Distant Barcodes

Distant Barcodes

What is this problem about?

The Distant Barcodes interview question asks you to rearrange an array of integers (barcodes) so that no two adjacent barcodes are equal. You are guaranteed that a valid arrangement exists. For example, if you have [1, 1, 1, 2, 2, 2], an arrangement like [1, 2, 1, 2, 1, 2] is valid, but [1, 1, 2, 2, 1, 2] is not.

Why is this asked in interviews?

Meta and Amazon use this problem to test your understanding of Greedy interview patterns and frequency management. It tests whether you can identify that the most frequent element is the most "constrained" and must be placed first to avoid violations. It evaluations your ability to use Priority Queues or interleaving strategies to maintain data integrity under constraints.

Algorithmic pattern used

This problem is typically solved using a Frequency-based Greedy approach.

1. Count the frequency of each barcode.
2. Sort the barcodes by frequency (most frequent first).
3. Place the barcodes into the result array at indices 0, 2, 4... (even indices).
4. Once you reach the end of the array, continue placing the remaining barcodes at indices 1, 3, 5... (odd indices). By placing the most frequent element in every other slot starting from index 0, you guarantee that it can never be adjacent to itself, as long as its count is $\le (n+1)/2$.

Example explanation

Barcodes: [1, 1, 1, 2, 2, 3]

1. Frequencies: {1: 3, 2: 2, 3: 1}.
2. Even Indices (0, 2, 4): Place the 1s. Result: [1, _, 1, _, 1, _].
3. Odd Indices (1, 3, 5): Place the 2s and 3. Result: [1, 2, 1, 2, 1, 3]. All adjacent elements are different.

Common mistakes candidates make

• Priority Queue overhead: Using a Max-Heap to pick the top two elements repeatedly. While this works ($O(N \log K)$), the "even-then-odd" interleaving is much faster ($O(N)$).
• Random Shuffling: Trying to solve the problem with randomization, which doesn't guarantee a valid result.
• Not handling the most frequent element first: If you start with a rare element, you might run out of "buffer" elements to separate the most frequent ones later.

Interview preparation tip

For any "rearrange so no two adjacent are the same" problem, the golden rule is: Always process the most frequent elements first. Whether you use interleaving or a Priority Queue, the most frequent element is the one that will cause a conflict if not handled early.