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.