feat: add solutions to lc problems: No.3120~3123

solution/0300-0399/0377.Combination Sum IV/README.md

@@ -131,7 +131,7 @@ func combinationSum4(nums []int, target int) int {
 
 ```ts
 function combinationSum4(nums: number[], target: number): number {
-    const f: number[] = new Array(target + 1).fill(0);
+    const f: number[] = Array(target + 1).fill(0);
     f[0] = 1;
     for (let i = 1; i <= target; ++i) {
         for (const x of nums) {
@@ -151,7 +151,7 @@ function combinationSum4(nums: number[], target: number): number {
  * @return {number}
  */
 var combinationSum4 = function (nums, target) {
-    const f = new Array(target + 1).fill(0);
+    const f = Array(target + 1).fill(0);
     f[0] = 1;
     for (let i = 1; i <= target; ++i) {
         for (const x of nums) {

solution/0300-0399/0377.Combination Sum IV/README_EN.md

@@ -50,7 +50,15 @@ Note that different sequences are counted as different combinations.
 
 ## Solutions
 
-### Solution 1
+### Solution 1: Dynamic Programming
+
+We define $f[i]$ as the number of combinations that sum up to $i$. Initially, $f[0] = 1$, and the rest $f[i] = 0$. The final answer is $f[target]$.
+
+For $f[i]$, we can enumerate each element $x$ in the array. If $i \ge x$, then $f[i] = f[i] + f[i - x]$.
+
+Finally, return $f[target]$.
+
+The time complexity is $O(n \times target)$, and the space complexity is $O(target)$, where $n$ is the length of the array.
 
 <!-- tabs:start -->
 
@@ -118,7 +126,7 @@ func combinationSum4(nums []int, target int) int {
 
 ```ts
 function combinationSum4(nums: number[], target: number): number {
-    const f: number[] = new Array(target + 1).fill(0);
+    const f: number[] = Array(target + 1).fill(0);
     f[0] = 1;
     for (let i = 1; i <= target; ++i) {
         for (const x of nums) {
@@ -138,7 +146,7 @@ function combinationSum4(nums: number[], target: number): number {
  * @return {number}
  */
 var combinationSum4 = function (nums, target) {
-    const f = new Array(target + 1).fill(0);
+    const f = Array(target + 1).fill(0);
     f[0] = 1;
     for (let i = 1; i <= target; ++i) {
         for (const x of nums) {

solution/0300-0399/0377.Combination Sum IV/Solution.js

@@ -4,7 +4,7 @@
  * @return {number}
  */
 var combinationSum4 = function (nums, target) {
-    const f = new Array(target + 1).fill(0);
+    const f = Array(target + 1).fill(0);
     f[0] = 1;
     for (let i = 1; i <= target; ++i) {
         for (const x of nums) {

solution/0300-0399/0377.Combination Sum IV/Solution.ts

@@ -1,5 +1,5 @@
 function combinationSum4(nums: number[], target: number): number {
-    const f: number[] = new Array(target + 1).fill(0);
+    const f: number[] = Array(target + 1).fill(0);
     f[0] = 1;
     for (let i = 1; i <= target; ++i) {
         for (const x of nums) {

solution/3100-3199/3120.Count the Number of Special Characters I/README.md

@@ -61,24 +61,87 @@
 
 ## 解法
 
-### 方法一
+### 方法一：哈希表或数组
+
+我们用一个哈希表或数组 $s$ 来记录字符串 $word$ 中出现的字符。然后遍历 $26$ 个字母，如果小写字母和大写字母都在 $s$ 中出现，则特殊字符的数量加一。
+
+最后返回特殊字符的数量即可。
+
+时间复杂度 $O(n + |\Sigma|)$，空间复杂度 $O(|\Sigma|)$。其中 $n$ 为字符串 $word$ 的长度；而 $|\Sigma|$ 为字符集大小，本题中 $|\Sigma| \leq 128$。
 
 <!-- tabs:start -->
 
 ```python
-
+class Solution:
+    def numberOfSpecialChars(self, word: str) -> int:
+        s = set(word)
+        return sum(a in s and b in s for a, b in zip(ascii_lowercase, ascii_uppercase))
 ```
 
 ```java
-
+class Solution {
+    public int numberOfSpecialChars(String word) {
+        boolean[] s = new boolean['z' + 1];
+        for (int i = 0; i < word.length(); ++i) {
+            s[word.charAt(i)] = true;
+        }
+        int ans = 0;
+        for (int i = 0; i < 26; ++i) {
+            if (s['a' + i] && s['A' + i]) {
+                ++ans;
+            }
+        }
+        return ans;
+    }
+}
 ```
 
 ```cpp
-
+class Solution {
+public:
+    int numberOfSpecialChars(string word) {
+        vector<bool> s('z' + 1);
+        for (char& c : word) {
+            s[c] = true;
+        }
+        int ans = 0;
+        for (int i = 0; i < 26; ++i) {
+            ans += s['a' + i] && s['A' + i];
+        }
+        return ans;
+    }
+};
 ```
 
 ```go
+func numberOfSpecialChars(word string) (ans int) {
+	s := make([]bool, 'z'+1)
+	for _, c := range word {
+		s[c] = true
+	}
+	for i := 0; i < 26; i++ {
+		if s['a'+i] && s['A'+i] {
+			ans++
+		}
+	}
+	return
+}
+```
 
+```ts
+function numberOfSpecialChars(word: string): number {
+    const s: boolean[] = Array.from({ length: 'z'.charCodeAt(0) + 1 }, () => false);
+    for (let i = 0; i < word.length; ++i) {
+        s[word.charCodeAt(i)] = true;
+    }
+    let ans: number = 0;
+    for (let i = 0; i < 26; ++i) {
+        if (s['a'.charCodeAt(0) + i] && s['A'.charCodeAt(0) + i]) {
+            ++ans;
+        }
+    }
+    return ans;
+}
 ```
 
 <!-- tabs:end -->

solution/3100-3199/3120.Count the Number of Special Characters I/README_EN.md

@@ -57,24 +57,87 @@
 
 ## Solutions
 
-### Solution 1
+### Solution 1: Hash Table or Array
+
+We use a hash table or array $s$ to record the characters that appear in the string $word$. Then we traverse the 26 letters. If both the lowercase and uppercase letters appear in $s$, the count of special characters is incremented by one.
+
+Finally, return the count of special characters.
+
+The time complexity is $O(n + |\Sigma|)$, and the space complexity is $O(|\Sigma|)$. Where $n$ is the length of the string $word$, and $|\Sigma|$ is the size of the character set. In this problem, $|\Sigma| \leq 128$.
 
 <!-- tabs:start -->
 
 ```python
-
+class Solution:
+    def numberOfSpecialChars(self, word: str) -> int:
+        s = set(word)
+        return sum(a in s and b in s for a, b in zip(ascii_lowercase, ascii_uppercase))
 ```
 
 ```java
-
+class Solution {
+    public int numberOfSpecialChars(String word) {
+        boolean[] s = new boolean['z' + 1];
+        for (int i = 0; i < word.length(); ++i) {
+            s[word.charAt(i)] = true;
+        }
+        int ans = 0;
+        for (int i = 0; i < 26; ++i) {
+            if (s['a' + i] && s['A' + i]) {
+                ++ans;
+            }
+        }
+        return ans;
+    }
+}
 ```
 
 ```cpp
-
+class Solution {
+public:
+    int numberOfSpecialChars(string word) {
+        vector<bool> s('z' + 1);
+        for (char& c : word) {
+            s[c] = true;
+        }
+        int ans = 0;
+        for (int i = 0; i < 26; ++i) {
+            ans += s['a' + i] && s['A' + i];
+        }
+        return ans;
+    }
+};
 ```
 
 ```go
+func numberOfSpecialChars(word string) (ans int) {
+	s := make([]bool, 'z'+1)
+	for _, c := range word {
+		s[c] = true
+	}
+	for i := 0; i < 26; i++ {
+		if s['a'+i] && s['A'+i] {
+			ans++
+		}
+	}
+	return
+}
+```
 
+```ts
+function numberOfSpecialChars(word: string): number {
+    const s: boolean[] = Array.from({ length: 'z'.charCodeAt(0) + 1 }, () => false);
+    for (let i = 0; i < word.length; ++i) {
+        s[word.charCodeAt(i)] = true;
+    }
+    let ans: number = 0;
+    for (let i = 0; i < 26; ++i) {
+        if (s['a'.charCodeAt(0) + i] && s['A'.charCodeAt(0) + i]) {
+            ++ans;
+        }
+    }
+    return ans;
+}
 ```
 
 <!-- tabs:end -->

solution/3100-3199/3120.Count the Number of Special Characters I/Solution.cpp

@@ -0,0 +1,14 @@
+class Solution {
+public:
+    int numberOfSpecialChars(string word) {
+        vector<bool> s('z' + 1);
+        for (char& c : word) {
+            s[c] = true;
+        }
+        int ans = 0;
+        for (int i = 0; i < 26; ++i) {
+            ans += s['a' + i] && s['A' + i];
+        }
+        return ans;
+    }
+};

solution/3100-3199/3120.Count the Number of Special Characters I/Solution.go

@@ -0,0 +1,12 @@
+func numberOfSpecialChars(word string) (ans int) {
+	s := make([]bool, 'z'+1)
+	for _, c := range word {
+		s[c] = true
+	}
+	for i := 0; i < 26; i++ {
+		if s['a'+i] && s['A'+i] {
+			ans++
+		}
+	}
+	return
+}

solution/3100-3199/3120.Count the Number of Special Characters I/Solution.java

@@ -0,0 +1,15 @@
+class Solution {
+    public int numberOfSpecialChars(String word) {
+        boolean[] s = new boolean['z' + 1];
+        for (int i = 0; i < word.length(); ++i) {
+            s[word.charAt(i)] = true;
+        }
+        int ans = 0;
+        for (int i = 0; i < 26; ++i) {
+            if (s['a' + i] && s['A' + i]) {
+                ++ans;
+            }
+        }
+        return ans;
+    }
+}

solution/3100-3199/3120.Count the Number of Special Characters I/Solution.py

@@ -0,0 +1,4 @@
+class Solution:
+    def numberOfSpecialChars(self, word: str) -> int:
+        s = set(word)
+        return sum(a in s and b in s for a, b in zip(ascii_lowercase, ascii_uppercase))

solution/3100-3199/3120.Count the Number of Special Characters I/Solution.ts

@@ -0,0 +1,13 @@
+function numberOfSpecialChars(word: string): number {
+    const s: boolean[] = Array.from({ length: 'z'.charCodeAt(0) + 1 }, () => false);
+    for (let i = 0; i < word.length; ++i) {
+        s[word.charCodeAt(i)] = true;
+    }
+    let ans: number = 0;
+    for (let i = 0; i < 26; ++i) {
+        if (s['a'.charCodeAt(0) + i] && s['A'.charCodeAt(0) + i]) {
+            ++ans;
+        }
+    }
+    return ans;
+}