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feat: add solutions to lc problem: No.1525
No.1525.Number of Good Ways to Split a String
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‎solution/1500-1599/1525.Number of Good Ways to Split a String/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**
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我们维护两个哈希表，分别记录左侧出现的字符、右侧的字符以及出现的次数。初始时，左侧的哈希表为空，右侧的哈希表为字符串 $s$ 中所有字符出现的次数。
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接下来，我们从左到右遍历字符串 $s$，对于遍历到的字符 $c$，我们将其加入左侧的哈希表，同时将其在右侧的哈希表中的出现次数减一。如果减一后，右侧哈希表中的出现次数为 0，则将其从右侧哈希表中移除。然后，我们判断左侧哈希表中的键值对数量是否与右侧哈希表中的键值对数量相等，如果相等，则将答案加一。
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最终，我们返回答案即可。
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时间复杂度为 $O(n)$，空间复杂度为 $O(n)$。其中 $n$ 为字符串 $s$ 的长度。
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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 numSplits(self, s: str) -> int:
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cnt = Counter(s)
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vis = set()
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ans = 0
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for c in s:
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vis.add(c)
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cnt[c] -= 1
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if cnt[c] == 0:
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cnt.pop(c)
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ans += len(vis) == len(cnt)
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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 numSplits(String s) {
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Map<Character, Integer> cnt = new HashMap<>();
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for (char c : s.toCharArray()) {
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cnt.merge(c, 1, Integer::sum);
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}
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Set<Character> vis = new HashSet<>();
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int ans = 0;
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for (char c : s.toCharArray()) {
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vis.add(c);
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if (cnt.merge(c, -1, Integer::sum) == 0) {
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cnt.remove(c);
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}
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if (vis.size() == cnt.size()) {
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++ans;
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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 numSplits(string s) {
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unordered_map<char, int> cnt;
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for (char& c : s) {
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++cnt[c];
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}
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unordered_set<char> vis;
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int ans = 0;
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for (char& c : s) {
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vis.insert(c);
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if (--cnt[c] == 0) {
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cnt.erase(c);
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}
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ans += vis.size() == cnt.size();
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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 numSplits(s string) (ans int) {
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cnt := map[rune]int{}
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for _, c := range s {
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cnt[c]++
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}
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vis := map[rune]bool{}
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for _, c := range s {
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vis[c] = true
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cnt[c]--
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if cnt[c] == 0 {
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delete(cnt, c)
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}
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if len(vis) == len(cnt) {
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ans++
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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/1500-1599/1525.Number of Good Ways to Split a String/README_EN.md

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### **Python3**
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```python
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class Solution:
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def numSplits(self, s: str) -> int:
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cnt = Counter(s)
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vis = set()
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ans = 0
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for c in s:
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vis.add(c)
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cnt[c] -= 1
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if cnt[c] == 0:
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cnt.pop(c)
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ans += len(vis) == len(cnt)
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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 numSplits(String s) {
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Map<Character, Integer> cnt = new HashMap<>();
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for (char c : s.toCharArray()) {
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cnt.merge(c, 1, Integer::sum);
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}
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Set<Character> vis = new HashSet<>();
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int ans = 0;
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for (char c : s.toCharArray()) {
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vis.add(c);
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if (cnt.merge(c, -1, Integer::sum) == 0) {
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cnt.remove(c);
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}
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if (vis.size() == cnt.size()) {
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++ans;
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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 numSplits(string s) {
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unordered_map<char, int> cnt;
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for (char& c : s) {
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++cnt[c];
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}
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unordered_set<char> vis;
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int ans = 0;
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for (char& c : s) {
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vis.insert(c);
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if (--cnt[c] == 0) {
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cnt.erase(c);
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}
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ans += vis.size() == cnt.size();
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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 numSplits(s string) (ans int) {
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cnt := map[rune]int{}
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for _, c := range s {
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cnt[c]++
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}
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vis := map[rune]bool{}
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for _, c := range s {
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vis[c] = true
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cnt[c]--
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if cnt[c] == 0 {
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delete(cnt, c)
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}
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if len(vis) == len(cnt) {
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ans++
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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/1500-1599/1525.Number of Good Ways to Split a String/Solution.cpp

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class Solution {
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public:
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int numSplits(string s) {
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unordered_map<char, int> cnt;
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for (char& c : s) {
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++cnt[c];
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}
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unordered_set<char> vis;
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int ans = 0;
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for (char& c : s) {
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vis.insert(c);
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if (--cnt[c] == 0) {
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cnt.erase(c);
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}
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ans += vis.size() == cnt.size();
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}
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return ans;
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}
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};

‎solution/1500-1599/1525.Number of Good Ways to Split a String/Solution.go

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func numSplits(s string) (ans int) {
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cnt := map[rune]int{}
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for _, c := range s {
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cnt[c]++
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}
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vis := map[rune]bool{}
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for _, c := range s {
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vis[c] = true
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cnt[c]--
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if cnt[c] == 0 {
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delete(cnt, c)
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}
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if len(vis) == len(cnt) {
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ans++
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}
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}
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return
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}

‎solution/1500-1599/1525.Number of Good Ways to Split a String/Solution.java

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class Solution {
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public int numSplits(String s) {
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Map<Character, Integer> cnt = new HashMap<>();
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for (char c : s.toCharArray()) {
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cnt.merge(c, 1, Integer::sum);
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}
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Set<Character> vis = new HashSet<>();
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int ans = 0;
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for (char c : s.toCharArray()) {
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vis.add(c);
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if (cnt.merge(c, -1, Integer::sum) == 0) {
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cnt.remove(c);
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}
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if (vis.size() == cnt.size()) {
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++ans;
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}
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}
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return ans;
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}
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}

‎solution/1500-1599/1525.Number of Good Ways to Split a String/Solution.py

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class Solution:
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def numSplits(self, s: str) -> int:
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cnt = Counter(s)
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vis = set()
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ans = 0
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for c in s:
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vis.add(c)
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cnt[c] -= 1
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if cnt[c] == 0:
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cnt.pop(c)
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ans += len(vis) == len(cnt)
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return ans

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