# [16.18. Pattern Matching](https://leetcode.cn/problems/pattern-matching-lcci)
[中文文档](/lcci/16.18.Pattern%20Matching/README.md)
## Description
<p>You are given two strings, pattern and value. The pattern string consists of just the letters a and b, describing a pattern within a string. For example, the string catcatgocatgo matches the pattern aabab (where cat is a and go is b). It also matches patterns like a, ab, and b. Write a method to determine if value matches pattern. a and b cannot be the same string.</p>
<p><strong>Example 1: </strong></p>
<pre>
<strong>Input: </strong> pattern = "abba", value = "dogcatcatdog"
<strong>Output: </strong> true
</pre>
<p><strong>Example 2: </strong></p>
<pre>
<strong>Input: </strong> pattern = "abba", value = "dogcatcatfish"
<strong>Output: </strong> false
</pre>
<p><strong>Example 3: </strong></p>
<pre>
<strong>Input: </strong> pattern = "aaaa", value = "dogcatcatdog"
<strong>Output: </strong> false
</pre>
<p><strong>Example 4: </strong></p>
<pre>
<strong>Input: </strong> pattern = "abba", value = "dogdogdogdog"
<strong>Output: </strong> true
<strong>Explanation: </strong> "a"="dogdog",b=""，vice versa.
</pre>
<p><strong>Note: </strong></p>
<ul>
<li><code>0 <= len(pattern) <= 1000</code></li>
<li><code>0 <= len(value) <= 1000</code></li>
<li><code>pattern</code> only contains <code>"a"</code> and <code>"b"</code>, <code>value</code> only contains lowercase letters.</li>
</ul>
## Solutions
### Solution 1: Enumeration
We first count the number of characters `'a'` and `'b'` in the pattern string $pattern$, denoted as $cnt[0]$ and $cnt[1]$, respectively. Let the length of the string $value$ be $n$.
If $cnt[0]=0$, it means that the pattern string only contains the character `'b'`. We need to check whether $n$ is a multiple of $cnt[1]$, and whether $value$ can be divided into $cnt[1]$ substrings of length $n/cnt[1]$, and all these substrings are the same. If not, return $false$ directly.
If $cnt[1]=0$, it means that the pattern string only contains the character `'a'`. We need to check whether $n$ is a multiple of $cnt[0]$, and whether $value$ can be divided into $cnt[0]$ substrings of length $n/cnt[0]$, and all these substrings are the same. If not, return $false$ directly.
Next, we denote the length of the string matched by the character `'a'` as $la$, and the length of the string matched by the character `'b'` as $lb$. Then we have $la \times cnt[0] + lb \times cnt[1] = n$. If we enumerate $la$, we can determine the value of $lb$. Therefore, we can enumerate $la$ and check whether there exists an integer $lb$ that satisfies the above equation.
The time complexity is $O(n^2)$, and the space complexity is $O(n)$. Here, $n$ is the length of the string $value$.
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```python
class Solution:
def patternMatching(self, pattern: str, value: str) -> bool:
def check(la: int, lb: int) -> bool:
i = 0
a, b = "", ""
for c in pattern:
if c == "a":
if a and value[i : i + la] != a:
return False
a = value[i : i + la]
i += la
else:
if b and value[i : i + lb] != b:
return False
b = value[i : i + lb]
i += lb
return a != b
n = len(value)
cnt = Counter(pattern)
if cnt["a"] == 0:
return n % cnt["b"] == 0 and value[: n // cnt["b"]] * cnt["b"] == value
if cnt["b"] == 0:
return n % cnt["a"] == 0 and value[: n // cnt["a"]] * cnt["a"] == value
for la in range(n + 1):
if la * cnt["a"] > n:
break
lb, mod = divmod(n - la * cnt["a"], cnt["b"])
if mod == 0 and check(la, lb):
return True
return False
```
```java
class Solution {
private String pattern;
private String value;
public boolean patternMatching(String pattern, String value) {
this.pattern = pattern;
this.value = value;
int[] cnt = new int[2];
for (char c : pattern.toCharArray()) {
++cnt[c - 'a'];
}
int n = value.length();
if (cnt[0] == 0) {
return n % cnt[1] == 0 && value.substring(0, n / cnt[1]).repeat(cnt[1]).equals(value);
}
if (cnt[1] == 0) {
return n % cnt[0] == 0 && value.substring(0, n / cnt[0]).repeat(cnt[0]).equals(value);
}
for (int la = 0; la <= n; ++la) {
if (la * cnt[0] > n) {
break;
}
if ((n - la * cnt[0]) % cnt[1] == 0) {
int lb = (n - la * cnt[0]) / cnt[1];
if (check(la, lb)) {
return true;
}
}
}
return false;
}
private boolean check(int la, int lb) {
int i = 0;
String a = null, b = null;
for (char c : pattern.toCharArray()) {
if (c == 'a') {
if (a != null && !a.equals(value.substring(i, i + la))) {
return false;
}
a = value.substring(i, i + la);
i += la;
} else {
if (b != null && !b.equals(value.substring(i, i + lb))) {
return false;
}
b = value.substring(i, i + lb);
i += lb;
}
}
return !a.equals(b);
}
}
```
```cpp
class Solution {
public:
bool patternMatching(string pattern, string value) {
int n = value.size();
int cnt[2]{};
for (char c : pattern) {
cnt[c - 'a']++;
}
if (cnt[0] == 0) {
return n % cnt[1] == 0 && repeat(value.substr(0, n / cnt[1]), cnt[1]) == value;
}
if (cnt[1] == 0) {
return n % cnt[0] == 0 && repeat(value.substr(0, n / cnt[0]), cnt[0]) == value;
}
auto check = [&](int la, int lb) {
int i = 0;
string a, b;
for (char c : pattern) {
if (c == 'a') {
if (!a.empty() && a != value.substr(i, la)) {
return false;
}
a = value.substr(i, la);
i += la;
} else {
if (!b.empty() && b != value.substr(i, lb)) {
return false;
}
b = value.substr(i, lb);
i += lb;
}
}
return a != b;
};
for (int la = 0; la <= n; ++la) {
if (la * cnt[0] > n) {
break;
}
if ((n - la * cnt[0]) % cnt[1] == 0) {
int lb = (n - la * cnt[0]) / cnt[1];
if (check(la, lb)) {
return true;
}
}
}
return false;
}
string repeat(string s, int n) {
string ans;
while (n--) {
ans += s;
}
return ans;
}
};
```
```go
func patternMatching(pattern string, value string) bool {
cnt := [2]int{}
for _, c := range pattern {
cnt[c-'a']++
}
n := len(value)
if cnt[0] == 0 {
return n%cnt[1] == 0 && strings.Repeat(value[:n/cnt[1]], cnt[1]) == value
}
if cnt[1] == 0 {
return n%cnt[0] == 0 && strings.Repeat(value[:n/cnt[0]], cnt[0]) == value
}
check := func(la, lb int) bool {
i := 0
a, b := "", ""
for _, c := range pattern {
if c == 'a' {
if a != "" && value[i:i+la] != a {
return false
}
a = value[i : i+la]
i += la
} else {
if b != "" && value[i:i+lb] != b {
return false
}
b = value[i : i+lb]
i += lb
}
}
return a != b
}
for la := 0; la <= n; la++ {
if la*cnt[0] > n {
break
}
if (n-la*cnt[0])%cnt[1] == 0 {
lb := (n - la*cnt[0]) / cnt[1]
if check(la, lb) {
return true
}
}
}
return false
}
```
```ts
function patternMatching(pattern: string, value: string): boolean {
const cnt: number[] = [0, 0];
for (const c of pattern) {
cnt[c === 'a' ? 0 : 1]++;
}
const n = value.length;
if (cnt[0] === 0) {
return n % cnt[1] === 0 && value.slice(0, (n / cnt[1]) | 0).repeat(cnt[1]) === value;
}
if (cnt[1] === 0) {
return n % cnt[0] === 0 && value.slice(0, (n / cnt[0]) | 0).repeat(cnt[0]) === value;
}
const check = (la: number, lb: number) => {
let i = 0;
let a = '';
let b = '';
for (const c of pattern) {
if (c === 'a') {
if (a && a !== value.slice(i, i + la)) {
return false;
}
a = value.slice(i, (i += la));
} else {
if (b && b !== value.slice(i, i + lb)) {
return false;
}
b = value.slice(i, (i += lb));
}
}
return a !== b;
};
for (let la = 0; la <= n; ++la) {
if (la * cnt[0] > n) {
break;
}
if ((n - la * cnt[0]) % cnt[1] === 0) {
const lb = ((n - la * cnt[0]) / cnt[1]) | 0;
if (check(la, lb)) {
return true;
}
}
}
return false;
}
```
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