# 767.Reorganize-String

## 767. Reorganize String

## 题目地址

<https://leetcode.com/problems/reorganize-string/>

## 题目描述

```
Given a string S, check if the letters can be rearranged so that two characters that are adjacent to each other are not the same.

If possible, output any possible result. If not possible, return the empty string.

Example 1:
Input: S = "aab"
Output: "aba"

Example 2:
Input: S = "aaab"
Output: ""

Note:
S will consist of lowercase letters and have length in range [1, 500]
```

## 代码

### Approach 1: Sort by Count

**Complexity Analysis**

1 3 5 7 9...

0 2 4 6 8...

* Time Complexity: &#x4F;*(A(\_N*+logA)), where N *is the length of \_S*, and A is the size of the alphabet. In Java, our implementation is O\_(\_N+AlogA). If A is fixed, this complexity is O(N).
* Space Complexity: O(N). In Java, our implementation is O\_(\_N+A).

```java
class Solution {
  public String reorganizeString(String S) {
    int N = S.length();
    int[] counts = new int[26];
    for (char c : S.toCharArray()) {
      counts[c - 'a'] += 100;
    }
    for (int i = 0; i < 26; i++) {
      counts[i] += i;
    }

    Arrays.sort(counts);

    char[] ans = new char[N];
    int t = 1;
    for (int code : counts) {
      int ct = code / 100;
      char ch = (char) ('a' + (code % 100));
      if (ct > (N + 1) / 2) return "";
      for (int i = 0; i < ct; i++) {
        if (t >= N) t = 0;
        ans[t] = ch;
        t += 2;
      }
    }

    return String.valueOf(ans);
  }
}
```

### Approach #2 Greedy with Heap

**Complexity Analysis**

* Time Complexity: &#x4F;*(\_N\_logA)), where N* is the length of S, and A is the size of the alphabet. If A is fixed, this complexity is O(N).
* Space Complexity: O(A). If A is fixed, this complexity is O(1).

```java
class Solution {
  public String reorganizeString(String S) {
    int N = S.length();
    int[] count = new int[26];
    for (char c : S.toCharArray()) {
      count[c - 'a']++;
    }
    PriorityQueue<MultiChar> pq = new PriorityQueue<MultiChar((a, b) -> a.count == b.count ? a.letter - b.letter : b.count - a.count);

    for (int i = 0; i < 26; i++) {
      if (count[i] > (N + 1) / 2) return "";

      pq.add(new MultiChar(count[i], (char) ('a' + i)));
    }

    StringBuilder ans = new StringBuilder();
    while (pq.size() >= 2) {
      MultiChar mc1 = pq.poll();
      MultiChar mc2 = pq.poll();

      ans.append(mc1.letter);
      ans.append(mc2.letter);
      if (--mc1.count > 0) pq.add(mc1);
      if (--mc2.count > 0) pq.add(mc2);
    }

    if (pq.size() > 0) ans.append(pq.poll().letter);
    return ans.toString();
  }
}

class MultiChar {
  int countP
  char letter;
  MultiChar(int ct, char ch) {
    count = ct;
    letter = ch;
  }
}
```