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feat: add solutions to lc problem: No.1297
No.1297.Maximum Number of Occurrences of a Substring
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‎solution/1200-1299/1297.Maximum Number of Occurrences of a Substring/README.md

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<!-- 这里可写通用的实现逻辑 -->
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**方法一：哈希表 + 枚举**
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根据题目描述，如果一个长串满足条件，那么这个长串的子串（长度至少为 `minSize`）也一定满足条件。因此，我们只需要枚举 $s$ 中所有长度为 `minSize` 的子串，然后利用哈希表记录所有子串的出现次数，找出最大的次数作为答案即可。
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时间复杂度 $O(n \times m)$，空间复杂度 $O(n \times m)$。其中 $n$ 和 $m$ 分别为字符串 $s$ 的长度以及 `minSize` 的大小。本题中 $m$ 不超过 $26$。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def maxFreq(self, s: str, maxLetters: int, minSize: int, maxSize: int) -> int:
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ans = 0
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cnt = Counter()
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for i in range(len(s) - minSize + 1):
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t = s[i: i + minSize]
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ss = set(t)
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if len(ss) <= maxLetters:
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cnt[t] += 1
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ans = max(ans, cnt[t])
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return ans
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public int maxFreq(String s, int maxLetters, int minSize, int maxSize) {
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int ans = 0;
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Map<String, Integer> cnt = new HashMap<>();
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for (int i = 0; i < s.length() - minSize + 1; ++i) {
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String t = s.substring(i, i + minSize);
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Set<Character> ss = new HashSet<>();
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for (int j = 0; j < minSize; ++j) {
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ss.add(t.charAt(j));
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}
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if (ss.size() <= maxLetters) {
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cnt.put(t, cnt.getOrDefault(t, 0) + 1);
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ans = Math.max(ans, cnt.get(t));
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}
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}
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return ans;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int maxFreq(string s, int maxLetters, int minSize, int maxSize) {
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int ans = 0;
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unordered_map<string, int> cnt;
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for (int i = 0; i < s.size() - minSize + 1; ++i) {
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string t = s.substr(i, minSize);
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unordered_set<char> ss(t.begin(), t.end());
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if (ss.size() <= maxLetters) {
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ans = max(ans, ++cnt[t]);
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}
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}
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return ans;
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}
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};
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```
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### **Go**
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```go
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func maxFreq(s string, maxLetters int, minSize int, maxSize int) (ans int) {
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cnt := map[string]int{}
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for i := 0; i < len(s)-minSize+1; i++ {
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t := s[i : i+minSize]
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ss := map[rune]bool{}
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for _, c := range t {
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ss[c] = true
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}
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if len(ss) <= maxLetters {
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cnt[t]++
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if ans < cnt[t] {
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ans = cnt[t]
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}
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}
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}
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return
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}
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```
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### **...**

‎solution/1200-1299/1297.Maximum Number of Occurrences of a Substring/README_EN.md

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@@ -46,13 +46,82 @@ It satisfies the conditions, 2 unique letters and size 3 (between minSize and ma
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### **Python3**
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```python
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class Solution:
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def maxFreq(self, s: str, maxLetters: int, minSize: int, maxSize: int) -> int:
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ans = 0
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cnt = Counter()
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for i in range(len(s) - minSize + 1):
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t = s[i: i + minSize]
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ss = set(t)
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if len(ss) <= maxLetters:
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cnt[t] += 1
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ans = max(ans, cnt[t])
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return ans
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```
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### **Java**
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```java
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class Solution {
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public int maxFreq(String s, int maxLetters, int minSize, int maxSize) {
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int ans = 0;
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Map<String, Integer> cnt = new HashMap<>();
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for (int i = 0; i < s.length() - minSize + 1; ++i) {
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String t = s.substring(i, i + minSize);
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Set<Character> ss = new HashSet<>();
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for (int j = 0; j < minSize; ++j) {
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ss.add(t.charAt(j));
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}
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if (ss.size() <= maxLetters) {
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cnt.put(t, cnt.getOrDefault(t, 0) + 1);
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ans = Math.max(ans, cnt.get(t));
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}
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}
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return ans;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int maxFreq(string s, int maxLetters, int minSize, int maxSize) {
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int ans = 0;
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unordered_map<string, int> cnt;
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for (int i = 0; i < s.size() - minSize + 1; ++i) {
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string t = s.substr(i, minSize);
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unordered_set<char> ss(t.begin(), t.end());
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if (ss.size() <= maxLetters) {
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ans = max(ans, ++cnt[t]);
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}
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}
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return ans;
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}
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};
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```
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### **Go**
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```go
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func maxFreq(s string, maxLetters int, minSize int, maxSize int) (ans int) {
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cnt := map[string]int{}
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for i := 0; i < len(s)-minSize+1; i++ {
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t := s[i : i+minSize]
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ss := map[rune]bool{}
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for _, c := range t {
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ss[c] = true
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}
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if len(ss) <= maxLetters {
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cnt[t]++
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if ans < cnt[t] {
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ans = cnt[t]
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}
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}
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}
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return
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}
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```
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### **...**

‎solution/1200-1299/1297.Maximum Number of Occurrences of a Substring/Solution.cpp

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class Solution {
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public:
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int maxFreq(string s, int maxLetters, int minSize, int maxSize) {
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int ans = 0;
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unordered_map<string, int> cnt;
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for (int i = 0; i < s.size() - minSize + 1; ++i) {
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string t = s.substr(i, minSize);
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unordered_set<char> ss(t.begin(), t.end());
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if (ss.size() <= maxLetters) {
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ans = max(ans, ++cnt[t]);
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}
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}
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return ans;
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}
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};

‎solution/1200-1299/1297.Maximum Number of Occurrences of a Substring/Solution.go

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func maxFreq(s string, maxLetters int, minSize int, maxSize int) (ans int) {
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cnt := map[string]int{}
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for i := 0; i < len(s)-minSize+1; i++ {
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t := s[i : i+minSize]
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ss := map[rune]bool{}
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for _, c := range t {
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ss[c] = true
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}
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if len(ss) <= maxLetters {
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cnt[t]++
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if ans < cnt[t] {
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ans = cnt[t]
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}
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}
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}
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return
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}

‎solution/1200-1299/1297.Maximum Number of Occurrences of a Substring/Solution.java

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class Solution {
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public int maxFreq(String s, int maxLetters, int minSize, int maxSize) {
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int ans = 0;
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Map<String, Integer> cnt = new HashMap<>();
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for (int i = 0; i < s.length() - minSize + 1; ++i) {
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String t = s.substring(i, i + minSize);
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Set<Character> ss = new HashSet<>();
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for (int j = 0; j < minSize; ++j) {
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ss.add(t.charAt(j));
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}
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if (ss.size() <= maxLetters) {
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cnt.put(t, cnt.getOrDefault(t, 0) + 1);
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ans = Math.max(ans, cnt.get(t));
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}
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}
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return ans;
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}
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}

‎solution/1200-1299/1297.Maximum Number of Occurrences of a Substring/Solution.py

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class Solution:
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def maxFreq(self, s: str, maxLetters: int, minSize: int, maxSize: int) -> int:
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ans = 0
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cnt = Counter()
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for i in range(len(s) - minSize + 1):
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t = s[i: i + minSize]
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ss = set(t)
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if len(ss) <= maxLetters:
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cnt[t] += 1
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ans = max(ans, cnt[t])
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return ans

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