Leetcode 358. Rearrange String k Distance Apart

Problem:

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.

Example 3:

Input: s = "aaadbbcc", k = 2
Output: "abacabcd"
Explanation: The same letters are at least distance 2 from each other.

 

Solution:

　　百度笔试原题，当时没做出来，现在来看这道题，依然做不出来。。。这道题用了哈希表，堆和贪心算法，感觉难度非常高，现在来看看这道题是如何解答的。首先用一个哈希表记录每个字符出现的次数，然后将哈系表中的键值对推入优先级队列中，现在的关键点在于如何使相同字符直接的距离为k。可以想象，出现次数多的字符应该优先考虑放置，所以这个优先级队列是以字符出现次数作为比较函数的，队列顶是出现次数最多的字符。所以我们每次从队列中取出k个键值对，对其字符进行组合，然后将其次数减一，如果不为0就重新加入优先级队列，由于队列中每个键值对必然不存在重复字符，所以这样做必然不会违背题意。接下来讲讲细节，如果可用字符串长度小于k，那么就全部添加字符串就行了，如果说优先级队列提前为空，说明剩下的字符不能够继续组合字符串，所以返回空。以第二个例子为例，第一次循环后res为“abc”，优先级队列中还有（2，‘a'），此时还需要放置两个元素，但放下一个a之后优先级队列中已经没有元素可以放置了，说明无法得到目标字符串，所以返回空。

Code:

 

class Solution {
public:
    string rearrangeString(string str, int k) {
        if (k == 0) return str;
        string res;
        int len = (int)str.size();
        unordered_map<char, int> m;
        priority_queue<pair<int, char>> q;
        for (auto a : str) ++m[a];
        for (auto it = m.begin(); it != m.end(); ++it) {
            q.push({it->second, it->first});
        }
        while (!q.empty()) {
            vector<pair<int, int>> v;
            int cnt = min(k, len);
            for (int i = 0; i < cnt; ++i) {
                if (q.empty()) return "";
                auto t = q.top(); q.pop();
                res.push_back(t.second);
                if (--t.first > 0) v.push_back(t);
                --len;
            }
            for (auto a : v) q.push(a);
        }
        return res;
    }
};