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feat: add solutions to lc problem: No.3114 #2592

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Apr 15, 2024
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feat: add solutions to lc problem: No.3114 #2592

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118 changes: 118 additions & 0 deletions .../3100-3199/3114.Latest Time You Can Obtain After Replacing Characters/README.md
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Expand Up @@ -157,4 +157,122 @@ function findLatestTime(s: string): string {

<!-- tabs:end -->

### 方法二：逐个判断

我们可以逐个判断 $s$ 的每一位字符，如果是 "?" 的话，我们就根据前后的字符来确定这一位字符的值。具体地，我们有以下规则：

- 如果 $s[0]$ 是 "?"，那么 $s[0]$ 的值应该是 "1" 或 "0"，具体取决于 $s[1]$ 的值。如果 $s[1]$ 是 "?" 或者 $s[1]$ 小于 "2"，那么 $s[0]$ 的值应该是 "1"，否则 $s[0]$ 的值应该是 "0"。
- 如果 $s[1]$ 是 "?"，那么 $s[1]$ 的值应该是 "1" 或 "9"，具体取决于 $s[0]$ 的值。如果 $s[0]$ 是 "1"，那么 $s[1]$ 的值应该是 "1"，否则 $s[1]$ 的值应该是 "9"。
- 如果 $s[3]$ 是 "?"，那么 $s[3]$ 的值应该是 "5"。
- 如果 $s[4]$ 是 "?"，那么 $s[4]$ 的值应该是 "9"。

时间复杂度 $O(1)$，空间复杂度 $O(1)$。

<!-- tabs:start -->

```python
class Solution:
def findLatestTime(self, s: str) -> str:
s = list(s)
if s[0] == "?":
s[0] = "1" if s[1] == "?" or s[1] < "2" else "0"
if s[1] == "?":
s[1] = "1" if s[0] == "1" else "9"
if s[3] == "?":
s[3] = "5"
if s[4] == "?":
s[4] = "9"
return "".join(s)
```

```java
class Solution {
public String findLatestTime(String s) {
char[] cs = s.toCharArray();
if (cs[0] == '?') {
cs[0] = cs[1] == '?' || cs[1] < '2' ? '1' : '0';
}
if (cs[1] == '?') {
cs[1] = cs[0] == '1' ? '1' : '9';
}
if (cs[3] == '?') {
cs[3] = '5';
}
if (cs[4] == '?') {
cs[4] = '9';
}
return new String(cs);
}
}
```

```cpp
class Solution {
public:
string findLatestTime(string s) {
if (s[0] == '?') {
s[0] = s[1] == '?' || s[1] < '2' ? '1' : '0';
}
if (s[1] == '?') {
s[1] = s[0] == '1' ? '1' : '9';
}
if (s[3] == '?') {
s[3] = '5';
}
if (s[4] == '?') {
s[4] = '9';
}
return s;
}
};
```

```go
func findLatestTime(s string) string {
cs := []byte(s)
if cs[0] == '?' {
if cs[1] == '?' || cs[1] < '2' {
cs[0] = '1'
} else {
cs[0] = '0'
}
}
if cs[1] == '?' {
if cs[0] == '1' {
cs[1] = '1'
} else {
cs[1] = '9'
}
}
if cs[3] == '?' {
cs[3] = '5'
}
if cs[4] == '?' {
cs[4] = '9'
}
return string(cs)
}
```

```ts
function findLatestTime(s: string): string {
const cs = s.split('');
if (cs[0] === '?') {
cs[0] = cs[1] === '?' || cs[1] < '2' ? '1' : '0';
}
if (cs[1] === '?') {
cs[1] = cs[0] === '1' ? '1' : '9';
}
if (cs[3] === '?') {
cs[3] = '5';
}
if (cs[4] === '?') {
cs[4] = '9';
}
return cs.join('');
}
```

<!-- tabs:end -->

<!-- end -->
118 changes: 118 additions & 0 deletions ...00-3199/3114.Latest Time You Can Obtain After Replacing Characters/README_EN.md
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Expand Up @@ -153,4 +153,122 @@ function findLatestTime(s: string): string {

<!-- tabs:end -->

### Solution 2: Judge Each Digit

We can judge each digit of $s$ one by one. If it is "?", we determine the value of this digit based on the characters before and after it. Specifically, we have the following rules:

- If $s[0]$ is "?", then the value of $s[0]$ should be "1" or "0", depending on the value of $s[1]$. If $s[1]$ is "?" or $s[1]$ is less than "2", then the value of $s[0]$ should be "1", otherwise the value of $s[0]$ should be "0".
- If $s[1]$ is "?", then the value of $s[1]$ should be "1" or "9", depending on the value of $s[0]$. If $s[0]$ is "1", then the value of $s[1]$ should be "1", otherwise the value of $s[1]$ should be "9".
- If $s[3]$ is "?", then the value of $s[3]$ should be "5".
- If $s[4]$ is "?", then the value of $s[4]$ should be "9".

The time complexity is $O(1)$, and the space complexity is $O(1)$.

<!-- tabs:start -->

```python
class Solution:
def findLatestTime(self, s: str) -> str:
s = list(s)
if s[0] == "?":
s[0] = "1" if s[1] == "?" or s[1] < "2" else "0"
if s[1] == "?":
s[1] = "1" if s[0] == "1" else "9"
if s[3] == "?":
s[3] = "5"
if s[4] == "?":
s[4] = "9"
return "".join(s)
```

```java
class Solution {
public String findLatestTime(String s) {
char[] cs = s.toCharArray();
if (cs[0] == '?') {
cs[0] = cs[1] == '?' || cs[1] < '2' ? '1' : '0';
}
if (cs[1] == '?') {
cs[1] = cs[0] == '1' ? '1' : '9';
}
if (cs[3] == '?') {
cs[3] = '5';
}
if (cs[4] == '?') {
cs[4] = '9';
}
return new String(cs);
}
}
```

```cpp
class Solution {
public:
string findLatestTime(string s) {
if (s[0] == '?') {
s[0] = s[1] == '?' || s[1] < '2' ? '1' : '0';
}
if (s[1] == '?') {
s[1] = s[0] == '1' ? '1' : '9';
}
if (s[3] == '?') {
s[3] = '5';
}
if (s[4] == '?') {
s[4] = '9';
}
return s;
}
};
```

```go
func findLatestTime(s string) string {
cs := []byte(s)
if cs[0] == '?' {
if cs[1] == '?' || cs[1] < '2' {
cs[0] = '1'
} else {
cs[0] = '0'
}
}
if cs[1] == '?' {
if cs[0] == '1' {
cs[1] = '1'
} else {
cs[1] = '9'
}
}
if cs[3] == '?' {
cs[3] = '5'
}
if cs[4] == '?' {
cs[4] = '9'
}
return string(cs)
}
```

```ts
function findLatestTime(s: string): string {
const cs = s.split('');
if (cs[0] === '?') {
cs[0] = cs[1] === '?' || cs[1] < '2' ? '1' : '0';
}
if (cs[1] === '?') {
cs[1] = cs[0] === '1' ? '1' : '9';
}
if (cs[3] === '?') {
cs[3] = '5';
}
if (cs[4] === '?') {
cs[4] = '9';
}
return cs.join('');
}
```

<!-- tabs:end -->

<!-- end -->
18 changes: 18 additions & 0 deletions solution/3100-3199/3114.Latest Time You Can Obtain After Replacing Characters/Solution2.cpp
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@@ -0,0 +1,18 @@
class Solution {
public:
string findLatestTime(string s) {
if (s[0] == '?') {
s[0] = s[1] == '?' || s[1] < '2' ? '1' : '0';
}
if (s[1] == '?') {
s[1] = s[0] == '1' ? '1' : '9';
}
if (s[3] == '?') {
s[3] = '5';
}
if (s[4] == '?') {
s[4] = '9';
}
return s;
}
};
24 changes: 24 additions & 0 deletions solution/3100-3199/3114.Latest Time You Can Obtain After Replacing Characters/Solution2.go
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Original file line number Diff line number Diff line change
@@ -0,0 +1,24 @@
func findLatestTime(s string) string {
cs := []byte(s)
if cs[0] == '?' {
if cs[1] == '?' || cs[1] < '2' {
cs[0] = '1'
} else {
cs[0] = '0'
}
}
if cs[1] == '?' {
if cs[0] == '1' {
cs[1] = '1'
} else {
cs[1] = '9'
}
}
if cs[3] == '?' {
cs[3] = '5'
}
if cs[4] == '?' {
cs[4] = '9'
}
return string(cs)
}
18 changes: 18 additions & 0 deletions solution/3100-3199/3114.Latest Time You Can Obtain After Replacing Characters/Solution2.java
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Original file line number Diff line number Diff line change
@@ -0,0 +1,18 @@
class Solution {
public String findLatestTime(String s) {
char[] cs = s.toCharArray();
if (cs[0] == '?') {
cs[0] = cs[1] == '?' || cs[1] < '2' ? '1' : '0';
}
if (cs[1] == '?') {
cs[1] = cs[0] == '1' ? '1' : '9';
}
if (cs[3] == '?') {
cs[3] = '5';
}
if (cs[4] == '?') {
cs[4] = '9';
}
return new String(cs);
}
}
12 changes: 12 additions & 0 deletions solution/3100-3199/3114.Latest Time You Can Obtain After Replacing Characters/Solution2.py
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Original file line number Diff line number Diff line change
@@ -0,0 +1,12 @@
class Solution:
def findLatestTime(self, s: str) -> str:
s = list(s)
if s[0] == "?":
s[0] = "1" if s[1] == "?" or s[1] < "2" else "0"
if s[1] == "?":
s[1] = "1" if s[0] == "1" else "9"
if s[3] == "?":
s[3] = "5"
if s[4] == "?":
s[4] = "9"
return "".join(s)
16 changes: 16 additions & 0 deletions solution/3100-3199/3114.Latest Time You Can Obtain After Replacing Characters/Solution2.ts
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Original file line number Diff line number Diff line change
@@ -0,0 +1,16 @@
function findLatestTime(s: string): string {
const cs = s.split('');
if (cs[0] === '?') {
cs[0] = cs[1] === '?' || cs[1] < '2' ? '1' : '0';
}
if (cs[1] === '?') {
cs[1] = cs[0] === '1' ? '1' : '9';
}
if (cs[3] === '?') {
cs[3] = '5';
}
if (cs[4] === '?') {
cs[4] = '9';
}
return cs.join('');
}
2 changes: 1 addition & 1 deletion solution/3100-3199/3117.Minimum Sum of Values by Dividing Array/README.md
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Expand Up @@ -86,7 +86,7 @@

### 方法一：记忆化搜索

我们设计一个函数 $dfs(i, j, a)$，表示从第 $i$ 个元素开始，且当前已经划分了 $j$ 个子数组，且当前待划分的子数组的按位与结果为 $a$ 的情况下，所能得到的可能的最小子数组值之和。那么答案就是 $dfs(0, 0, -1)$。
我们设计一个函数 $dfs(i, j, a)$，表示从第 $i$ 个元素开始，当前已经划分了 $j$ 个子数组，且当前待划分的子数组的按位与结果为 $a$ 的情况下，所能得到的可能的最小子数组值之和。那么答案就是 $dfs(0, 0, -1)$。

函数 $dfs(i, j, a)$ 的执行过程如下：

Expand Down
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