LeetCode 358. Rearrange String k Distance Apart

1: First thought: brute force. Get all the permutations and check permutations one by one.

/*  Given a non-empty str 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 "".  For example  str = "aabbcc", k = 3, return "abcabc"  str = "aaabc", k = 3, return ""  str = "aaadbbcc", k = 2, return "abacabcd", another possible "abcabcda"*/#include <string>#include <iostream>#include <unordered_map>using namespace std;// first thought, to get the permutation of the string and check validation.// once we found one, we can say that there is one.bool checkValid(string str, int k) {  unordered_map<char, int> charDistance;  for(int i = 0; i < str.size(); ++i) {    if(charDistance.find(str[i]) == charDistance.end()) {      charDistance[str[i]] = i;    } else {      if(i - charDistance[str[i]] >= k) charDistance[str[i]] = i;      else return false;    }  }  return true;}void rearrangeString(string str, int k, string path, string& target, bool& found) {  if(found) return;  if(str.size() == 0) {    if(checkValid(path, k)) {      found = true;      target = path;      return;    }  }  for(int i = 0; i < str.size(); ++i) {    string remaining = str.substr(0, i) + str.substr(i + 1);    rearrangeString(remaining, k, path + str[i], target, found);  }}string rearrangeString(string str, int k) {  string target = "";  string path = "";  bool found = false;  rearrangeString(str, k, path, target, found);  return target;}int main(void) {  cout << rearrangeString("aaabc", 3) << endl;}

Second Method:

// second version, using heap and greedy.string rearrangeString(string str, int k) {  int len = str.size();  unordered_map<char, int> charToCount;  for(int i = 0; i < str.size(); ++i) {    charToCount[str[i]]++;  }  priority_queue< pair<int, char> > maxHeap;  for(auto it = charToCount.begin(); it != charToCount.end(); ++it) {    maxHeap.push({it->second, it->first});  }  string res;  while(!maxHeap.empty()) {    vector< pair<int, char> > v;    int cnt = min(k, len); // get the minimum length.    for(int i = 0; i < cnt; ++i) {      if(maxHeap.empty()) return ""; // if left different chars numbers are smaller than the required, return false.      auto q = maxHeap.top();      maxHeap.pop();      res.push_back(q.second);      if(--q.first > 0) v.push_back(t);      len--;    }    for(auto a : v) maxHeap.push(a);  }  return res;}