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Merged
merged 9 commits into from
Feb 4, 2024
Merged

feat: add solutions to lc problems: No.3029 #2316

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Merge branch 'doocs:main' into main
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103 changes: 101 additions & 2 deletions solution/3000-3099/3029.Minimum Time to Revert Word to Initial State I/README.md
View file
Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -70,19 +70,118 @@
<!-- tabs:start -->

```python
class Solution:
def minimumTimeToInitialState(self, word: str, k: int) -> int:
n = len(word)
for i in range(1, 10001):
re = i * k
if re >= n:
return i
if word[re:] == word[:n - re]:
return i
return 0

```

```java

class Solution {
public int minimumTimeToInitialState(String word, int k) {
int n = word.length();
for (int i = 1; i <= 10000; i++) {
int re = i * k;
if (re >= n) {
return i;
}
String str = word.substring(re);
boolean flag = true;
for (int j = 0; j < str.length(); j++) {
if (str.charAt(j) != word.charAt(j)) {
flag = false;
break;
}
}
if (flag) {
return i;
}
}
return 0;
}
}
```

```cpp

class Solution {
public:
int minimumTimeToInitialState(string word, int k) {
int n = word.length();
for (int i = 1; i <= 10000; i++) {
int re = i * k;
if (re >= n) {
return i;
}
string str = word.substr(re);
bool flag = true;
for (int j = 0; j < str.length(); j++) {
if (str[j] != word[j]) {
flag = false;
break;
}
}
if (flag) {
return i;
}
}
return 0;
}
};
```

```go
func minimumTimeToInitialState(word string, k int) int {
n := len(word)
for i := 1; i <= 10000; i++ {
re := i * k
if re >= n {
return i
}
str := word[re:]
flag := true
for j := 0; j < len(str); j++ {
if str[j] != word[j] {
flag = false
break
}
}
if flag {
return i
}
}
return 0
}
```

```ts
function minimumTimeToInitialState(word: string, k: number): number {
const n = word.length;
for (let i = 1; i <= 10000; i++) {
const re = i * k;
if (re >= n) {
return i;
}
const str = word.substring(re);
let flag = true;
for (let j = 0; j < str.length; j++) {
if (str[j] !== word[j]) {
flag = false;
break;
}
}
if (flag) {
return i;
}
}
return 0;
}
```

<!-- tabs:end -->
Expand Down
103 changes: 101 additions & 2 deletions ...tion/3000-3099/3029.Minimum Time to Revert Word to Initial State I/README_EN.md
View file
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Original file line number Diff line number Diff line change
Expand Up @@ -63,19 +63,118 @@ It can be shown that 4 seconds is the minimum time greater than zero required fo
<!-- tabs:start -->

```python
class Solution:
def minimumTimeToInitialState(self, word: str, k: int) -> int:
n = len(word)
for i in range(1, 10001):
re = i * k
if re >= n:
return i
if word[re:] == word[:n - re]:
return i
return 0

```

```java

class Solution {
public int minimumTimeToInitialState(String word, int k) {
int n = word.length();
for (int i = 1; i <= 10000; i++) {
int re = i * k;
if (re >= n) {
return i;
}
String str = word.substring(re);
boolean flag = true;
for (int j = 0; j < str.length(); j++) {
if (str.charAt(j) != word.charAt(j)) {
flag = false;
break;
}
}
if (flag) {
return i;
}
}
return 0;
}
}
```

```cpp

class Solution {
public:
int minimumTimeToInitialState(string word, int k) {
int n = word.length();
for (int i = 1; i <= 10000; i++) {
int re = i * k;
if (re >= n) {
return i;
}
string str = word.substr(re);
bool flag = true;
for (int j = 0; j < str.length(); j++) {
if (str[j] != word[j]) {
flag = false;
break;
}
}
if (flag) {
return i;
}
}
return 0;
}
};
```

```go
func minimumTimeToInitialState(word string, k int) int {
n := len(word)
for i := 1; i <= 10000; i++ {
re := i * k
if re >= n {
return i
}
str := word[re:]
flag := true
for j := 0; j < len(str); j++ {
if str[j] != word[j] {
flag = false
break
}
}
if flag {
return i
}
}
return 0
}
```

```ts
function minimumTimeToInitialState(word: string, k: number): number {
const n = word.length;
for (let i = 1; i <= 10000; i++) {
const re = i * k;
if (re >= n) {
return i;
}
const str = word.substring(re);
let flag = true;
for (let j = 0; j < str.length; j++) {
if (str[j] !== word[j]) {
flag = false;
break;
}
}
if (flag) {
return i;
}
}
return 0;
}
```

<!-- tabs:end -->
Expand Down
24 changes: 24 additions & 0 deletions solution/3000-3099/3029.Minimum Time to Revert Word to Initial State I/Solution.cpp
View file
Open in desktop
Original file line number Diff line number Diff line change
@@ -0,0 +1,24 @@
class Solution {
public:
int minimumTimeToInitialState(string word, int k) {
int n = word.length();
for (int i = 1; i <= 10000; i++) {
int re = i * k;
if (re >= n) {
return i;
}
string str = word.substr(re);
bool flag = true;
for (int j = 0; j < str.length(); j++) {
if (str[j] != word[j]) {
flag = false;
break;
}
}
if (flag) {
return i;
}
}
return 0;
}
};
21 changes: 21 additions & 0 deletions solution/3000-3099/3029.Minimum Time to Revert Word to Initial State I/Solution.go
View file
Open in desktop
Original file line number Diff line number Diff line change
@@ -0,0 +1,21 @@
func minimumTimeToInitialState(word string, k int) int {
n := len(word)
for i := 1; i <= 10000; i++ {
re := i * k
if re >= n {
return i
}
str := word[re:]
flag := true
for j := 0; j < len(str); j++ {
if str[j] != word[j] {
flag = false
break
}
}
if flag {
return i
}
}
return 0
}
23 changes: 23 additions & 0 deletions solution/3000-3099/3029.Minimum Time to Revert Word to Initial State I/Solution.java
View file
Open in desktop
Original file line number Diff line number Diff line change
@@ -0,0 +1,23 @@
class Solution {
public int minimumTimeToInitialState(String word, int k) {
int n = word.length();
for (int i = 1; i <= 10000; i++) {
int re = i * k;
if (re >= n) {
return i;
}
String str = word.substring(re);
boolean flag = true;
for (int j = 0; j < str.length(); j++) {
if (str.charAt(j) != word.charAt(j)) {
flag = false;
break;
}
}
if (flag) {
return i;
}
}
return 0;
}
}
10 changes: 10 additions & 0 deletions solution/3000-3099/3029.Minimum Time to Revert Word to Initial State I/Solution.py
View file
Open in desktop
Original file line number Diff line number Diff line change
@@ -0,0 +1,10 @@
class Solution:
def minimumTimeToInitialState(self, word: str, k: int) -> int:
n = len(word)
for i in range(1, 10001):
re = i * k
if re >= n:
return i
if word[re:] == word[: n - re]:
return i
return 0
21 changes: 21 additions & 0 deletions solution/3000-3099/3029.Minimum Time to Revert Word to Initial State I/Solution.ts
View file
Open in desktop
Original file line number Diff line number Diff line change
@@ -0,0 +1,21 @@
function minimumTimeToInitialState(word: string, k: number): number {
const n = word.length;
for (let i = 1; i <= 10000; i++) {
const re = i * k;
if (re >= n) {
return i;
}
const str = word.substring(re);
let flag = true;
for (let j = 0; j < str.length; j++) {
if (str[j] !== word[j]) {
flag = false;
break;
}
}
if (flag) {
return i;
}
}
return 0;
}