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Commit 97fc76c

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feat: add solutions to lc problems: No.3120~3123 (doocs#2637)
* No.3120.Count the Number of Special Characters I * No.3121.Count the Number of Special Characters II * No.3122.Minimum Number of Operations to Satisfy Conditions * No.3123.Find Edges in Shortest Paths
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31 files changed

+1523
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lines changed

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

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@@ -131,7 +131,7 @@ func combinationSum4(nums []int, target int) int {
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```ts
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function combinationSum4(nums: number[], target: number): number {
134-
const f: number[] = newArray(target + 1).fill(0);
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const f: number[] = Array(target + 1).fill(0);
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f[0] = 1;
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for (let i = 1; i <= target; ++i) {
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for (const x of nums) {
@@ -151,7 +151,7 @@ function combinationSum4(nums: number[], target: number): number {
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* @return {number}
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*/
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var combinationSum4 = function (nums, target) {
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const f = newArray(target + 1).fill(0);
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const f = Array(target + 1).fill(0);
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f[0] = 1;
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for (let i = 1; i <= target; ++i) {
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for (const x of nums) {

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

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@@ -50,7 +50,15 @@ Note that different sequences are counted as different combinations.
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## Solutions
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### Solution 1
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### Solution 1: Dynamic Programming
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We define $f[i]$ as the number of combinations that sum up to $i$. Initially, $f[0] = 1,ドル and the rest $f[i] = 0$. The final answer is $f[target]$.
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For $f[i],ドル we can enumerate each element $x$ in the array. If $i \ge x,ドル then $f[i] = f[i] + f[i - x]$.
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Finally, return $f[target]$.
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The time complexity is $O(n \times target),ドル and the space complexity is $O(target),ドル where $n$ is the length of the array.
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<!-- tabs:start -->
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@@ -118,7 +126,7 @@ func combinationSum4(nums []int, target int) int {
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```ts
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function combinationSum4(nums: number[], target: number): number {
121-
const f: number[] = newArray(target + 1).fill(0);
129+
const f: number[] = Array(target + 1).fill(0);
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f[0] = 1;
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for (let i = 1; i <= target; ++i) {
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for (const x of nums) {
@@ -138,7 +146,7 @@ function combinationSum4(nums: number[], target: number): number {
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* @return {number}
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*/
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var combinationSum4 = function (nums, target) {
141-
const f = newArray(target + 1).fill(0);
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const f = Array(target + 1).fill(0);
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f[0] = 1;
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for (let i = 1; i <= target; ++i) {
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for (const x of nums) {

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

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

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

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

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

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@@ -61,24 +61,87 @@
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## 解法
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### 方法一
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### 方法一:哈希表或数组
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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ドル$ 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。
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最后返回特殊字符的数量即可。
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时间复杂度 $O(n + |\Sigma|),ドル空间复杂度 $O(|\Sigma|)$。其中 $n$ 为字符串 $word$ 的长度;而 $|\Sigma|$ 为字符集大小,本题中 $|\Sigma| \leq 128$。
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<!-- tabs:start -->
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```python
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class Solution:
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def numberOfSpecialChars(self, word: str) -> int:
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s = set(word)
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return sum(a in s and b in s for a, b in zip(ascii_lowercase, ascii_uppercase))
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```
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```java
73-
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class Solution {
83+
public int numberOfSpecialChars(String word) {
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boolean[] s = new boolean['z' + 1];
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for (int i = 0; i < word.length(); ++i) {
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s[word.charAt(i)] = true;
87+
}
88+
int ans = 0;
89+
for (int i = 0; i < 26; ++i) {
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if (s['a' + i] && s['A' + i]) {
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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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```cpp
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class Solution {
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public:
102+
int numberOfSpecialChars(string word) {
103+
vector<bool> s('z' + 1);
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for (char& c : word) {
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s[c] = true;
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}
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int ans = 0;
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for (int i = 0; i < 26; ++i) {
109+
ans += s['a' + i] && s['A' + i];
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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
117+
func numberOfSpecialChars(word string) (ans int) {
118+
s := make([]bool, 'z'+1)
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for _, c := range word {
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s[c] = true
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}
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for i := 0; i < 26; i++ {
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if s['a'+i] && s['A'+i] {
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ans++
125+
}
126+
}
127+
return
128+
}
129+
```
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```ts
132+
function numberOfSpecialChars(word: string): number {
133+
const s: boolean[] = Array.from({ length: 'z'.charCodeAt(0) + 1 }, () => false);
134+
for (let i = 0; i < word.length; ++i) {
135+
s[word.charCodeAt(i)] = true;
136+
}
137+
let ans: number = 0;
138+
for (let i = 0; i < 26; ++i) {
139+
if (s['a'.charCodeAt(0) + i] && s['A'.charCodeAt(0) + i]) {
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++ans;
141+
}
142+
}
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return ans;
144+
}
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```
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<!-- tabs:end -->

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

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## Solutions
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### Solution 1
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### Solution 1: Hash Table or Array
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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,ドル the count of special characters is incremented by one.
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Finally, return the count of special characters.
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The time complexity is $O(n + |\Sigma|),ドル and the space complexity is $O(|\Sigma|)$. Where $n$ is the length of the string $word,ドル and $|\Sigma|$ is the size of the character set. In this problem, $|\Sigma| \leq 128$.
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<!-- tabs:start -->
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```python
65-
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class Solution:
72+
def numberOfSpecialChars(self, word: str) -> int:
73+
s = set(word)
74+
return sum(a in s and b in s for a, b in zip(ascii_lowercase, ascii_uppercase))
6675
```
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```java
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78+
class Solution {
79+
public int numberOfSpecialChars(String word) {
80+
boolean[] s = new boolean['z' + 1];
81+
for (int i = 0; i < word.length(); ++i) {
82+
s[word.charAt(i)] = true;
83+
}
84+
int ans = 0;
85+
for (int i = 0; i < 26; ++i) {
86+
if (s['a' + i] && s['A' + i]) {
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++ans;
88+
}
89+
}
90+
return ans;
91+
}
92+
}
7093
```
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7295
```cpp
73-
96+
class Solution {
97+
public:
98+
int numberOfSpecialChars(string word) {
99+
vector<bool> s('z' + 1);
100+
for (char& c : word) {
101+
s[c] = true;
102+
}
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int ans = 0;
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for (int i = 0; i < 26; ++i) {
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ans += s['a' + i] && s['A' + i];
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}
107+
return ans;
108+
}
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};
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```
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```go
113+
func numberOfSpecialChars(word string) (ans int) {
114+
s := make([]bool, 'z'+1)
115+
for _, c := range word {
116+
s[c] = true
117+
}
118+
for i := 0; i < 26; i++ {
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if s['a'+i] && s['A'+i] {
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ans++
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}
122+
}
123+
return
124+
}
125+
```
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```ts
128+
function numberOfSpecialChars(word: string): number {
129+
const s: boolean[] = Array.from({ length: 'z'.charCodeAt(0) + 1 }, () => false);
130+
for (let i = 0; i < word.length; ++i) {
131+
s[word.charCodeAt(i)] = true;
132+
}
133+
let ans: number = 0;
134+
for (let i = 0; i < 26; ++i) {
135+
if (s['a'.charCodeAt(0) + i] && s['A'.charCodeAt(0) + i]) {
136+
++ans;
137+
}
138+
}
139+
return ans;
140+
}
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```
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<!-- tabs:end -->
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class Solution {
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public:
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int numberOfSpecialChars(string word) {
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vector<bool> s('z' + 1);
5+
for (char& c : word) {
6+
s[c] = true;
7+
}
8+
int ans = 0;
9+
for (int i = 0; i < 26; ++i) {
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ans += s['a' + i] && s['A' + i];
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}
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return ans;
13+
}
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};
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func numberOfSpecialChars(word string) (ans int) {
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s := make([]bool, 'z'+1)
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for _, c := range word {
4+
s[c] = true
5+
}
6+
for i := 0; i < 26; i++ {
7+
if s['a'+i] && s['A'+i] {
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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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class Solution {
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public int numberOfSpecialChars(String word) {
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boolean[] s = new boolean['z' + 1];
4+
for (int i = 0; i < word.length(); ++i) {
5+
s[word.charAt(i)] = true;
6+
}
7+
int ans = 0;
8+
for (int i = 0; i < 26; ++i) {
9+
if (s['a' + i] && s['A' + i]) {
10+
++ans;
11+
}
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}
13+
return ans;
14+
}
15+
}
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class Solution:
2+
def numberOfSpecialChars(self, word: str) -> int:
3+
s = set(word)
4+
return sum(a in s and b in s for a, b in zip(ascii_lowercase, ascii_uppercase))

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