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feat: add solutions to lcci problem: No.16.15
No.16.15.Master Mind
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‎lcci/16.15.Master Mind/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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同时遍历两个字符串，算出对应位置字符相同的个数，累加到 $x$ 中，然后将两个字符串出现的字符以及出现的次数分别记录在哈希表 `cnt1` 和 `cnt2` 中。
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接着遍历两个哈希表，算出有多少共同出现的字符，累加到 $y$ 中。那么答案就是 $[x, y - x]$。
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时间复杂度 $O(C)$，空间复杂度 $O(C)$。本题中 $C=4$。
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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 masterMind(self, solution: str, guess: str) -> List[int]:
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x = sum(a == b for a, b in zip(solution, guess))
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y = sum((Counter(solution) & Counter(guess)).values())
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return [x, y - x]
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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[] masterMind(String solution, String guess) {
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int x = 0, y = 0;
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Map<Character, Integer> cnt1 = new HashMap<>();
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Map<Character, Integer> cnt2 = new HashMap<>();
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for (int i = 0; i < 4; ++i) {
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char a = solution.charAt(i), b = guess.charAt(i);
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x += a == b ? 1 : 0;
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cnt1.put(a, cnt1.getOrDefault(a, 0) + 1);
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cnt2.put(b, cnt2.getOrDefault(b, 0) + 1);
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}
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for (char c : "RYGB".toCharArray()) {
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y += Math.min(cnt1.getOrDefault(c, 0), cnt2.getOrDefault(c, 0));
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}
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return new int[] {x, y - x};
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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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vector<int> masterMind(string solution, string guess) {
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int x = 0, y = 0;
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unordered_map<char, int> cnt1;
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unordered_map<char, int> cnt2;
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for (int i = 0; i < 4; ++i) {
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x += solution[i] == guess[i];
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cnt1[solution[i]]++;
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cnt2[guess[i]]++;
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}
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for (char c : "RYGB") y += min(cnt1[c], cnt2[c]);
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return vector<int>{x, y - x};
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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 masterMind(solution string, guess string) []int {
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var x, y int
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cnt1 := map[byte]int{}
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cnt2 := map[byte]int{}
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for i := range solution {
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a, b := solution[i], guess[i]
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if a == b {
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x++
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}
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cnt1[a]++
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cnt2[b]++
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}
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for _, c := range []byte("RYGB") {
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y += min(cnt1[c], cnt2[c])
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}
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return []int{x, y - x}
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}
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func min(a, b int) int {
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if a < b {
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return a
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}
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return b
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}
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```
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### **JavaScript**

‎lcci/16.15.Master Mind/README_EN.md

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### **Python3**
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```python
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class Solution:
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def masterMind(self, solution: str, guess: str) -> List[int]:
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x = sum(a == b for a, b in zip(solution, guess))
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y = sum((Counter(solution) & Counter(guess)).values())
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return [x, y - x]
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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[] masterMind(String solution, String guess) {
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int x = 0, y = 0;
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Map<Character, Integer> cnt1 = new HashMap<>();
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Map<Character, Integer> cnt2 = new HashMap<>();
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for (int i = 0; i < 4; ++i) {
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char a = solution.charAt(i), b = guess.charAt(i);
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x += a == b ? 1 : 0;
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cnt1.put(a, cnt1.getOrDefault(a, 0) + 1);
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cnt2.put(b, cnt2.getOrDefault(b, 0) + 1);
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}
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for (char c : "RYGB".toCharArray()) {
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y += Math.min(cnt1.getOrDefault(c, 0), cnt2.getOrDefault(c, 0));
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}
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return new int[] {x, y - x};
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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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vector<int> masterMind(string solution, string guess) {
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int x = 0, y = 0;
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unordered_map<char, int> cnt1;
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unordered_map<char, int> cnt2;
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for (int i = 0; i < 4; ++i) {
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x += solution[i] == guess[i];
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cnt1[solution[i]]++;
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cnt2[guess[i]]++;
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}
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for (char c : "RYGB") y += min(cnt1[c], cnt2[c]);
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return vector<int>{x, y - x};
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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 masterMind(solution string, guess string) []int {
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var x, y int
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cnt1 := map[byte]int{}
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cnt2 := map[byte]int{}
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for i := range solution {
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a, b := solution[i], guess[i]
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if a == b {
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x++
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}
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cnt1[a]++
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cnt2[b]++
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}
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for _, c := range []byte("RYGB") {
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y += min(cnt1[c], cnt2[c])
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}
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return []int{x, y - x}
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}
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func min(a, b int) int {
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if a < b {
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return a
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}
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return b
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}
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```
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### **JavaScript**

‎lcci/16.15.Master Mind/Solution.cpp

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class Solution {
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public:
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vector<int> masterMind(string solution, string guess) {
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int x = 0, y = 0;
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unordered_map<char, int> cnt1;
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unordered_map<char, int> cnt2;
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for (int i = 0; i < 4; ++i) {
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x += solution[i] == guess[i];
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cnt1[solution[i]]++;
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cnt2[guess[i]]++;
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}
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for (char c : "RYGB") y += min(cnt1[c], cnt2[c]);
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return vector<int>{x, y - x};
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}
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};

‎lcci/16.15.Master Mind/Solution.go

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func masterMind(solution string, guess string) []int {
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var x, y int
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cnt1 := map[byte]int{}
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cnt2 := map[byte]int{}
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for i := range solution {
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a, b := solution[i], guess[i]
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if a == b {
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x++
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}
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cnt1[a]++
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cnt2[b]++
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}
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for _, c := range []byte("RYGB") {
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y += min(cnt1[c], cnt2[c])
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}
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return []int{x, y - x}
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}
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func min(a, b int) int {
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if a < b {
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return a
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}
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return b
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}

‎lcci/16.15.Master Mind/Solution.java

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class Solution {
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public int[] masterMind(String solution, String guess) {
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int x = 0, y = 0;
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Map<Character, Integer> cnt1 = new HashMap<>();
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Map<Character, Integer> cnt2 = new HashMap<>();
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for (int i = 0; i < 4; ++i) {
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char a = solution.charAt(i), b = guess.charAt(i);
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x += a == b ? 1 : 0;
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cnt1.put(a, cnt1.getOrDefault(a, 0) + 1);
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cnt2.put(b, cnt2.getOrDefault(b, 0) + 1);
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}
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for (char c : "RYGB".toCharArray()) {
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y += Math.min(cnt1.getOrDefault(c, 0), cnt2.getOrDefault(c, 0));
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}
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return new int[] {x, y - x};
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}
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}

‎lcci/16.15.Master Mind/Solution.py

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class Solution:
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def masterMind(self, solution: str, guess: str) -> List[int]:
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x = sum(a == b for a, b in zip(solution, guess))
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y = sum((Counter(solution) & Counter(guess)).values())
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return [x, y - x]

‎solution/1700-1799/1781.Sum of Beauty of All Substrings/README.md

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**方法一：枚举 + 计数**
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枚举每个子串的起点，找到以该起点为左端点的所有子串，然后计算每个子串的美丽值，最后将所有子串的美丽值相加即可。
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枚举每个子串的起点位置 $i$，找到以该起点位置的字符为左端点的所有子串，然后计算每个子串的美丽值，累加到答案中。
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时间复杂度 O(n^2 \times C)，空间复杂度 $O(C)$。其中 $n$ 为字符串的长度，而 $C$ 为字符集的大小。本题中 $C = 26$。
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时间复杂度 $O(n^2 \times C)$，空间复杂度 $O(C)$。其中 $n$ 为字符串的长度，而 $C$ 为字符集的大小。本题中 $C = 26$。
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<!-- tabs:start -->
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