Rearrange String k Distance Apart

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:

Input: s = "aabbcc", k = 3
Output: "abcabc" 
Explanation: The same letters are at least distance 3 from each other.

Example 2:

Input: s = "aaabc", k = 3
Output: "" 
Explanation: It is not possible to rearrange the string.

思路：這題其實是 Reorganize String 的升級版；這裏需要跳過K個了。這裏用個queue來存儲wait的node；

class Solution {
    public class Node {
        public char c;
        public int fre;
        public Node(char c, int fre) {
            this.c = c;
            this.fre = fre;
        }
    }
    
    public class NodeComparator implements Comparator<Node> {
        @Override
        public int compare(Node a, Node b) {
            return b.fre - a.fre;
        }
    }
    
    public String rearrangeString(String s, int k) {
        HashMap<Character, Integer> countMap = new HashMap<>();
        for(int i = 0; i < s.length(); i++) {
            char c = s.charAt(i);
            countMap.put(c, countMap.getOrDefault(c, 0) + 1);
        }
        
        PriorityQueue<Node> pq = new PriorityQueue<Node>(new NodeComparator());
        for(Character c : countMap.keySet()) {
            pq.offer(new Node(c, countMap.get(c)));
        }
        
        StringBuilder sb = new StringBuilder();
        Queue<Node> waitqueue = new LinkedList<Node>();
        while(!pq.isEmpty()) {
            Node node = pq.poll();
            sb.append(node.c);
            node.fre--;
            countMap.put(node.c, countMap.get(node.c) - 1);
            waitqueue.offer(node);
            
            // 這裏不滿足就繼續，滿足了k，就要加入之前的node了；
            // 比如abc了，那麼已經有3個了，那麼接下來就應該把a加入隊列了
            if(waitqueue.size() < k) {
                continue;
            }
            
            Node waitnode = waitqueue.poll();
            if(waitnode.fre > 0) {
                pq.offer(waitnode);
            }
        }
        return sb.length() == s.length() ? sb.toString() : "";
    }
}