#1054

Medium

auto_awesomeArray scanning plus hash lookup

LeetCode Problem Workspace

Distant Barcodes

Rearrange barcodes in an array so that no two adjacent elements are equal, using a greedy approach and hash table for efficient lookups.

array hash table greedy sorting heap priority queue

Problem Statement

Given an array of barcodes, rearrange the array such that no two adjacent barcodes are the same. It is guaranteed that a solution exists for the given input.

Your task is to return any valid arrangement of the barcodes satisfying the condition. The input array has a length between 1 and 10,000, and each element in the array represents a barcode with an integer value between 1 and 10,000.

Examples

Example 1

Input: barcodes = [1,1,1,2,2,2]

Output: [2,1,2,1,2,1]

Example details omitted.

Example 2

Input: barcodes = [1,1,1,1,2,2,3,3]

Output: [1,3,1,3,1,2,1,2]

Example details omitted.

Constraints

1 <= barcodes.length <= 10000
1 <= barcodes[i] <= 10000

Solution Approach

Greedy Approach with Hash Table

Use a hash table to count the frequency of each barcode, then select the most frequent barcode to place next. This ensures the barcode with the highest availability is placed in the sequence first. Adjust the frequency count after each placement.

Heap for Optimal Selection

A max heap (priority queue) can be used to always select the most frequent barcode that hasn’t been placed next to a duplicate. The heap stores barcodes along with their frequencies, and after placing a barcode, the heap is updated to reflect the new frequency.

Array Scanning for Efficient Rearrangement

After selecting the next barcode, scan the array to find an appropriate position to place the barcode, ensuring that no adjacent barcodes are the same. This step is repeated until all elements are placed.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

Time complexity depends on the method used for selecting barcodes. Using a hash table and heap for selection results in O(n log k) time, where n is the length of the array and k is the number of unique barcodes. Space complexity is O(k) for the storage of frequencies and heap.

What Interviewers Usually Probe

Candidates who use a greedy approach with heap selection and hash tables are likely on the right track.
Look for candidates who efficiently update barcode counts after placing each element in the sequence.
Candidates who rely on brute force without using efficient data structures may struggle with larger inputs.

Common Pitfalls or Variants

Common pitfalls

Not updating the barcode frequency after each placement, which can lead to incorrect placements of barcodes.
Ignoring edge cases where fewer barcode types are present in the array.
Inefficient approaches that do not utilize data structures like heaps or hash tables can lead to slower solutions.

Follow-up variants

Try with arrays containing only one type of barcode to test if the algorithm handles this edge case.
Consider arrays with large numbers of unique barcode types to evaluate the efficiency of the heap-based solution.
Test cases with arrays that require precise placement of barcodes at specific indices.

FAQ

How do I solve the 'Distant Barcodes' problem?

The problem can be solved using a greedy approach with a hash table to count frequencies, followed by a heap for optimal barcode selection.

What is the time complexity of solving the Distant Barcodes problem?

The time complexity is O(n log k), where n is the number of barcodes, and k is the number of unique barcodes.

What data structure should be used for the Distant Barcodes problem?

A hash table for counting frequencies and a heap for selecting the most frequent barcode are optimal data structures.

Are there any edge cases I should be aware of?

Consider arrays with only one barcode type or arrays where the barcode count is low compared to the total array length.

How does GhostInterview help with this problem?

GhostInterview provides step-by-step guidance on how to apply greedy algorithms, hash tables, and heaps to solve the problem efficiently.

terminal

Solution

Solution 1: Counting + Sorting

First, we use a hash table or array $cnt$ to count the number of occurrences of each number in the array $barcodes$. Then, we sort the numbers in $barcodes$ according to their occurrence times in $cnt$ from large to small. If the occurrence times are the same, we sort them from small to large (to ensure the same numbers are adjacent).

1

2

3

4

5

6

7

8

9

class Solution:
    def rearrangeBarcodes(self, barcodes: List[int]) -> List[int]:
        cnt = Counter(barcodes)
        barcodes.sort(key=lambda x: (-cnt[x], x))
        n = len(barcodes)
        ans = [0] * len(barcodes)
        ans[::2] = barcodes[: (n + 1) // 2]
        ans[1::2] = barcodes[(n + 1) // 2 :]
        return ans

Continue Practicing

#621 task scheduler #1090 largest values from labels #1338 reduce array size to the half