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feat: add solutions to lc problem: No.1585
No.1585.Check If String Is Transformable With Substring Sort Operations
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‎solution/1500-1599/1585.Check If String Is Transformable With Substring Sort Operations/README.md

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<!-- 这里可写通用的实现逻辑 -->
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**方法一：冒泡排序**
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题目实际上等价于判断：将字符串 $s$ 中任意长度为 $2$ 的子字符串采用冒泡排序交换，是否能得到 $t$。
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因此我们用一个长度为 $10$ 的数组 $pos$ 记录字符串 $s$ 中每个字符数字的下标，其中 $pos[i]$ 表示数字 $i$ 出现的下标列表，按从小到大排序。
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接下来，我们遍历字符串 $t$，对于 $t$ 中的每个字符 $t[i]$，我们转为数字 $x$，我们判断 $pos[x]$ 是否为空，若是，说明字符串 $s$ 中不存在 $t$ 中的数字，直接返回 `false`。否则，若要将 $pos[x]$ 的第一个位置下标的字符交换到下标 $i$ 的位置，需要满足小于 $x$ 的所有数字的下标均不小于 $pos[x]$ 的第一个位置下标，若不满足，返回 `false`。否则，我们将 $pos[x]$ 的第一个位置下标弹出，然后继续遍历字符串 $t$。
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遍历结束，返回 `true`。
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时间复杂度 $O(n \times C)$，空间复杂度 $O(n)$。其中 $n$ 为字符串 $s$ 的长度，而 $C$ 是数字集的大小，本题中 $C=10$。
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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 isTransformable(self, s: str, t: str) -> bool:
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pos = defaultdict(deque)
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for i, c in enumerate(s):
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pos[int(c)].append(i)
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for c in t:
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x = int(c)
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if not pos[x] or any(pos[i] and pos[i][0] < pos[x][0] for i in range(x)):
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return False
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pos[x].popleft()
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return True
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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 boolean isTransformable(String s, String t) {
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Deque<Integer>[] pos = new Deque[10];
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Arrays.setAll(pos, k -> new ArrayDeque<>());
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for (int i = 0; i < s.length(); ++i) {
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pos[s.charAt(i) - '0'].offer(i);
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}
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for (int i = 0; i < t.length(); ++i) {
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int x = t.charAt(i) - '0';
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if (pos[x].isEmpty()) {
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return false;
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}
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for (int j = 0; j < x; ++j) {
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if (!pos[j].isEmpty() && pos[j].peek() < pos[x].peek()) {
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return false;
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}
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}
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pos[x].poll();
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}
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return true;
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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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bool isTransformable(string s, string t) {
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queue<int> pos[10];
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for (int i = 0; i < s.size(); ++i) {
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pos[s[i] - '0'].push(i);
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}
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for (char& c : t) {
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int x = c - '0';
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if (pos[x].empty()) {
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return false;
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}
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for (int j = 0; j < x; ++j) {
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if (!pos[j].empty() && pos[j].front() < pos[x].front()) {
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return false;
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}
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}
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pos[x].pop();
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}
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return true;
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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 isTransformable(s string, t string) bool {
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pos := [10][]int{}
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for i, c := range s {
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pos[c-'0'] = append(pos[c-'0'], i)
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}
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for _, c := range t {
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x := int(c - '0')
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if len(pos[x]) == 0 {
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return false
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}
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for j := 0; j < x; j++ {
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if len(pos[j]) > 0 && pos[j][0] < pos[x][0] {
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return false
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}
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}
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pos[x] = pos[x][1:]
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}
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return true
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}
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```
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### **...**

‎solution/1500-1599/1585.Check If String Is Transformable With Substring Sort Operations/README_EN.md

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### **Python3**
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```python
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class Solution:
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def isTransformable(self, s: str, t: str) -> bool:
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pos = defaultdict(deque)
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for i, c in enumerate(s):
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pos[int(c)].append(i)
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for c in t:
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x = int(c)
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if not pos[x] or any(pos[i] and pos[i][0] < pos[x][0] for i in range(x)):
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return False
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pos[x].popleft()
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return True
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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 boolean isTransformable(String s, String t) {
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Deque<Integer>[] pos = new Deque[10];
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Arrays.setAll(pos, k -> new ArrayDeque<>());
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for (int i = 0; i < s.length(); ++i) {
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pos[s.charAt(i) - '0'].offer(i);
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}
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for (int i = 0; i < t.length(); ++i) {
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int x = t.charAt(i) - '0';
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if (pos[x].isEmpty()) {
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return false;
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}
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for (int j = 0; j < x; ++j) {
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if (!pos[j].isEmpty() && pos[j].peek() < pos[x].peek()) {
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return false;
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}
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}
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pos[x].poll();
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}
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return true;
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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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bool isTransformable(string s, string t) {
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queue<int> pos[10];
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for (int i = 0; i < s.size(); ++i) {
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pos[s[i] - '0'].push(i);
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}
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for (char& c : t) {
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int x = c - '0';
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if (pos[x].empty()) {
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return false;
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}
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for (int j = 0; j < x; ++j) {
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if (!pos[j].empty() && pos[j].front() < pos[x].front()) {
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return false;
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}
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}
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pos[x].pop();
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}
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return true;
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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 isTransformable(s string, t string) bool {
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pos := [10][]int{}
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for i, c := range s {
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pos[c-'0'] = append(pos[c-'0'], i)
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}
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for _, c := range t {
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x := int(c - '0')
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if len(pos[x]) == 0 {
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return false
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}
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for j := 0; j < x; j++ {
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if len(pos[j]) > 0 && pos[j][0] < pos[x][0] {
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return false
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}
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}
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pos[x] = pos[x][1:]
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}
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return true
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}
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```
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### **...**

‎solution/1500-1599/1585.Check If String Is Transformable With Substring Sort Operations/Solution.cpp

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class Solution {
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public:
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bool isTransformable(string s, string t) {
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queue<int> pos[10];
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for (int i = 0; i < s.size(); ++i) {
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pos[s[i] - '0'].push(i);
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}
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for (char& c : t) {
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int x = c - '0';
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if (pos[x].empty()) {
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return false;
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}
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for (int j = 0; j < x; ++j) {
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if (!pos[j].empty() && pos[j].front() < pos[x].front()) {
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return false;
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}
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}
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pos[x].pop();
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}
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return true;
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}
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};

‎solution/1500-1599/1585.Check If String Is Transformable With Substring Sort Operations/Solution.go

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func isTransformable(s string, t string) bool {
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pos := [10][]int{}
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for i, c := range s {
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pos[c-'0'] = append(pos[c-'0'], i)
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}
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for _, c := range t {
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x := int(c - '0')
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if len(pos[x]) == 0 {
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return false
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}
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for j := 0; j < x; j++ {
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if len(pos[j]) > 0 && pos[j][0] < pos[x][0] {
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return false
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}
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}
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pos[x] = pos[x][1:]
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}
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return true
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}

‎solution/1500-1599/1585.Check If String Is Transformable With Substring Sort Operations/Solution.java

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class Solution {
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public boolean isTransformable(String s, String t) {
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Deque<Integer>[] pos = new Deque[10];
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Arrays.setAll(pos, k -> new ArrayDeque<>());
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for (int i = 0; i < s.length(); ++i) {
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pos[s.charAt(i) - '0'].offer(i);
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}
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for (int i = 0; i < t.length(); ++i) {
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int x = t.charAt(i) - '0';
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if (pos[x].isEmpty()) {
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return false;
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}
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for (int j = 0; j < x; ++j) {
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if (!pos[j].isEmpty() && pos[j].peek() < pos[x].peek()) {
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return false;
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}
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}
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pos[x].poll();
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}
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return true;
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}
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}

‎solution/1500-1599/1585.Check If String Is Transformable With Substring Sort Operations/Solution.py

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class Solution:
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def isTransformable(self, s: str, t: str) -> bool:
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pos = defaultdict(deque)
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for i, c in enumerate(s):
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pos[int(c)].append(i)
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for c in t:
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x = int(c)
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if not pos[x] or any(pos[i] and pos[i][0] < pos[x][0] for i in range(x)):
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return False
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pos[x].popleft()
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return True

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