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package com.fishercoder.solutions;
import java.util.HashMap;
import java.util.LinkedList;
import java.util.Map;
import java.util.PriorityQueue;
import java.util.Queue;
/**
* 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:
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.
*/
public class _358 {
public static class Solution1 {
public String rearrangeString(String s, int k) {
Map<Character, Integer> count = new HashMap<>();
for (char c : s.toCharArray()) {
count.put(c, count.getOrDefault(c, 0) + 1);
}
PriorityQueue<Map.Entry<Character, Integer>> heap =
new PriorityQueue<>((a, b) -> b.getValue() - a.getValue());
heap.addAll(count.entrySet());
Queue<Map.Entry<Character, Integer>> waitQueue = new LinkedList<>();
StringBuilder stringBuilder = new StringBuilder();
while (!heap.isEmpty()) {
Map.Entry<Character, Integer> entry = heap.poll();
stringBuilder.append(entry.getKey());
entry.setValue(entry.getValue() - 1);
waitQueue.offer(entry);
if (waitQueue.size() < k) {
continue; //there's only k-1 chars in the waitHeap, not full yet
}
Map.Entry<Character, Integer> front = waitQueue.poll();
if (front.getValue() > 0) {
heap.offer(front);
}
}
return stringBuilder.length() == s.length() ? stringBuilder.toString() : "";
}
}
}
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