Is Reorganize String easy or hard?

Reorganize String is generally classified as a medium difficulty problem. The challenge lies in recognizing the greedy pattern and correctly handling character frequencies to prevent adjacent duplicates.

Reorganize String Python/Java solution

Python and Java solutions commonly implement a max heap using heapq (with negative counts) in Python or PriorityQueue in Java. The algorithm repeatedly pops the two highest-frequency characters, appends them to the result, and pushes them back if their counts remain.

How to solve Reorganize String in O(n)?

Use a frequency array to count characters, then place the character with the highest frequency at even indices of the result array (0, 2, 4...). Once its count is exhausted, fill the remaining positions with other characters. This works as long as the maximum frequency does not exceed (n + 1) / 2, otherwise a valid arrangement is impossible.

What is the best approach for Reorganize String?

The most common approach uses a greedy strategy with a max heap (priority queue). Characters are stored by frequency, and the two most frequent characters are placed alternately to avoid adjacent duplicates. This method runs in O(n log k) time where k is the number of distinct characters and is widely accepted in coding interviews.

Is Reorganize String asked at Google/Amazon/Meta?

Reorganize String is a common greedy and heap interview problem that has appeared in coding interviews at companies such as Amazon, Google, and Meta. Interviewers use it to evaluate understanding of priority queues, frequency counting, and greedy decision making.

What data structure is used in Reorganize String?

The typical solution uses a max heap (priority queue) to repeatedly select the characters with the highest remaining frequency. A hash map or array is used for counting frequencies before pushing them into the heap.

What is the time complexity of Reorganize String?

The heap-based greedy solution runs in O(n log k) time because each insertion and removal from the priority queue costs log k. An optimized counting and greedy placement approach can achieve O(n) time by placing the most frequent character at even indices and filling the remaining characters afterward.