A. Single Wildcard Pattern Matching

Codeforces

IDCF1023A

Time1000ms

Memory256MB

Difficulty—

brute forceimplementationstrings

English · Original

Chinese · Translation

Formal · Original

You are given two strings $s$ and $t$. The string $s$ consists of lowercase Latin letters and **at most one** wildcard character '_*_', the string $t$ consists only of lowercase Latin letters. The length of the string $s$ equals $n$, the length of the string $t$ equals $m$. The wildcard character '_*_' in the string $s$ (if any) can be replaced with an arbitrary sequence (possibly empty) of lowercase Latin letters. No other character of $s$ can be replaced with anything. If it is possible to replace a wildcard character '_*_' in $s$ to obtain a string $t$, then the string $t$ matches the pattern $s$. For example, if $s=$"_aba*aba_" then the following strings match it "_abaaba_", "_abacaba_" and "_abazzzaba_", but the following strings do not match: "_ababa_", "_abcaaba_", "_codeforces_", "_aba1aba_", "_aba?aba_". If the given string $t$ matches the given string $s$, print "_YES_", otherwise print "_NO_". ## Input The first line contains two integers $n$ and $m$ ($1 \le n, m \le 2 \cdot 10^5$) — the length of the string $s$ and the length of the string $t$, respectively. The second line contains string $s$ of length $n$, which consists of lowercase Latin letters and **at most one** wildcard character '_*_'. The third line contains string $t$ of length $m$, which consists only of lowercase Latin letters. ## Output Print "_YES_" (without quotes), if you can obtain the string $t$ from the string $s$. Otherwise print "_NO_" (without quotes). [samples] ## Note In the first example a wildcard character '_*_' can be replaced with a string "_force_". So the string $s$ after this replacement is "_codeforces_" and the answer is "_YES_". In the second example a wildcard character '_*_' can be replaced with an empty string. So the string $s$ after this replacement is "_vkcup_" and the answer is "_YES_". There is no wildcard character '_*_' in the third example and the strings "_v_" and "_k_" are different so the answer is "_NO_". In the fourth example there is no such replacement of a wildcard character '_*_' that you can obtain the string $t$ so the answer is "_NO_".

你被给定两个字符串 $s$ 和 $t$。字符串 $s$ 由小写拉丁字母和至多一个通配符 '_*_' 组成，字符串 $t$ 仅由小写拉丁字母组成。字符串 $s$ 的长度为 $n$，字符串 $t$ 的长度为 $m$。字符串 $s$ 中的通配符 '_*_'（如果存在）可以被任意长度（包括空）的小写拉丁字母序列替换。$s$ 中的其他字符不能被替换。如果可以通过替换 $s$ 中的通配符 '_*_' 得到字符串 $t$，则称字符串 $t$ 匹配模式 $s$。例如，若 $s =$"_aba*aba_"，则以下字符串与之匹配："_abaaba_"、"_abacaba_" 和 "_abazzzaba_"，但以下字符串不匹配："_ababa_"、"_abcaaba_"、"_codeforces_"、"_aba1aba_"、"_aba?aba_"。如果给定的字符串 $t$ 匹配给定的字符串 $s$，请输出 "_YES_"，否则输出 "_NO_"。第一行包含两个整数 $n$ 和 $m$ ($1 lt.eq n, m lt.eq 2 dot.op 10^5$) —— 分别表示字符串 $s$ 和 $t$ 的长度。第二行包含长度为 $n$ 的字符串 $s$，由小写拉丁字母和至多一个通配符 '_*_' 组成。第三行包含长度为 $m$ 的字符串 $t$，仅由小写拉丁字母组成。如果你能从字符串 $s$ 得到字符串 $t$，请输出 "_YES_"（不含引号），否则输出 "_NO_"（不含引号）。在第一个例子中，通配符 '_*_' 可以被字符串 "_force_" 替换。因此替换后的字符串 $s$ 为 "_codeforces_"，答案为 "_YES_"。在第二个例子中，通配符 '_*_' 可以被空字符串替换。因此替换后的字符串 $s$ 为 "_vkcup_"，答案为 "_YES_"。在第三个例子中，字符串中没有通配符 '_*_'，且字符串 "_v_" 和 "_k_" 不同，因此答案为 "_NO_"。在第四个例子中，不存在任何对通配符 '_*_' 的替换方式能获得字符串 $t$，因此答案为 "_NO_"。 ## Input 第一行包含两个整数 $n$ 和 $m$ ($1 lt.eq n, m lt.eq 2 dot.op 10^5$) —— 分别表示字符串 $s$ 和 $t$ 的长度。第二行包含长度为 $n$ 的字符串 $s$，由小写拉丁字母和至多一个通配符 '_*_' 组成。第三行包含长度为 $m$ 的字符串 $t$，仅由小写拉丁字母组成。 ## Output 如果你能从字符串 $s$ 得到字符串 $t$，请输出 "_YES_"（不含引号），否则输出 "_NO_"（不含引号）。 [samples] ## Note 在第一个例子中，通配符 '_*_' 可以被字符串 "_force_" 替换。因此替换后的字符串 $s$ 为 "_codeforces_"，答案为 "_YES_"。在第二个例子中，通配符 '_*_' 可以被空字符串替换。因此替换后的字符串 $s$ 为 "_vkcup_"，答案为 "_YES_"。在第三个例子中，字符串中没有通配符 '_*_'，且字符串 "_v_" 和 "_k_" 不同，因此答案为 "_NO_"。在第四个例子中，不存在任何对通配符 '_*_' 的替换方式能获得字符串 $t$，因此答案为 "_NO_"。

**Definitions** Let $ s \in (\Sigma \cup \{*\})^n $ be a pattern string of length $ n $, where $ \Sigma $ is the set of lowercase Latin letters, and $ * $ denotes at most one wildcard character. Let $ t \in \Sigma^m $ be a target string of length $ m $. Let $ p \in \{0, 1\} $ indicate the presence of the wildcard: - $ p = 1 $ if $ * $ appears exactly once in $ s $, - $ p = 0 $ if $ * $ does not appear in $ s $. Let $ s_{\text{left}} \in \Sigma^* $ be the prefix of $ s $ before the wildcard (if any), and $ s_{\text{right}} \in \Sigma^* $ be the suffix of $ s $ after the wildcard (if any). If $ p = 0 $, then $ s_{\text{left}} = s $ and $ s_{\text{right}} = \varepsilon $ (empty string). **Constraints** 1. $ 1 \le n, m \le 2 \cdot 10^5 $ 2. $ s $ contains at most one occurrence of $ * $ 3. $ t $ contains only characters from $ \Sigma $ **Objective** Determine whether there exists a string $ w \in \Sigma^* $ such that: $$ s = s_{\text{left}} + * + s_{\text{right}} \quad \Rightarrow \quad s_{\text{left}} + w + s_{\text{right}} = t $$ Equivalently: - If $ p = 0 $: $ s = t $ - If $ p = 1 $: $ t = s_{\text{left}} + w + s_{\text{right}} $ for some $ w \in \Sigma^* $, i.e., $$ t[0:|s_{\text{left}}|] = s_{\text{left}} \quad \text{and} \quad t[m - |s_{\text{right}}|:m] = s_{\text{right}} \quad \text{and} \quad |s_{\text{left}}| + |s_{\text{right}}| \le m $$ **Output** $$ \begin{cases} \text{YES} & \text{if such } w \text{ exists} \\ \text{NO} & \text{otherwise} \end{cases} $$

Samples

Input #1

6 10
code*s
codeforces

Output #1

YES

Input #2

6 5
vk*cup
vkcup

Output #2

YES

Input #3

1 1
v
k

Output #3

NO

Input #4

9 6
gfgf*gfgf
gfgfgf

Output #4

NO

API Response (JSON)

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    "platform": "Codeforces",
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