# Solving Leetcode Interviews in Seconds with AI: Distant Barcodes


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "1054" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> In a warehouse, there is a row of barcodes, where the ith barcode is barcodes[i]. Rearrange the barcodes so that no two adjacent barcodes are equal. You may return any answer, and it is guaranteed an answer exists.   Example 1: Input: barcodes = [1,1,1,2,2,2] Output: [2,1,2,1,2,1] Example 2: Input: barcodes = [1,1,1,1,2,2,3,3] Output: [1,3,1,3,1,2,1,2]    Constraints:  1 <= barcodes.length <= 10000 1 <= barcodes[i] <= 10000  

	# Explanation
	Here's the solution:

*   **Frequency Counting:** Count the frequency of each barcode using a dictionary or array.
*   **Prioritize Frequent Elements:** Use a max heap (priority queue) to store the barcodes and their frequencies. The barcode with the highest frequency will be at the top of the heap.
*   **Greedy Placement:** Repeatedly extract the two most frequent barcodes from the heap and place them in the result array. To avoid adjacent duplicates, alternate between the two extracted barcodes.

*   **Runtime Complexity:** O(N log K), where N is the number of barcodes and K is the number of unique barcodes. In the worst case, K can be equal to N.
    **Storage Complexity:** O(K), where K is the number of unique barcodes (for storing frequency counts and the heap).

	
	# Code
	```python
	import heapq
from collections import Counter

def rearrangeBarcodes(barcodes):
    """
    Rearranges barcodes such that no two adjacent barcodes are equal.

    Args:
        barcodes: A list of integers representing the barcodes.

    Returns:
        A list of integers representing the rearranged barcodes.
    """

    counts = Counter(barcodes)
    max_heap = []

    for barcode, count in counts.items():
        heapq.heappush(max_heap, (-count, barcode))  # Negate count for max heap

    result = []
    while max_heap:
        count1, barcode1 = heapq.heappop(max_heap)
        result.append(barcode1)

        if max_heap:
            count2, barcode2 = heapq.heappop(max_heap)
            result.append(barcode2)

            if count2 + 1 < 0:  # Push back if frequency is not zero
                heapq.heappush(max_heap, (count2 + 1, barcode2))

        if count1 + 1 < 0:  # Push back if frequency is not zero
            heapq.heappush(max_heap, (count1 + 1, barcode1))

    return result
	```