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Distant Barcodes

Difficulty: Medium

Problem Description

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.

Key Insights

  • The problem requires a rearrangement of barcodes such that no two adjacent elements are the same.
  • A greedy approach can be beneficial, where the most frequently occurring barcode is placed first to maximize spacing.
  • Using a max heap (or priority queue) can help efficiently manage and retrieve the most frequent barcodes.
  • A counting mechanism will help determine the frequency of each barcode.

Space and Time Complexity

Time Complexity: O(n log n) - due to the heap operations for sorting and rearranging the barcodes.
Space Complexity: O(n) - for storing the frequency counts and the result.

Solution

To solve the problem, we can follow these steps:

  1. Count the frequency of each barcode using a hash table.
  2. Use a max heap to store the barcodes based on their frequency.
  3. Pop the most frequent barcode from the heap and place it into the result array, ensuring that no two adjacent positions contain the same barcode.
  4. Repeat the process until all barcodes are placed in the result array.

Code Solutions

from collections import Counter
import heapq

def rearrangeBarcodes(barcodes):
    # Count the frequency of each barcode
    count = Counter(barcodes)
    # Create a max heap based on the frequency
    max_heap = [(-freq, barcode) for barcode, freq in count.items()]
    heapq.heapify(max_heap)
    
    result = []
    
    # Previous barcode and its frequency
    prev_barcode = None
    prev_freq = 0
    
    while max_heap:
        freq, barcode = heapq.heappop(max_heap)
        result.append(barcode)
        
        # If the previous barcode still has a frequency left, push it back to the heap
        if prev_freq < 0:
            heapq.heappush(max_heap, (prev_freq, prev_barcode))
        
        # Update the previous barcode and its frequency
        prev_barcode = barcode
        prev_freq = freq + 1  # Decrease the frequency since we used it
        
    return result
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