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feat: add solutions to lc problems: No.2937,2938 (#1985)
* No.2937.Make Three Strings Equal * No.2938.Separate Black and White Balls
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‎solution/2900-2999/2937.Make Three Strings Equal/README.md

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Original file line numberDiff line numberDiff line change
@@ -42,34 +42,97 @@
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<!-- 这里可写通用的实现逻辑 -->
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**方法一：枚举**
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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 $1$ 的公共前缀。因此，我们可以枚举公共前缀的位置 $i$，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 $i$，此时，我们判断 $i$ 是否为 $0$，若是，返回 $-1$，否则返回 $s - 3 \times i$，其中 $s$ 为三个字符串的长度和。
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时间复杂度 $O(n)$，其中 $n$ 为三个字符串的最小长度。空间复杂度 $O(1)$。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def findMinimumOperations(self, s1: str, s2: str, s3: str) -> int:
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s = len(s1) + len(s2) + len(s3)
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n = min(len(s1), len(s2), len(s3))
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for i in range(n):
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if not s1[i] == s2[i] == s3[i]:
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return -1 if i == 0 else s - 3 * i
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return s - 3 * n
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public int findMinimumOperations(String s1, String s2, String s3) {
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int s = s1.length() + s2.length() + s3.length();
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int n = Math.min(Math.min(s1.length(), s2.length()), s3.length());
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for (int i = 0; i < n; ++i) {
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if (!(s1.charAt(i) == s2.charAt(i) && s2.charAt(i) == s3.charAt(i))) {
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return i == 0 ? -1 : s - 3 * i;
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}
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}
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return s - 3 * n;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int findMinimumOperations(string s1, string s2, string s3) {
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int s = s1.size() + s2.size() + s3.size();
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int n = min({s1.size(), s2.size(), s3.size()});
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for (int i = 0; i < n; ++i) {
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if (!(s1[i] == s2[i] && s2[i] == s3[i])) {
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return i == 0 ? -1 : s - 3 * i;
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}
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}
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return s - 3 * n;
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}
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};
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```
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### **Go**
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```go
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func findMinimumOperations(s1 string, s2 string, s3 string) int {
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s := len(s1) + len(s2) + len(s3)
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n := min(len(s1), len(s2), len(s3))
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for i := range s1[:n] {
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if !(s1[i] == s2[i] && s2[i] == s3[i]) {
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if i == 0 {
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return -1
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}
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return s - 3*i
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}
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}
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return s - 3*n
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}
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```
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### **TypeScript**
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```ts
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function findMinimumOperations(s1: string, s2: string, s3: string): number {
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const s = s1.length + s2.length + s3.length;
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const n = Math.min(s1.length, s2.length, s3.length);
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for (let i = 0; i < n; ++i) {
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if (!(s1[i] === s2[i] && s2[i] === s3[i])) {
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return i === 0 ? -1 : s - 3 * i;
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}
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}
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return s - 3 * n;
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}
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```
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### **...**

‎solution/2900-2999/2937.Make Three Strings Equal/README_EN.md

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@@ -37,30 +37,93 @@ It can be shown that there is no way to make them equal with less than two opera
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## Solutions
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**Solution 1: Enumeration**
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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 $1$. 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 $0$. If it is, return $-1$. Otherwise, return $s - 3 \times i$, where $s$ is the sum of the lengths of the three strings.
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The time complexity is $O(n)$, where $n$ is the minimum length of the three strings. The space complexity is $O(1)$.
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<!-- tabs:start -->
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### **Python3**
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```python
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class Solution:
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def findMinimumOperations(self, s1: str, s2: str, s3: str) -> int:
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s = len(s1) + len(s2) + len(s3)
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n = min(len(s1), len(s2), len(s3))
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for i in range(n):
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if not s1[i] == s2[i] == s3[i]:
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return -1 if i == 0 else s - 3 * i
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return s - 3 * n
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```
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### **Java**
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```java
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class Solution {
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public int findMinimumOperations(String s1, String s2, String s3) {
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int s = s1.length() + s2.length() + s3.length();
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int n = Math.min(Math.min(s1.length(), s2.length()), s3.length());
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for (int i = 0; i < n; ++i) {
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if (!(s1.charAt(i) == s2.charAt(i) && s2.charAt(i) == s3.charAt(i))) {
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return i == 0 ? -1 : s - 3 * i;
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}
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}
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return s - 3 * n;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int findMinimumOperations(string s1, string s2, string s3) {
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int s = s1.size() + s2.size() + s3.size();
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int n = min({s1.size(), s2.size(), s3.size()});
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for (int i = 0; i < n; ++i) {
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if (!(s1[i] == s2[i] && s2[i] == s3[i])) {
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return i == 0 ? -1 : s - 3 * i;
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}
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}
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return s - 3 * n;
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}
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};
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```
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### **Go**
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```go
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func findMinimumOperations(s1 string, s2 string, s3 string) int {
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s := len(s1) + len(s2) + len(s3)
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n := min(len(s1), len(s2), len(s3))
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for i := range s1[:n] {
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if !(s1[i] == s2[i] && s2[i] == s3[i]) {
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if i == 0 {
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return -1
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}
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return s - 3*i
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}
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}
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return s - 3*n
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}
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```
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### **TypeScript**
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```ts
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function findMinimumOperations(s1: string, s2: string, s3: string): number {
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const s = s1.length + s2.length + s3.length;
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const n = Math.min(s1.length, s2.length, s3.length);
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for (let i = 0; i < n; ++i) {
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if (!(s1[i] === s2[i] && s2[i] === s3[i])) {
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return i === 0 ? -1 : s - 3 * i;
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}
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}
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return s - 3 * n;
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}
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```
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### **...**

‎solution/2900-2999/2937.Make Three Strings Equal/Solution.cpp

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

‎solution/2900-2999/2937.Make Three Strings Equal/Solution.go

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

‎solution/2900-2999/2937.Make Three Strings Equal/Solution.java

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

‎solution/2900-2999/2937.Make Three Strings Equal/Solution.py

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

‎solution/2900-2999/2937.Make Three Strings Equal/Solution.ts

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

‎solution/2900-2999/2938.Separate Black and White Balls/README.md

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<!-- 这里可写通用的实现逻辑 -->
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**方法一：计数模拟**
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我们考虑将所有的 $1$ 移到最右边，用一个变量 $cnt$ 记录当前已经移动到最右边的 $1$ 的个数，用一个变量 $ans$ 记录移动的次数。
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我们从右往左遍历字符串，如果当前位置是 $1$，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。
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时间复杂度 $O(n)$，其中 $n$ 是字符串的长度。空间复杂度 $O(1)$。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def minimumSteps(self, s: str) -> int:
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n = len(s)
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ans = cnt = 0
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for i in range(n - 1, -1, -1):
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if s[i] == '1':
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cnt += 1
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ans += n - i - cnt
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return ans
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public long minimumSteps(String s) {
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long ans = 0;
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int cnt = 0;
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int n = s.length();
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for (int i = n - 1; i >= 0; --i) {
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if (s.charAt(i) == '1') {
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++cnt;
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ans += n - i - cnt;
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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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### **C++**
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```cpp
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class Solution {
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public:
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long long minimumSteps(string s) {
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long long ans = 0;
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int cnt = 0;
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int n = s.size();
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for (int i = n - 1; i >= 0; --i) {
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if (s[i] == '1') {
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++cnt;
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ans += n - i - cnt;
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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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### **Go**
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```go
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func minimumSteps(s string) (ans int64) {
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n := len(s)
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cnt := 0
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for i := n - 1; i >= 0; i-- {
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if s[i] == '1' {
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cnt++
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ans += int64(n - i - cnt)
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}
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}
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return
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}
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```
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### **TypeScript**
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```ts
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function minimumSteps(s: string): number {
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const n = s.length;
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let [ans, cnt] = [0, 0];
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for (let i = n - 1; ~i; --i) {
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if (s[i] === '1') {
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++cnt;
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ans += n - i - cnt;
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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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