LC 358. 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.

Example 3:

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

Runtime: 8 ms, faster than 99.59% of C++ online submissions for Rearrange String k Distance Apart.

和之前的一道题很类似

bool cmp(pair<char, int> a, pair<char, int> b) {
    if (a.second != b.second) return a.second < b.second;
    return a.first < b.first;
}

class Solution {
public:
    string rearrangeString(string s, int k) {
        vector<pair<char, int>> map(26, pair<char,int>('a',0));
        for (int i = 0; i < s.size(); i++) {
            int idx = s[i] - 'a';
            if (map[idx].second == 0) {
                map[idx].first = s[i];
                map[idx].second = 1;
            }
            else {
                map[idx].second++;
            }
        }
        sort(map.begin(), map.end(), cmp);
        //for(auto m : map) cout << m.first << m.second << endl;
        int idx = map.size() - 1;
        int maxrep = map[idx].second;
        string ret = "";
        while (idx >= 0 && map[idx].second == maxrep) {
            string tmpstr = string(1,map[idx].first);
            ret += tmpstr;
            idx--;
        }
        //cout << ret << endl;
        vector<string> retvec(map.back().second - 1, ret);
        int cnt = 0;
        while (idx >= 0 && map[idx].second != 0) {
            int tmp = idx;
            string tmpstr =string(1, map[tmp].first);
            while (map[tmp].second != 0) {
                retvec[cnt] += tmpstr;
                map[tmp].second--;
                cnt++;
                if(cnt >= retvec.size()) cnt = 0;
            }
            idx--;
        }
        for (auto s : retvec) {
            if (s.size() < k) return "";
        }
        string finalret = "";
        for (auto s : retvec) finalret += s;
        finalret += ret;
        return finalret;
    }
};