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feat: add solutions to lc problems: No.0358,0767
* No.0358.Rearrange String k Distance Apart * No.0767.Reorganize String
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‎solution/0300-0399/0358.Rearrange String k Distance Apart/README.md

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<!-- 这里可写通用的实现逻辑 -->
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**方法一：贪心 + 哈希表 + 优先队列（大根堆）**
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先用哈希表 `cnt` 统计每个字母出现的次数，然后构建一个大根堆 `pq`，其中每个元素是一个 `(v, c)` 的元组，其中 `c` 是字母，`v` 是字母出现的次数。
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重排字符串时，我们每次从堆顶弹出一个元素 `(v, c)`，将 `c` 添加到结果字符串中，并将 `(v-1, c)` 放入队列 `q` 中。当队列 `q` 的长度达到 $k$ 及以上时，弹出队首元素，若此时 `v` 大于 0，则将队首元素放入堆中。循环，直至堆为空。
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最后判断结果字符串的长度，若与 `s` 长度相等，则返回结果字符串，否则返回空串。
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时间复杂度 $O(n\log n)$，其中 $n$ 是字符串 `s` 的长度。
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相似题目：[767. 重构字符串](/solution/0700-0799/0767.Reorganize%20String/README.md)
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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 rearrangeString(self, s: str, k: int) -> str:
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h = [(-v, c) for c, v in Counter(s).items()]
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heapify(h)
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q = deque()
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ans = []
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while h:
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v, c = heappop(h)
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v *= -1
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ans.append(c)
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q.append((v - 1, c))
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if len(q) >= k:
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w, c = q.popleft()
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if w:
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heappush(h, (-w, c))
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return "" if len(ans) != len(s) else "".join(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 String rearrangeString(String s, int k) {
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int n = s.length();
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int[] cnt = new int[26];
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for (char c : s.toCharArray()) {
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++cnt[c - 'a'];
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}
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PriorityQueue<int[]> pq = new PriorityQueue<>((a, b) -> b[0] - a[0]);
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for (int i = 0; i < 26; ++i) {
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if (cnt[i] > 0) {
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pq.offer(new int[] {cnt[i], i});
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}
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}
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Deque<int[]> q = new ArrayDeque<>();
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StringBuilder ans = new StringBuilder();
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while (!pq.isEmpty()) {
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var p = pq.poll();
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int v = p[0], c = p[1];
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ans.append((char) ('a' + c));
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q.offer(new int[] {v - 1, c});
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if (q.size() >= k) {
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p = q.pollFirst();
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if (p[0] > 0) {
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pq.offer(p);
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}
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}
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}
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return ans.length() == n ? ans.toString() : "";
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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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string rearrangeString(string s, int k) {
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unordered_map<char, int> cnt;
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for (char c : s) ++cnt[c];
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priority_queue<pair<int, char>> pq;
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for (auto& [c, v] : cnt) pq.push({v, c});
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queue<pair<int, char>> q;
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string ans;
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while (!pq.empty()) {
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auto [v, c] = pq.top();
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pq.pop();
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ans += c;
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q.push({v - 1, c});
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if (q.size() >= k) {
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auto p = q.front();
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q.pop();
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if (p.first) {
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pq.push(p);
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}
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}
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}
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return ans.size() == s.size() ? 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 rearrangeString(s string, k int) string {
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cnt := map[byte]int{}
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for i := range s {
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cnt[s[i]]++
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}
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pq := hp{}
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for c, v := range cnt {
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heap.Push(&pq, pair{v, c})
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}
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ans := []byte{}
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q := []pair{}
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for len(pq) > 0 {
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p := heap.Pop(&pq).(pair)
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v, c := p.v, p.c
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ans = append(ans, c)
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q = append(q, pair{v - 1, c})
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if len(q) >= k {
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p = q[0]
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q = q[1:]
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if p.v > 0 {
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heap.Push(&pq, p)
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}
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}
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}
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if len(ans) == len(s) {
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return string(ans)
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}
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return ""
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}
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type pair struct {
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v int
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c byte
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}
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type hp []pair
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func (h hp) Len() int { return len(h) }
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func (h hp) Less(i, j int) bool {
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a, b := h[i], h[j]
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return a.v > b.v
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}
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func (h hp) Swap(i, j int) { h[i], h[j] = h[j], h[i] }
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func (h *hp) Push(v interface{}) { *h = append(*h, v.(pair)) }
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func (h *hp) Pop() interface{} { a := *h; v := a[len(a)-1]; *h = a[:len(a)-1]; return v }
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```
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### **...**

‎solution/0300-0399/0358.Rearrange String k Distance Apart/README_EN.md

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### **Python3**
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```python
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class Solution:
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def rearrangeString(self, s: str, k: int) -> str:
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h = [(-v, c) for c, v in Counter(s).items()]
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heapify(h)
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q = deque()
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ans = []
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while h:
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v, c = heappop(h)
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v *= -1
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ans.append(c)
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q.append((v - 1, c))
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if len(q) >= k:
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w, c = q.popleft()
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if w:
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heappush(h, (-w, c))
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return "" if len(ans) != len(s) else "".join(ans)
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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 String rearrangeString(String s, int k) {
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int n = s.length();
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int[] cnt = new int[26];
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for (char c : s.toCharArray()) {
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++cnt[c - 'a'];
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}
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PriorityQueue<int[]> pq = new PriorityQueue<>((a, b) -> b[0] - a[0]);
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for (int i = 0; i < 26; ++i) {
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if (cnt[i] > 0) {
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pq.offer(new int[] {cnt[i], i});
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}
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}
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Deque<int[]> q = new ArrayDeque<>();
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StringBuilder ans = new StringBuilder();
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while (!pq.isEmpty()) {
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var p = pq.poll();
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int v = p[0], c = p[1];
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ans.append((char) ('a' + c));
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q.offer(new int[] {v - 1, c});
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if (q.size() >= k) {
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p = q.pollFirst();
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if (p[0] > 0) {
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pq.offer(p);
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}
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}
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}
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return ans.length() == n ? ans.toString() : "";
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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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string rearrangeString(string s, int k) {
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unordered_map<char, int> cnt;
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for (char c : s) ++cnt[c];
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priority_queue<pair<int, char>> pq;
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for (auto& [c, v] : cnt) pq.push({v, c});
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queue<pair<int, char>> q;
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string ans;
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while (!pq.empty()) {
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auto [v, c] = pq.top();
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pq.pop();
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ans += c;
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q.push({v - 1, c});
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if (q.size() >= k) {
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auto p = q.front();
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q.pop();
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if (p.first) {
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pq.push(p);
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}
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}
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}
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return ans.size() == s.size() ? 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 rearrangeString(s string, k int) string {
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cnt := map[byte]int{}
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for i := range s {
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cnt[s[i]]++
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}
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pq := hp{}
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for c, v := range cnt {
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heap.Push(&pq, pair{v, c})
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}
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ans := []byte{}
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q := []pair{}
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for len(pq) > 0 {
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p := heap.Pop(&pq).(pair)
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v, c := p.v, p.c
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ans = append(ans, c)
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q = append(q, pair{v - 1, c})
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if len(q) >= k {
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p = q[0]
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q = q[1:]
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if p.v > 0 {
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heap.Push(&pq, p)
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}
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}
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}
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if len(ans) == len(s) {
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return string(ans)
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}
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return ""
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}
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type pair struct {
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v int
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c byte
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}
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type hp []pair
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func (h hp) Len() int { return len(h) }
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func (h hp) Less(i, j int) bool {
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a, b := h[i], h[j]
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return a.v > b.v
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}
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func (h hp) Swap(i, j int) { h[i], h[j] = h[j], h[i] }
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func (h *hp) Push(v interface{}) { *h = append(*h, v.(pair)) }
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func (h *hp) Pop() interface{} { a := *h; v := a[len(a)-1]; *h = a[:len(a)-1]; return v }
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```
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### **...**

‎solution/0300-0399/0358.Rearrange String k Distance Apart/Solution.cpp

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class Solution {
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public:
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string rearrangeString(string s, int k) {
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unordered_map<char, int> cnt;
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for (char c : s) ++cnt[c];
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priority_queue<pair<int, char>> pq;
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for (auto& [c, v] : cnt) pq.push({v, c});
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queue<pair<int, char>> q;
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string ans;
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while (!pq.empty()) {
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auto [v, c] = pq.top();
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pq.pop();
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ans += c;
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q.push({v - 1, c});
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if (q.size() >= k) {
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auto p = q.front();
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q.pop();
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if (p.first) {
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pq.push(p);
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}
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}
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}
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return ans.size() == s.size() ? ans : "";
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}
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};

‎solution/0300-0399/0358.Rearrange String k Distance Apart/Solution.go

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func rearrangeString(s string, k int) string {
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cnt := map[byte]int{}
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for i := range s {
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cnt[s[i]]++
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}
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pq := hp{}
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for c, v := range cnt {
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heap.Push(&pq, pair{v, c})
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}
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ans := []byte{}
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q := []pair{}
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for len(pq) > 0 {
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p := heap.Pop(&pq).(pair)
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v, c := p.v, p.c
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ans = append(ans, c)
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q = append(q, pair{v - 1, c})
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if len(q) >= k {
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p = q[0]
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q = q[1:]
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if p.v > 0 {
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heap.Push(&pq, p)
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}
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}
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}
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if len(ans) == len(s) {
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return string(ans)
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}
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return ""
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}
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type pair struct {
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v int
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c byte
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}
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type hp []pair
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func (h hp) Len() int { return len(h) }
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func (h hp) Less(i, j int) bool {
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a, b := h[i], h[j]
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return a.v > b.v
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}
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func (h hp) Swap(i, j int) { h[i], h[j] = h[j], h[i] }
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func (h *hp) Push(v interface{}) { *h = append(*h, v.(pair)) }
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func (h *hp) Pop() interface{} { a := *h; v := a[len(a)-1]; *h = a[:len(a)-1]; return v }

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