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feat: add solutions to lc problems: No.2937,2938 #1985

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yanglbme merged 1 commit into main from dev
Nov 19, 2023
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69 changes: 66 additions & 3 deletions solution/2900-2999/2937.Make Three Strings Equal/README.md
View file Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -42,34 +42,97 @@

<!-- 这里可写通用的实现逻辑 -->

**方法一:枚举**

根据题目描述,我们知道,如果删除字符后的三个字符串相等,那么它们存在一个长度大于 1ドル$ 的公共前缀。因此,我们可以枚举公共前缀的位置 $i,ドル如果当前下标 $i$ 对应的三个字符不完全相等,那么公共前缀长度为 $i,ドル此时,我们判断 $i$ 是否为 0ドル,ドル若是,返回 $-1,ドル否则返回 $s - 3 \times i,ドル其中 $s$ 为三个字符串的长度和。

时间复杂度 $O(n),ドル其中 $n$ 为三个字符串的最小长度。空间复杂度 $O(1)$。

<!-- tabs:start -->

### **Python3**

<!-- 这里可写当前语言的特殊实现逻辑 -->

```python

class Solution:
def findMinimumOperations(self, s1: str, s2: str, s3: str) -> int:
s = len(s1) + len(s2) + len(s3)
n = min(len(s1), len(s2), len(s3))
for i in range(n):
if not s1[i] == s2[i] == s3[i]:
return -1 if i == 0 else s - 3 * i
return s - 3 * n
```

### **Java**

<!-- 这里可写当前语言的特殊实现逻辑 -->

```java

class Solution {
public int findMinimumOperations(String s1, String s2, String s3) {
int s = s1.length() + s2.length() + s3.length();
int n = Math.min(Math.min(s1.length(), s2.length()), s3.length());
for (int i = 0; i < n; ++i) {
if (!(s1.charAt(i) == s2.charAt(i) && s2.charAt(i) == s3.charAt(i))) {
return i == 0 ? -1 : s - 3 * i;
}
}
return s - 3 * n;
}
}
```

### **C++**

```cpp

class Solution {
public:
int findMinimumOperations(string s1, string s2, string s3) {
int s = s1.size() + s2.size() + s3.size();
int n = min({s1.size(), s2.size(), s3.size()});
for (int i = 0; i < n; ++i) {
if (!(s1[i] == s2[i] && s2[i] == s3[i])) {
return i == 0 ? -1 : s - 3 * i;
}
}
return s - 3 * n;
}
};
```

### **Go**

```go
func findMinimumOperations(s1 string, s2 string, s3 string) int {
s := len(s1) + len(s2) + len(s3)
n := min(len(s1), len(s2), len(s3))
for i := range s1[:n] {
if !(s1[i] == s2[i] && s2[i] == s3[i]) {
if i == 0 {
return -1
}
return s - 3*i
}
}
return s - 3*n
}
```

### **TypeScript**

```ts
function findMinimumOperations(s1: string, s2: string, s3: string): number {
const s = s1.length + s2.length + s3.length;
const n = Math.min(s1.length, s2.length, s3.length);
for (let i = 0; i < n; ++i) {
if (!(s1[i] === s2[i] && s2[i] === s3[i])) {
return i === 0 ? -1 : s - 3 * i;
}
}
return s - 3 * n;
}
```

### **...**
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69 changes: 66 additions & 3 deletions solution/2900-2999/2937.Make Three Strings Equal/README_EN.md
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Original file line number Diff line number Diff line change
Expand Up @@ -37,30 +37,93 @@ It can be shown that there is no way to make them equal with less than two opera

## Solutions

**Solution 1: Enumeration**

According to the problem description, we know that if the three strings are equal after deleting characters, then they have a common prefix of length greater than 1ドル$. Therefore, we can enumerate the position $i$ of the common prefix. If the three characters at the current index $i$ are not all equal, then the length of the common prefix is $i$. At this point, we check if $i$ is 0ドル$. If it is, return $-1$. Otherwise, return $s - 3 \times i,ドル where $s$ is the sum of the lengths of the three strings.

The time complexity is $O(n),ドル where $n$ is the minimum length of the three strings. The space complexity is $O(1)$.

<!-- tabs:start -->

### **Python3**

```python

class Solution:
def findMinimumOperations(self, s1: str, s2: str, s3: str) -> int:
s = len(s1) + len(s2) + len(s3)
n = min(len(s1), len(s2), len(s3))
for i in range(n):
if not s1[i] == s2[i] == s3[i]:
return -1 if i == 0 else s - 3 * i
return s - 3 * n
```

### **Java**

```java

class Solution {
public int findMinimumOperations(String s1, String s2, String s3) {
int s = s1.length() + s2.length() + s3.length();
int n = Math.min(Math.min(s1.length(), s2.length()), s3.length());
for (int i = 0; i < n; ++i) {
if (!(s1.charAt(i) == s2.charAt(i) && s2.charAt(i) == s3.charAt(i))) {
return i == 0 ? -1 : s - 3 * i;
}
}
return s - 3 * n;
}
}
```

### **C++**

```cpp

class Solution {
public:
int findMinimumOperations(string s1, string s2, string s3) {
int s = s1.size() + s2.size() + s3.size();
int n = min({s1.size(), s2.size(), s3.size()});
for (int i = 0; i < n; ++i) {
if (!(s1[i] == s2[i] && s2[i] == s3[i])) {
return i == 0 ? -1 : s - 3 * i;
}
}
return s - 3 * n;
}
};
```

### **Go**

```go
func findMinimumOperations(s1 string, s2 string, s3 string) int {
s := len(s1) + len(s2) + len(s3)
n := min(len(s1), len(s2), len(s3))
for i := range s1[:n] {
if !(s1[i] == s2[i] && s2[i] == s3[i]) {
if i == 0 {
return -1
}
return s - 3*i
}
}
return s - 3*n
}
```

### **TypeScript**

```ts
function findMinimumOperations(s1: string, s2: string, s3: string): number {
const s = s1.length + s2.length + s3.length;
const n = Math.min(s1.length, s2.length, s3.length);
for (let i = 0; i < n; ++i) {
if (!(s1[i] === s2[i] && s2[i] === s3[i])) {
return i === 0 ? -1 : s - 3 * i;
}
}
return s - 3 * n;
}
```

### **...**
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13 changes: 13 additions & 0 deletions solution/2900-2999/2937.Make Three Strings Equal/Solution.cpp
View file Open in desktop
Original file line number Diff line number Diff line change
@@ -0,0 +1,13 @@
class Solution {
public:
int findMinimumOperations(string s1, string s2, string s3) {
int s = s1.size() + s2.size() + s3.size();
int n = min({s1.size(), s2.size(), s3.size()});
for (int i = 0; i < n; ++i) {
if (!(s1[i] == s2[i] && s2[i] == s3[i])) {
return i == 0 ? -1 : s - 3 * i;
}
}
return s - 3 * n;
}
};
13 changes: 13 additions & 0 deletions solution/2900-2999/2937.Make Three Strings Equal/Solution.go
View file Open in desktop
Original file line number Diff line number Diff line change
@@ -0,0 +1,13 @@
func findMinimumOperations(s1 string, s2 string, s3 string) int {
s := len(s1) + len(s2) + len(s3)
n := min(len(s1), len(s2), len(s3))
for i := range s1[:n] {
if !(s1[i] == s2[i] && s2[i] == s3[i]) {
if i == 0 {
return -1
}
return s - 3*i
}
}
return s - 3*n
}
12 changes: 12 additions & 0 deletions solution/2900-2999/2937.Make Three Strings Equal/Solution.java
View file Open in desktop
Original file line number Diff line number Diff line change
@@ -0,0 +1,12 @@
class Solution {
public int findMinimumOperations(String s1, String s2, String s3) {
int s = s1.length() + s2.length() + s3.length();
int n = Math.min(Math.min(s1.length(), s2.length()), s3.length());
for (int i = 0; i < n; ++i) {
if (!(s1.charAt(i) == s2.charAt(i) && s2.charAt(i) == s3.charAt(i))) {
return i == 0 ? -1 : s - 3 * i;
}
}
return s - 3 * n;
}
}
8 changes: 8 additions & 0 deletions solution/2900-2999/2937.Make Three Strings Equal/Solution.py
View file Open in desktop
Original file line number Diff line number Diff line change
@@ -0,0 +1,8 @@
class Solution:
def findMinimumOperations(self, s1: str, s2: str, s3: str) -> int:
s = len(s1) + len(s2) + len(s3)
n = min(len(s1), len(s2), len(s3))
for i in range(n):
if not s1[i] == s2[i] == s3[i]:
return -1 if i == 0 else s - 3 * i
return s - 3 * n
10 changes: 10 additions & 0 deletions solution/2900-2999/2937.Make Three Strings Equal/Solution.ts
View file Open in desktop
Original file line number Diff line number Diff line change
@@ -0,0 +1,10 @@
function findMinimumOperations(s1: string, s2: string, s3: string): number {
const s = s1.length + s2.length + s3.length;
const n = Math.min(s1.length, s2.length, s3.length);
for (let i = 0; i < n; ++i) {
if (!(s1[i] === s2[i] && s2[i] === s3[i])) {
return i === 0 ? -1 : s - 3 * i;
}
}
return s - 3 * n;
}
75 changes: 72 additions & 3 deletions solution/2900-2999/2938.Separate Black and White Balls/README.md
View file Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -57,34 +57,103 @@

<!-- 这里可写通用的实现逻辑 -->

**方法一:计数模拟**

我们考虑将所有的 1ドル$ 移到最右边,用一个变量 $cnt$ 记录当前已经移动到最右边的 1ドル$ 的个数,用一个变量 $ans$ 记录移动的次数。

我们从右往左遍历字符串,如果当前位置是 1ドル,ドル那么我们将 $cnt$ 加一,同时将 $n - i - cnt$ 加到 $ans$ 中,其中 $n$ 是字符串的长度。最后返回 $ans$ 即可。

时间复杂度 $O(n),ドル其中 $n$ 是字符串的长度。空间复杂度 $O(1)$。

<!-- tabs:start -->

### **Python3**

<!-- 这里可写当前语言的特殊实现逻辑 -->

```python

class Solution:
def minimumSteps(self, s: str) -> int:
n = len(s)
ans = cnt = 0
for i in range(n - 1, -1, -1):
if s[i] == '1':
cnt += 1
ans += n - i - cnt
return ans
```

### **Java**

<!-- 这里可写当前语言的特殊实现逻辑 -->

```java

class Solution {
public long minimumSteps(String s) {
long ans = 0;
int cnt = 0;
int n = s.length();
for (int i = n - 1; i >= 0; --i) {
if (s.charAt(i) == '1') {
++cnt;
ans += n - i - cnt;
}
}
return ans;
}
}
```

### **C++**

```cpp

class Solution {
public:
long long minimumSteps(string s) {
long long ans = 0;
int cnt = 0;
int n = s.size();
for (int i = n - 1; i >= 0; --i) {
if (s[i] == '1') {
++cnt;
ans += n - i - cnt;
}
}
return ans;
}
};
```

### **Go**

```go
func minimumSteps(s string) (ans int64) {
n := len(s)
cnt := 0
for i := n - 1; i >= 0; i-- {
if s[i] == '1' {
cnt++
ans += int64(n - i - cnt)
}
}
return
}
```

### **TypeScript**

```ts
function minimumSteps(s: string): number {
const n = s.length;
let [ans, cnt] = [0, 0];
for (let i = n - 1; ~i; --i) {
if (s[i] === '1') {
++cnt;
ans += n - i - cnt;
}
}
return ans;
}
```

### **...**
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