LeetCode 358: Rearrange String k Distance Apart Solution

Master LeetCode problem 358 (Rearrange String k Distance Apart), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

358. Rearrange String k Distance Apart

Hard

Hash Table String Greedy Sorting Heap (Priority Queue)Counting

Problem Explanation

Explanation:

The problem asks us to rearrange a string such that no two same characters are K distance apart. If it is not possible to rearrange the string, return an empty string.

To solve this problem, we can use a greedy approach. We will start by constructing a frequency map of characters in the input string. Then, we will use a priority queue to keep track of the characters that need to be placed in the output string based on their frequencies. At each step, we will pop the top element from the priority queue, append it to the output string, and decrease its frequency. We will also keep track of the characters that are waiting to be placed back into the priority queue after K distance. Solution:

Solution Code

import java.util.*;

class Solution {
    public String rearrangeString(String s, int k) {
        if (k <= 1) return s;

        Map<Character, Integer> freqMap = new HashMap<>();
        for (char c : s.toCharArray()) {
            freqMap.put(c, freqMap.getOrDefault(c, 0) + 1);
        }

        PriorityQueue<Map.Entry<Character, Integer>> maxHeap = new PriorityQueue<>(
            (a, b) -> b.getValue() - a.getValue()
        );
        maxHeap.addAll(freqMap.entrySet());

        Queue<Map.Entry<Character, Integer>> waitQueue = new LinkedList<>();
        StringBuilder rearranged = new StringBuilder();

        while (!maxHeap.isEmpty()) {
            Map.Entry<Character, Integer> curr = maxHeap.poll();
            rearranged.append(curr.getKey());
            curr.setValue(curr.getValue() - 1);
            waitQueue.offer(curr);

            if (waitQueue.size() < k) continue;

            Map.Entry<Character, Integer> front = waitQueue.poll();
            if (front.getValue() > 0) {
                maxHeap.offer(front);
            }
        }

        return rearranged.length() == s.length() ? rearranged.toString() : "";
    }
}

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