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.

#1054

Medium

auto_awesome数组·哈希·扫描

LeetCode 题解工作台

距离相等的条形码

在一个仓库里，有一排条形码，其中第 i 个条形码为 barcodes[i] 。请你重新排列这些条形码，使其中任意两个相邻的条形码不能相等。你可以返回任何满足该要求的答案，此题保证存在答案。示例 1：输入： barcodes = [1,1,1,2,2,2] 输出： [2,1,2,1,2,1] …

数组哈希表贪心排序堆（优先队列）

题目描述

在一个仓库里，有一排条形码，其中第 i 个条形码为 barcodes[i]。

请你重新排列这些条形码，使其中任意两个相邻的条形码不能相等。你可以返回任何满足该要求的答案，此题保证存在答案。

示例 1：

输入：barcodes = [1,1,1,2,2,2]
输出：[2,1,2,1,2,1]

示例 2：

输入：barcodes = [1,1,1,1,2,2,3,3]
输出：[1,3,1,3,2,1,2,1]

提示：

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

lightbulb

解题思路

方法一：计数 + 排序

我们先用哈希表或数组 $cnt$ 统计数组 $barcodes$ 中各个数出现的次数，然后将 $barcodes$ 中的数按照它们在 $cnt$ 中出现的次数从大到小排序，如果出现次数相同，那么就按照数的大小从小到大排序（确保相同的数相邻）。

接下来，我们创建一个长度为 $n$ 的答案数组 $ans$ ，然后遍历排好序的 $barcodes$ ，将元素依次填入答案数组的 $0, 2, 4, \cdots$ 等偶数下标位置，然后将剩余元素依次填入答案数组的 $1, 3, 5, \cdots$ 等奇数下标位置即可。

时间复杂度 $O(n \times \log n)$ ，空间复杂度 $O(M)$ 。其中 $n$ 和 $M$ 分别是数组 $barcodes$ 的长度以及数组 $barcodes$ 中的最大值。

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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

speed

复杂度分析

指标	值
时间	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.
空间	Depends on the final approach

psychology

面试官常问的追问

外企场景

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

warning

常见陷阱

外企场景

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

swap_horiz

进阶变体

外企场景

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

help

常见问题

外企场景

继续练习

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