LeetCode 358. Rearrange String k Distance Apart

原题链接在这里：https://leetcode.com/problems/rearrange-string-k-distance-apart/description/

题目：

Given a non-empty string s and an integer k, rearrange the string such that the same characters are at least distance k from each other.

All input strings are given in lowercase letters. If it is not possible to rearrange the string, return an empty string "".

Example 1:

s = "aabbcc", k = 3

Result: "abcabc"

The same letters are at least distance 3 from each other.

Example 2:

s = "aaabc", k = 3 

Answer: ""

It is not possible to rearrange the string.

Example 3:

s = "aaadbbcc", k = 2

Answer: "abacabcd"

Another possible answer is: "abcabcda"

The same letters are at least distance 2 from each other.

题解：

Greedy问题. 感觉上是应该先排剩余frequency 最多的Character.

用maxHeap来维护剩余的Character, 根据剩余的count.

那么如何保持断开的距离大于k呢, 用queue来存放已经加过的Character, 只有当queue的size等于k时, 才允许把头上的Character放回到maxHeap中.

Time Complexity: O(nlogn). n = s.length(). 都加入进maxHeap用时O(nlogn).

Space: O(n).

AC Java:

 class Solution {
     public String rearrangeString(String s, int k) {
         if(s == null || s.length() == 0){
             return s;
         }

         HashMap<Character, Integer> hm = new HashMap<Character, Integer>();
         for(int i = 0; i<s.length(); i++){
             hm.put(s.charAt(i), hm.getOrDefault(s.charAt(i), 0)+1);
         }

         PriorityQueue<Map.Entry<Character, Integer>> maxHeap = new PriorityQueue<Map.Entry<Character, Integer>>(
             (a, b) -> b.getValue() - a.getValue()
         );
         maxHeap.addAll(hm.entrySet());

         LinkedList<Map.Entry<Character, Integer>> que = new LinkedList<Map.Entry<Character, Integer>>();
         StringBuilder sb = new StringBuilder();
         while(!maxHeap.isEmpty()){
             Map.Entry<Character, Integer> cur = maxHeap.poll();
             sb.append(cur.getKey());
             cur.setValue(cur.getValue()-1);
             que.add(cur);

             if(que.size() < k){
                 continue;
             }

             Map.Entry<Character, Integer> head = que.poll();
             if(head.getValue() > 0){
                 maxHeap.add(head);
             }
         }
         return sb.length() == s.length() ? sb.toString() : "";
     }
 }

时间上可以优化.

利用两个int array, count计数剩余frequency, validPo代表这个字符能出现的最早位置.

找到最大frequency的合法字符, 更新其对应的frequency 和 再次出现的位置.

Time Complexity: O(s.length()). findMaxValidCount走了遍长度为i26的count array.

Space: O(s.length()). StringBuilder size.

AC Java:

 class Solution {
     public String rearrangeString(String s, int k) {
         if(s == null || s.length() == 0){
             return s;
         }

         int [] count = new int[26];
         int [] validPo = new int[26];
         for(int i = 0; i<s.length(); i++){
             count[s.charAt(i)-'a']++;
         }

         StringBuilder sb = new StringBuilder();
         for(int i = 0; i<s.length(); i++){
             int po = findMaxValidCount(count, validPo, i);
             if(po == -1){
                 return "";
             }

             sb.append((char)('a'+po));
             count[po]--;
             validPo[po] = i+k;
         }
         return sb.toString();
     }

     private int findMaxValidCount(int [] count, int [] validPo, int ind){
         int po = -1;
         int max = Integer.MIN_VALUE;
         for(int i = 0; i<count.length; i++){
             if(count[i]>0 && count[i]>max && ind>=validPo[i]){
                 max = count[i];
                 po = i;
             }
         }

         return po;
     }
 }

类似Task Scheduler, Reorganize String.