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feat: add solutions to lc problems: No.0292,0293 (doocs#2308)
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‎solution/0200-0299/0292.Nim Game/README.md

+6-5
Original file line numberDiff line numberDiff line change
@@ -55,19 +55,20 @@
5555

5656
## 解法
5757

58-
### 方法一：数学推理
58+
### 方法一：找规律
5959

6060
第一个得到 $4$ 的倍数（即 $n$ 能被 $4$ 整除）的将会输掉比赛。
6161

6262
证明：
6363

64-
1. 当 $n=4$，无论第一个玩家选择 $1/2/3$ 哪个数字，第二个玩家总能选择剩下的数字，**第一个玩家将会输掉比赛**。
65-
1. 当 $4<n<8$，即 ($n=5,6,7$)，第一个玩家可以相应地将数字减少为 $4$，那么 $4$ 这个死亡数字给到了第二个玩家，第二个玩家将会输掉比赛。
66-
1. 当 $n=8$，无论第一个玩家选择 $1/2/3$ 哪个数字，都会把 $4<n<8$ 的数字留给第二个，**第一个玩家将会输掉比赛**。
64+
1. 当 $n \lt 4$ 时，第一个玩家可以直接拿走所有的石头，所以第一个玩家将会赢得比赛。
65+
1. 当 $n = 4$，无论第一个玩家选择 $1, 2, 3$ 哪个数字，第二个玩家总能选择剩下的数字，所以第一个玩家将会输掉比赛。
66+
1. 当 $4 \lt n \lt 8$ 时，即 $n = 5, 6, 7$，第一个玩家可以相应地将数字减少为 $4$，那么 $4$ 这个死亡数字给到了第二个玩家，第二个玩家将会输掉比赛。
67+
1. 当 $n = 8$，无论第一个玩家选择 $1, 2, 3$ 哪个数字，都会把 $4 \lt n \lt 8$ 的数字留给第二个，所以第一个玩家将会输掉比赛。
6768
1. ...
6869
1. 依次类推，当玩家拿到 $n$ 这个数字，且 $n$ 能被 $4$ 整除，他将会输掉比赛，否则他将赢得比赛。
6970

70-
时间复杂度 $O(1)$。
71+
时间复杂度 $O(1)$，空间复杂度 $O(1)$。
7172

7273
<!-- tabs:start -->
7374

‎solution/0200-0299/0292.Nim Game/README_EN.md

+14-1
Original file line numberDiff line numberDiff line change
@@ -51,7 +51,20 @@ In all outcomes, your friend wins.
5151

5252
## Solutions
5353

54-
### Solution 1
54+
### Solution 1: Finding the Pattern
55+
56+
The first player who gets a multiple of $4$ (i.e., $n$ can be divided by $4$) will lose the game.
57+
58+
Proof:
59+
60+
1. When $n \lt 4$, the first player can directly take all the stones, so the first player will win the game.
61+
1. When $n = 4$, no matter whether the first player chooses $1, 2, 3$, the second player can always choose the remaining number, so the first player will lose the game.
62+
1. When $4 \lt n \lt 8$, i.e., $n = 5, 6, 7$, the first player can correspondingly reduce the number to $4$, then the "death number" $4$ is given to the second player, and the second player will lose the game.
63+
1. When $n = 8$, no matter whether the first player chooses $1, 2, 3$, it will leave a number between $4 \lt n \lt 8$ to the second player, so the first player will lose the game.
64+
1. ...
65+
1. By induction, when a player gets the number $n$, and $n$ can be divided by $4$, he will lose the game, otherwise, he will win the game.
66+
67+
The time complexity is $O(1)$, and the space complexity is $O(1)$.
5568

5669
<!-- tabs:start -->
5770

‎solution/0200-0299/0293.Flip Game/README.md

+46-28
Original file line numberDiff line numberDiff line change
@@ -39,7 +39,13 @@
3939

4040
## 解法
4141

42-
### 方法一
42+
### 方法一：遍历 + 模拟
43+
44+
我们遍历字符串，如果当前字符和下一个字符都是 `+`，那么我们就将这两个字符变成 `-`，然后将结果加入到结果数组中，再将这两个字符变回 `+`。
45+
46+
遍历结束后，返回结果数组即可。
47+
48+
时间复杂度 $O(n^2)$，其中 $n$ 是字符串长度。忽略答案数组的空间复杂度，空间复杂度 $O(n)$ 或 $O(1)$。
4349

4450
<!-- tabs:start -->
4551

@@ -48,8 +54,8 @@ class Solution:
4854
def generatePossibleNextMoves(self, currentState: str) -> List[str]:
4955
s = list(currentState)
5056
ans = []
51-
for i, c in enumerate(s[:-1]):
52-
if c == "+" and s[i + 1] == "+":
57+
for i, (a, b) in enumerate(pairwise(s)):
58+
if a == b == "+":
5359
s[i] = s[i + 1] = "-"
5460
ans.append("".join(s))
5561
s[i] = s[i + 1] = "+"
@@ -59,15 +65,15 @@ class Solution:
5965
```java
6066
class Solution {
6167
public List<String> generatePossibleNextMoves(String currentState) {
62-
char[] cs = currentState.toCharArray();
6368
List<String> ans = new ArrayList<>();
64-
for (int i = 0; i < cs.length - 1; ++i) {
65-
if (cs[i] == '+' && cs[i + 1] == '+') {
66-
cs[i] = '-';
67-
cs[i + 1] = '-';
68-
ans.add(String.valueOf(cs));
69-
cs[i] = '+';
70-
cs[i + 1] = '+';
69+
char[] s = currentState.toCharArray();
70+
for (int i = 0; i < s.length - 1; ++i) {
71+
if (s[i] == '+' && s[i + 1] == '+') {
72+
s[i] = '-';
73+
s[i + 1] = '-';
74+
ans.add(new String(s));
75+
s[i] = '+';
76+
s[i + 1] = '+';
7177
}
7278
}
7379
return ans;
@@ -78,15 +84,13 @@ class Solution {
7884
```cpp
7985
class Solution {
8086
public:
81-
vector<string> generatePossibleNextMoves(string currentState) {
87+
vector<string> generatePossibleNextMoves(string s) {
8288
vector<string> ans;
83-
for (int i = 0; i < currentState.size() - 1; ++i) {
84-
if (currentState[i] == '+' && currentState[i + 1] == '+') {
85-
currentState[i] = '-';
86-
currentState[i + 1] = '-';
87-
ans.push_back(currentState);
88-
currentState[i] = '+';
89-
currentState[i + 1] = '+';
89+
for (int i = 0; i < s.size() - 1; ++i) {
90+
if (s[i] == '+' && s[i + 1] == '+') {
91+
s[i] = s[i + 1] = '-';
92+
ans.emplace_back(s);
93+
s[i] = s[i + 1] = '+';
9094
}
9195
}
9296
return ans;
@@ -95,17 +99,31 @@ public:
9599
```
96100
97101
```go
98-
func generatePossibleNextMoves(currentState string) []string {
99-
ans := []string{}
100-
cs := []byte(currentState)
101-
for i, c := range cs[1:] {
102-
if c == '+' && cs[i] == '+' {
103-
cs[i], cs[i+1] = '-', '-'
104-
ans = append(ans, string(cs))
105-
cs[i], cs[i+1] = '+', '+'
102+
func generatePossibleNextMoves(currentState string) (ans []string) {
103+
s := []byte(currentState)
104+
for i := 0; i < len(s)-1; i++ {
105+
if s[i] == '+' && s[i+1] == '+' {
106+
s[i], s[i+1] = '-', '-'
107+
ans = append(ans, string(s))
108+
s[i], s[i+1] = '+', '+'
106109
}
107110
}
108-
return ans
111+
return
112+
}
113+
```
114+
115+
```ts
116+
function generatePossibleNextMoves(currentState: string): string[] {
117+
const s = currentState.split('');
118+
const ans: string[] = [];
119+
for (let i = 0; i < s.length - 1; ++i) {
120+
if (s[i] === '+' && s[i + 1] === '+') {
121+
s[i] = s[i + 1] = '-';
122+
ans.push(s.join(''));
123+
s[i] = s[i + 1] = '+';
124+
}
125+
}
126+
return ans;
109127
}
110128
```
111129

‎solution/0200-0299/0293.Flip Game/README_EN.md

+46-28
Original file line numberDiff line numberDiff line change
@@ -35,7 +35,13 @@
3535

3636
## Solutions
3737

38-
### Solution 1
38+
### Solution 1: Traversal + Simulation
39+
40+
We traverse the string. If the current character and the next character are both `+`, we change these two characters to `-`, add the result to the result array, and then change these two characters back to `+`.
41+
42+
After the traversal ends, we return the result array.
43+
44+
The time complexity is $O(n^2)$, where $n$ is the length of the string. Ignoring the space complexity of the result array, the space complexity is $O(n)$ or $O(1)$.
3945

4046
<!-- tabs:start -->
4147

@@ -44,8 +50,8 @@ class Solution:
4450
def generatePossibleNextMoves(self, currentState: str) -> List[str]:
4551
s = list(currentState)
4652
ans = []
47-
for i, c in enumerate(s[:-1]):
48-
if c == "+" and s[i + 1] == "+":
53+
for i, (a, b) in enumerate(pairwise(s)):
54+
if a == b == "+":
4955
s[i] = s[i + 1] = "-"
5056
ans.append("".join(s))
5157
s[i] = s[i + 1] = "+"
@@ -55,15 +61,15 @@ class Solution:
5561
```java
5662
class Solution {
5763
public List<String> generatePossibleNextMoves(String currentState) {
58-
char[] cs = currentState.toCharArray();
5964
List<String> ans = new ArrayList<>();
60-
for (int i = 0; i < cs.length - 1; ++i) {
61-
if (cs[i] == '+' && cs[i + 1] == '+') {
62-
cs[i] = '-';
63-
cs[i + 1] = '-';
64-
ans.add(String.valueOf(cs));
65-
cs[i] = '+';
66-
cs[i + 1] = '+';
65+
char[] s = currentState.toCharArray();
66+
for (int i = 0; i < s.length - 1; ++i) {
67+
if (s[i] == '+' && s[i + 1] == '+') {
68+
s[i] = '-';
69+
s[i + 1] = '-';
70+
ans.add(new String(s));
71+
s[i] = '+';
72+
s[i + 1] = '+';
6773
}
6874
}
6975
return ans;
@@ -74,15 +80,13 @@ class Solution {
7480
```cpp
7581
class Solution {
7682
public:
77-
vector<string> generatePossibleNextMoves(string currentState) {
83+
vector<string> generatePossibleNextMoves(string s) {
7884
vector<string> ans;
79-
for (int i = 0; i < currentState.size() - 1; ++i) {
80-
if (currentState[i] == '+' && currentState[i + 1] == '+') {
81-
currentState[i] = '-';
82-
currentState[i + 1] = '-';
83-
ans.push_back(currentState);
84-
currentState[i] = '+';
85-
currentState[i + 1] = '+';
85+
for (int i = 0; i < s.size() - 1; ++i) {
86+
if (s[i] == '+' && s[i + 1] == '+') {
87+
s[i] = s[i + 1] = '-';
88+
ans.emplace_back(s);
89+
s[i] = s[i + 1] = '+';
8690
}
8791
}
8892
return ans;
@@ -91,17 +95,31 @@ public:
9195
```
9296
9397
```go
94-
func generatePossibleNextMoves(currentState string) []string {
95-
ans := []string{}
96-
cs := []byte(currentState)
97-
for i, c := range cs[1:] {
98-
if c == '+' && cs[i] == '+' {
99-
cs[i], cs[i+1] = '-', '-'
100-
ans = append(ans, string(cs))
101-
cs[i], cs[i+1] = '+', '+'
98+
func generatePossibleNextMoves(currentState string) (ans []string) {
99+
s := []byte(currentState)
100+
for i := 0; i < len(s)-1; i++ {
101+
if s[i] == '+' && s[i+1] == '+' {
102+
s[i], s[i+1] = '-', '-'
103+
ans = append(ans, string(s))
104+
s[i], s[i+1] = '+', '+'
102105
}
103106
}
104-
return ans
107+
return
108+
}
109+
```
110+
111+
```ts
112+
function generatePossibleNextMoves(currentState: string): string[] {
113+
const s = currentState.split('');
114+
const ans: string[] = [];
115+
for (let i = 0; i < s.length - 1; ++i) {
116+
if (s[i] === '+' && s[i + 1] === '+') {
117+
s[i] = s[i + 1] = '-';
118+
ans.push(s.join(''));
119+
s[i] = s[i + 1] = '+';
120+
}
121+
}
122+
return ans;
105123
}
106124
```
107125

‎solution/0200-0299/0293.Flip Game/Solution.cpp

+6-8
Original file line numberDiff line numberDiff line change
@@ -1,14 +1,12 @@
11
class Solution {
22
public:
3-
vector<string> generatePossibleNextMoves(string currentState) {
3+
vector<string> generatePossibleNextMoves(string s) {
44
vector<string> ans;
5-
for (int i = 0; i < currentState.size() - 1; ++i) {
6-
if (currentState[i] == '+' && currentState[i + 1] == '+') {
7-
currentState[i] = '-';
8-
currentState[i + 1] = '-';
9-
ans.push_back(currentState);
10-
currentState[i] = '+';
11-
currentState[i + 1] = '+';
5+
for (int i = 0; i < s.size() - 1; ++i) {
6+
if (s[i] == '+' && s[i + 1] == '+') {
7+
s[i] = s[i + 1] = '-';
8+
ans.emplace_back(s);
9+
s[i] = s[i + 1] = '+';
1210
}
1311
}
1412
return ans;

‎solution/0200-0299/0293.Flip Game/Solution.go

+8-9
Original file line numberDiff line numberDiff line change
@@ -1,12 +1,11 @@
1-
func generatePossibleNextMoves(currentState string) []string {
2-
ans := []string{}
3-
cs := []byte(currentState)
4-
for i, c := range cs[1:] {
5-
if c == '+' && cs[i] == '+' {
6-
cs[i], cs[i+1] = '-', '-'
7-
ans = append(ans, string(cs))
8-
cs[i], cs[i+1] = '+', '+'
1+
func generatePossibleNextMoves(currentState string) (ans []string) {
2+
s := []byte(currentState)
3+
for i := 0; i < len(s)-1; i++ {
4+
if s[i] == '+' && s[i+1] == '+' {
5+
s[i], s[i+1] = '-', '-'
6+
ans = append(ans, string(s))
7+
s[i], s[i+1] = '+', '+'
98
}
109
}
11-
return ans
10+
return
1211
}

‎solution/0200-0299/0293.Flip Game/Solution.java

+8-8
Original file line numberDiff line numberDiff line change
@@ -1,14 +1,14 @@
11
class Solution {
22
public List<String> generatePossibleNextMoves(String currentState) {
3-
char[] cs = currentState.toCharArray();
43
List<String> ans = new ArrayList<>();
5-
for (int i = 0; i < cs.length - 1; ++i) {
6-
if (cs[i] == '+' && cs[i + 1] == '+') {
7-
cs[i] = '-';
8-
cs[i + 1] = '-';
9-
ans.add(String.valueOf(cs));
10-
cs[i] = '+';
11-
cs[i + 1] = '+';
4+
char[] s = currentState.toCharArray();
5+
for (int i = 0; i < s.length - 1; ++i) {
6+
if (s[i] == '+' && s[i + 1] == '+') {
7+
s[i] = '-';
8+
s[i + 1] = '-';
9+
ans.add(new String(s));
10+
s[i] = '+';
11+
s[i + 1] = '+';
1212
}
1313
}
1414
return ans;

‎solution/0200-0299/0293.Flip Game/Solution.py

+2-2
Original file line numberDiff line numberDiff line change
@@ -2,8 +2,8 @@ class Solution:
22
def generatePossibleNextMoves(self, currentState: str) -> List[str]:
33
s = list(currentState)
44
ans = []
5-
for i, c in enumerate(s[:-1]):
6-
if c == "+" and s[i + 1] == "+":
5+
for i, (a, b) in enumerate(pairwise(s)):
6+
if a == b == "+":
77
s[i] = s[i + 1] = "-"
88
ans.append("".join(s))
99
s[i] = s[i + 1] = "+"

‎solution/0200-0299/0293.Flip Game/Solution.ts

+12
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,12 @@
1+
function generatePossibleNextMoves(currentState: string): string[] {
2+
const s = currentState.split('');
3+
const ans: string[] = [];
4+
for (let i = 0; i < s.length - 1; ++i) {
5+
if (s[i] === '+' && s[i + 1] === '+') {
6+
s[i] = s[i + 1] = '-';
7+
ans.push(s.join(''));
8+
s[i] = s[i + 1] = '+';
9+
}
10+
}
11+
return ans;
12+
}

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