Rearrange String k Distance Apart

Given a non-empty string 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"".

Example 1:

str = "aabbcc", k = 3

Result: "abcabc"

The same letters are at least distance 3 from each other.

Example 2:

str = "aaabc", k = 3 

Answer: ""

It is not possible to rearrange the string.

Example 3:

str = "aaadbbcc", k = 2

Answer: "abacabcd"

Another possible answer is: "abcabcda"

The same letters are at least distance 2 from each other.

Note: 2016-11-27
Why do we need to put sb.length() < s.length() in if condition?
Because if we reach the end of the string and i is less than 0, queue is guaranteed to be empty, and the code will falsely return empty string.

Using a greedy approach. Every time we add the character with the highest frequency to the string. If there is no k distinct characters left in the queue, we return "". For all characters that have remaining frequency larger than 0, we add them back to queue.

public String rerrange(String s, int k) {
        Map<character, integer> tempMap = new HashMap<>();
        for (char c : s.toCharArray()) {
            if (!tempMap.containsKey(c)) {
                tempMap.put(c, 1);
            } else {
                tempMap.put(c, tempMap.get(c) + 1);
            }
        }

        PriorityQueue<frequency> queue = new PriorityQueue<>();
        for (Map.Entry<character, integer> entry : tempMap.entrySet()) {
            queue.add(new Frequency(entry.getValue(), entry.getKey()));
        }

        StringBuilder sb = new StringBuilder();
        while (!queue.isEmpty()) {
            Queue<frequency> tempQueue = new LinkedList<>();
            for (int i = 0; i < k && sb.length() < s.length(); i++) {
                if (queue.isEmpty()) {
                    return "";
                }
                Frequency curr = queue.poll();
                sb.append(curr.c);
                curr.freq--;
                if (curr.freq > 0) {
                    tempQueue.add(curr);
                }
            }
            while (!tempQueue.isEmpty()) {
                queue.add(tempQueue.poll());
            }
        }
        return sb.toString();
    }

    private class Frequency implements Comparable<frequency> {
        int freq;
        char c;

        public Frequency(int freq, char c) {
            this.c = c;
            this.freq = freq;
        }

        @Override public int compareTo(Frequency o) {
            return o.freq - this.freq;
        }
    }