解码方法。
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因此，"2*" 共有 (6 * 2) + (3 * 1) = 12 + 3 = 15 种解码方法。
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</pre>
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#### 提示:
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* <code>1 <= s.length <= 10<sup>5</sup></code>
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* `s[i]` 是 `0 - 9` 中的一位数字或字符 `'*'`
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## 题解 (Rust)
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### 1. 题解
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```Rust
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impl Solution {
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pub fn num_decodings(s: String) -> i32 {
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let s = s.as_bytes();
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let mut dp = vec![[0_i64; 2]; s.len()];
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match s[0] {
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b'0' => return 0,
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b'*' => dp[0][0] = 9,
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_ => dp[0][0] = 1,
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}
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if s.len() > 1 {
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dp[1] = match (s[0], s[1]) {
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(b'1', b'0') => [0, 1],
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(b'1', b'*') => [9, 9],
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(b'1', _) => [1, 1],
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(b'2', b'0') => [0, 1],
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(b'2', b'*') => [9, 6],
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(b'2', x) if x < b'7' => [1, 1],
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(b'2', _) => [1, 0],
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(b'*', b'0') => [0, 2],
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(b'*', b'*') => [81, 15],
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(b'*', x) if x < b'7' => [9, 2],
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(b'*', _) => [9, 1],
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(_, b'0') => [0, 0],
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(_, b'*') => [9, 0],
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_ => [1, 0],
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};
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}
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for i in 2..s.len() {
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dp[i][0] = match s[i] {
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b'0' => 0,
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b'*' => 9 * (dp[i - 1][0] + dp[i - 1][1]),
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_ => dp[i - 1][0] + dp[i - 1][1],
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} % 1_000_000_007;
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dp[i][1] = match (s[i - 1], s[i]) {
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(b'1', b'*') => 9 * (dp[i - 2][0] + dp[i - 2][1]),
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(b'1', _) => dp[i - 2][0] + dp[i - 2][1],
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(b'2', b'*') => 6 * (dp[i - 2][0] + dp[i - 2][1]),
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(b'2', x) if x < b'7' => dp[i - 2][0] + dp[i - 2][1],
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(b'*', b'*') => 15 * (dp[i - 2][0] + dp[i - 2][1]),
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(b'*', x) if x < b'7' => 2 * (dp[i - 2][0] + dp[i - 2][1]),
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(b'*', _) => dp[i - 2][0] + dp[i - 2][1],
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_ => 0,
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} % 1_000_000_007;
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}
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((dp[dp.len() - 1][0] + dp[dp.len() - 1][1]) % 1_000_000_007) as i32
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}
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}
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```

‎Problemset/0639-Decode Ways II/Solution.rs

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impl Solution {
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pub fn num_decodings(s: String) -> i32 {
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let s = s.as_bytes();
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let mut dp = vec![[0_i64; 2]; s.len()];
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match s[0] {
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b'0' => return 0,
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b'*' => dp[0][0] = 9,
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_ => dp[0][0] = 1,
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}
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if s.len() > 1 {
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dp[1] = match (s[0], s[1]) {
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(b'1', b'0') => [0, 1],
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(b'1', b'*') => [9, 9],
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(b'1', _) => [1, 1],
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(b'2', b'0') => [0, 1],
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(b'2', b'*') => [9, 6],
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(b'2', x) if x < b'7' => [1, 1],
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(b'2', _) => [1, 0],
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(b'*', b'0') => [0, 2],
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(b'*', b'*') => [81, 15],
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(b'*', x) if x < b'7' => [9, 2],
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(b'*', _) => [9, 1],
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(_, b'0') => [0, 0],
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(_, b'*') => [9, 0],
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_ => [1, 0],
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};
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}
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for i in 2..s.len() {
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dp[i][0] = match s[i] {
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b'0' => 0,
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b'*' => 9 * (dp[i - 1][0] + dp[i - 1][1]),
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_ => dp[i - 1][0] + dp[i - 1][1],
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} % 1_000_000_007;
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dp[i][1] = match (s[i - 1], s[i]) {
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(b'1', b'*') => 9 * (dp[i - 2][0] + dp[i - 2][1]),
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(b'1', _) => dp[i - 2][0] + dp[i - 2][1],
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(b'2', b'*') => 6 * (dp[i - 2][0] + dp[i - 2][1]),
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(b'2', x) if x < b'7' => dp[i - 2][0] + dp[i - 2][1],
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(b'*', b'*') => 15 * (dp[i - 2][0] + dp[i - 2][1]),
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(b'*', x) if x < b'7' => 2 * (dp[i - 2][0] + dp[i - 2][1]),
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(b'*', _) => dp[i - 2][0] + dp[i - 2][1],
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_ => 0,
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} % 1_000_000_007;
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}
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((dp[dp.len() - 1][0] + dp[dp.len() - 1][1]) % 1_000_000_007) as i32
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}
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}

‎README.md

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[633][633l] |[Sum of Square Numbers][633] |![rs]
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[636][636l] |[Exclusive Time of Functions][636] |![rs]
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[637][637l] |[Average of Levels in Binary Tree][637] |![py]
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[639][639l] |[Decode Ways II][639] |![rs]
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[640][640l] |[Solve the Equation][640] |![rs]
390391
[641][641l] |[Design Circular Deque][641] |![rs]
391392
[643][643l] |[Maximum Average Subarray I][643] |![rs]
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17841785
[633]:Problemset/0633-Sum%20of%20Square%20Numbers/README.md#633-sum-of-square-numbers
17851786
[636]:Problemset/0636-Exclusive%20Time%20of%20Functions/README.md#636-exclusive-time-of-functions
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[637]:Problemset/0637-Average%20of%20Levels%20in%20Binary%20Tree/README.md#637-average-of-levels-in-binary-tree
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[639]:Problemset/0639-Decode%20Ways%20II/README.md#639-decode-ways-ii
17871789
[640]:Problemset/0640-Solve%20the%20Equation/README.md#640-solve-the-equation
17881790
[641]:Problemset/0641-Design%20Circular%20Deque/README.md#641-design-circular-deque
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[643]:Problemset/0643-Maximum%20Average%20Subarray%20I/README.md#643-maximum-average-subarray-i
@@ -3180,6 +3182,7 @@
31803182
[633l]:https://leetcode.com/problems/sum-of-square-numbers/
31813183
[636l]:https://leetcode.com/problems/exclusive-time-of-functions/
31823184
[637l]:https://leetcode.com/problems/average-of-levels-in-binary-tree/
3185+
[639l]:https://leetcode.com/problems/decode-ways-ii/
31833186
[640l]:https://leetcode.com/problems/solve-the-equation/
31843187
[641l]:https://leetcode.com/problems/design-circular-deque/
31853188
[643l]:https://leetcode.com/problems/maximum-average-subarray-i/

‎README_CN.md

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[633][633l] |[平方数之和][633] |![rs]
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[636][636l] |[函数的独占时间][636] |![rs]
388388
[637][637l] |[二叉树的层平均值][637] |![py]
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[639][639l] |[解码方法 II][639] |![rs]
389390
[640][640l] |[求解方程][640] |![rs]
390391
[641][641l] |[设计循环双端队列][641] |![rs]
391392
[643][643l] |[子数组最大平均数 I][643] |![rs]
@@ -1784,6 +1785,7 @@
17841785
[633]:Problemset/0633-Sum%20of%20Square%20Numbers/README_CN.md#633-平方数之和
17851786
[636]:Problemset/0636-Exclusive%20Time%20of%20Functions/README_CN.md#636-函数的独占时间
17861787
[637]:Problemset/0637-Average%20of%20Levels%20in%20Binary%20Tree/README_CN.md#637-二叉树的层平均值
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[639]:Problemset/0639-Decode%20Ways%20II/README_CN.md#639-解码方法-ii
17871789
[640]:Problemset/0640-Solve%20the%20Equation/README_CN.md#640-求解方程
17881790
[641]:Problemset/0641-Design%20Circular%20Deque/README_CN.md#641-设计循环双端队列
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[643]:Problemset/0643-Maximum%20Average%20Subarray%20I/README_CN.md#643-子数组最大平均数-i
@@ -3180,6 +3182,7 @@
31803182
[633l]:https://leetcode.cn/problems/sum-of-square-numbers/
31813183
[636l]:https://leetcode.cn/problems/exclusive-time-of-functions/
31823184
[637l]:https://leetcode.cn/problems/average-of-levels-in-binary-tree/
3185+
[639l]:https://leetcode.cn/problems/decode-ways-ii/
31833186
[640l]:https://leetcode.cn/problems/solve-the-equation/
31843187
[641l]:https://leetcode.cn/problems/design-circular-deque/
31853188
[643l]:https://leetcode.cn/problems/maximum-average-subarray-i/

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