Solution Explanation:

This problem requires rearranging a given string s such that any two occurrences of the same character are at least k distance apart. The solution uses a greedy approach with a priority queue (heap) to manage character frequencies.

Approach:

1. Character Frequency Count: First, we count the frequency of each character in the string s. This is done using a Counter (Python), HashMap (Java), unordered_map (C++), or a map (Go).

2. Priority Queue: We create a priority queue (PriorityQueue in Java, priority_queue in C++, hp in Go, or heapq in Python) to store character frequencies. The priority queue is ordered by character frequencies in descending order, ensuring that the most frequent characters are processed first.

3. Greedy Rearrangement: We iterate while the priority queue is not empty. In each iteration:

   • We extract the character with the highest frequency from the priority queue.
   • We append this character to the result string ans.
   • We decrement its frequency by 1 and put it into a queue q to track characters that are being "cooled down" before they can be used again. This queue is used to enforce the k distance constraint.
   • If the size of the q queue reaches k, we check if the character at the beginning of q still has a non-zero frequency. If it does, we re-insert it back into the priority queue so it can be processed again later.

4. Result: If the length of the resulting string ans is equal to the length of the input string s, it means a valid rearrangement has been found, and we return ans. Otherwise, no valid rearrangement exists, and we return an empty string.

Time and Space Complexity Analysis:

• Time Complexity: O(N log K), where N is the length of the string and K is the number of unique characters. Building the priority queue takes O(K log K) time. The loop iterates N times, and each iteration involves heap operations (push and pop), which take O(log K) time. The overall time complexity is dominated by the loop, resulting in O(N log K).

• Space Complexity: O(K) to store the character counts and the priority queue. The space used by the q queue is at most k. Therefore, space complexity is linear with respect to the number of unique characters in the string.

Code Examples (with detailed comments):

The code examples in Python, Java, C++, and Go provided in the original response effectively implement this approach. The comments within each code snippet explain the logic at each step. The use of data structures like priority queues and queues is crucial for efficiency. The handling of the k-distance constraint is achieved through the q queue, which ensures that the same character doesn't appear too frequently within a short window.