Rearrange Distant Barcodes

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.

Examples:

Input: barcodes = {1, 1, 1, 2, 2, 2}
Output: {2, 1, 2, 1, 2, 1}
Explanation: In {2, 1, 2, 1, 2, 1}, no two adjacent barcodes are equal.

Input: barcodes = {1, 1, 1, 1, 2, 2, 3, 3}
Output: {1, 3, 1, 3, 1, 2, 1, 2}
Explanation: In {1, 3, 1, 3, 1, 2, 1, 2}, no two adjacent barcodes are equal.

Approach: To solve the problem, follow the below idea:

The idea is to first use the map to get the frequency of characters. Then add them all into a heap , and if possible each iteration in while loop , pop out 2 elements from the heap. We have to pop out 2 elements because in case we just pop out one and put back in the heap, then there is a chance that the same element is still occurring the most number of times and hence it will become adjacent. so pop out 2 at a time to avoid this.

Step-by-step algorithm:

• Count the frequency of each barcode in the input vector using an unordered_map.
• Create a priority queue (max-heap) to store the barcodes sorted by their frequencies in descending order.
• Iterate through the priority queue, and for each barcode:
  • Add the barcode to the rearranged vector, skipping every other position.
  • Decrement the frequency of the barcode and add it back to the priority queue if its frequency is still greater than 0.
  • If there's another barcode in the priority queue, add it to the next position in the rearranged vector.
• Return the rearranged vector.

Below is the implementation of the algorithm:

Rearranged barcodes: 2 1 2 1 2 1 

Time complexity: O(n log m), where n is the size of the input vector and m is the number of unique barcodes.
Auxiliary space: O(m + n), where m is the number of unique barcodes and n is the size of the input vector.