Rearrange String k Distance Apart in C++

Suppose we have a non-empty string s and an integer k; we have to rearrange the string such that the same characters are at least distance k from each other. Given strings are in lowercase letters. If there is no way to rearrange the strings, then we will an empty string.

So, if the input is like s = "aabbcc", k = 3, then the output will be "abcabc" this is because same letters are at least distance 3 from each other.

To solve this, we will follow these steps −

• ret := an empty string

• Define one map m

• n := size of s

• for initialize i := 0, when i < n, update (increase i by 1), do −

  • (increase m[s[i]] by 1)

• Define one priority queue pq

• for each key-value pair it in m, do −

  • temp := a pair with key and value of it

  • insert temp into pq

  • (increase it by 1)

• Define one deque dq

• while (not pq is empty), do −

  • curr := top of pq

  • delete element from pq

  • ret := ret + curr.first

  • (decrease curr.second by 1)

  • insert curr at the end of dq

  • if size of dq >= k, then −

    • curr := first element of dq

    • delete front element from dq

    • if curr.second > 0, then −

      • insert curr into pq

• while (not dq is empty and first element of dq is same as 0), do −

  • delete front element from dq

• return (if dq is empty, then ret, otherwise blank string)

Example  

Let us see the following implementation to get better understanding −

 Live Demo

#include <bits/stdc++.h>
using namespace std;
struct Comparator {
   bool operator()(pair<char, int> a, pair<char, int> b) {
      return !(a.second > b.second);
   }
};
class Solution {
public:
   string rearrangeString(string s, int k) {
      string ret = "";
      unordered_map<char, int> m;
      int n = s.size();
      for (int i = 0; i < n; i++) {
         m[s[i]]++;
      }
      unordered_map<char, int>::iterator it = m.begin();
      priority_queue<pair<char, int<, vector<pair<char, int>>,
      Comparator> pq;
      while (it != m.end()) {
         pair<char, int> temp = {it->first, it->second};
         pq.push(temp);
         it++;
      }
      deque<pair<char, int>> dq;
      while (!pq.empty()) {
         pair<char, int> curr = pq.top();
         pq.pop();
         ret += curr.first;
         curr.second--;
         dq.push_back(curr);
         // cout << ret << " " << dq.size() << endl;
         if (dq.size() >= k) {
            curr = dq.front();
            dq.pop_front();
            if (curr.second > 0)
            pq.push(curr);
         }
      }
      while (!dq.empty() && dq.front().second == 0)
         dq.pop_front();
      return dq.empty() ? ret : "";
   }
};
main() {
   Solution ob;
   cout << (ob.rearrangeString("aabbcc", 3));
}

Input

"aabbcc", 3

Output

bacbac