Rearrange a string so that all same characters become d distance away minDistance priority queue

最新推荐文章于 2024-08-28 11:31:52 发布

转载最新推荐文章于 2024-08-28 11:31:52 发布 · 686 阅读

Google面试专栏收录该内容

22 篇文章

订阅专栏

本文介绍一种算法，用于重新排列字符串中的字符，确保相同字符之间的最小距离为指定值。通过使用哈希表和优先队列实现最优解，本文提供了一个有效的解决方案，包括实例演示和复杂性分析。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Given a string and a positive integer d. Some characters may be repeated in the given string. Rearrange characters of the given string such that the same characters become d distance away from each other. Note that there can be many possible rearrangements, the output should be one of the possible rearrangements. If no such arrangement is possible, that should also be reported.
Expected time complexity is O(n) where n is length of input string.

**Examples:
Input: “abb”, d = 2
Output: “bab”
Input: “aacbbc”, d = 3
Output: “abcabc”
Input: “geeksforgeeks”, d = 3
Output: egkegkesfesor
Input: “aaa”, d = 2
Output: Cannot be rearranged**

Solution: The idea is to count frequencies of all characters and consider the most frequent character first and place all occurrences of it as close as possible. After the most frequent character is placed, repeat the same process for remaining characters.

Use HashMap for storing appearance, then priority queue of map.entry, comparator looks like this (max heap, descending order):

private class CompareByValue implements Comparator<Map.Entry<Cell, Integer>> {

    @Override
    public int compare(Entry<Cell, Integer> lhs,
            Entry<Cell, Integer> rhs) {

        return rhs.getValue().compareTo(lhs.getValue());
    }
}

http://www.geeksforgeeks.org/rearrange-a-string-so-that-all-same-characters-become-at-least-d-distance-away/

max heap is implemented with priority queue

priority Q example:

// Test.java
import java.util.Comparator;
import java.util.PriorityQueue;

public class Test
{
    public static void main(String[] args)
    {
        Comparator<String> comparator = new StringLengthComparator();
        PriorityQueue<String> queue = 
            new PriorityQueue<String>(10, comparator);
        queue.add("short");
        queue.add("very long indeed");
        queue.add("medium");
        while (queue.size() != 0)
        {
            System.out.println(queue.remove());
        }
    }
}

// StringLengthComparator.java
import java.util.Comparator;

public class StringLengthComparator implements Comparator<String>
{
    @Override
    public int compare(String x, String y)
    {
        // Assume neither string is null. Real code should
        // probably be more robust
        // You could also just return x.length() - y.length(),
        // which would be more efficient.
        if (x.length() < y.length())
        {
            return -1;
        }
        if (x.length() > y.length())
        {
            return 1;
        }
        return 0;
    }
}