在已排序的生成数组中查找第 K 个最小元素

给定一个包含N 个元素的数组arr[]和一个整数K ，任务是使用以下规则生成一个B[] ：

1. 将元素arr[1…N]复制N次到数组B[] 。
2. 复制元素arr[1…N/2] 2*N次到数组B[] 。
3. 复制元素arr[1…N/4] 3*N次到数组B[] 。
4. 类似地，直到没有元素被复制到数组 B[]。

最后打印数组B[] 中^第K^个最小的元素。如果K超出B[]的范围，则返回-1 。
例子：

Input: arr[] = {1, 2, 3}, K = 4 
Output: 1 
{1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 1, 1, 1, 1, 1} is the required array B[] 
{1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 3, 3, 3} in the sorted form where 
1 is the 4th smallest element.
Input: arr[] = {2, 4, 5, 1}, K = 13 
Output: 2 
 

方法：

1. 维护一个 Count_Array，我们必须在其中存储数组 B[] 中每个元素出现的次数。可以通过在开始索引处添加计数并在结束索引 + 1 位置减去相同计数来完成元素范围。
2. 取计数数组的累积和。
3. 维护 arr[] 的所有元素及其在数组 B[] 中的计数及其计数，并根据元素值对它们进行排序。
4. 遍历向量并查看哪个元素根据其各自的计数在 B[] 中具有第 K 个位置。
5. 如果 K 超出 B[] 的范围，则返回 -1。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to return the Kth element in B[]
int solve(int Array[], int N, int K)
{
 
    // Initialize the count Array
    int count_Arr[N + 1] = { 0 };
    int factor = 1;
    int size = N;
 
    // Reduce N repeatedly to half its value
    while (size) {
        int start = 1;
        int end = size;
 
        // Add count to start
        count_Arr[1] += factor * N;
 
        // Subtract same count after end index
        count_Arr[end + 1] -= factor * N;
        factor++;
        size /= 2;
    }
 
    for (int i = 2; i <= N; i++)
        count_Arr[i] += count_Arr[i - 1];
 
    // Store each element of Array[] with their count
    vector > element;
    for (int i = 0; i < N; i++) {
        element.push_back({ Array[i], count_Arr[i + 1] });
    }
 
    // Sort the elements wrt value
    sort(element.begin(), element.end());
 
    int start = 1;
    for (int i = 0; i < N; i++) {
        int end = start + element[i].second - 1;
 
        // If Kth element is in range of element[i]
        // return element[i]
        if (K >= start && K <= end) {
            return element[i].first;
        }
 
        start += element[i].second;
    }
 
    // If K is out of bound
    return -1;
}
 
// Driver code
int main()
{
    int arr[] = { 2, 4, 5, 1 };
    int N = sizeof(arr) / sizeof(arr[0]);
    int K = 13;
 
    cout << solve(arr, N, K);
 
    return 0;
}

Java

// Java implementation of the approach
import java.util.Vector;
 
class GFG
{
 
    // Pair class implementation to use Pair
    static class Pair
    {
        private int first;
        private int second;
 
        Pair(int first, int second)
        {
            this.first = first;
            this.second = second;
        }
 
        public int getFirst()
        {
            return first;
        }
 
        public int getSecond()
        {
            return second;
        }
    }
 
    // Function to return the Kth element in B[]
    static int solve(int[] Array, int N, int K)
    {
 
        // Initialize the count Array
        int[] count_Arr = new int[N + 2];
        int factor = 1;
        int size = N;
 
        // Reduce N repeatedly to half its value
        while (size > 0)
        {
            int start = 1;
            int end = size;
 
            // Add count to start
            count_Arr[1] += factor * N;
 
            // Subtract same count after end index
            count_Arr[end + 1] -= factor * N;
            factor++;
            size /= 2;
        }
 
        for (int i = 2; i <= N; i++)
            count_Arr[i] += count_Arr[i - 1];
 
        // Store each element of Array[]
        // with their count
        Vector element = new Vector<>();
        for (int i = 0; i < N; i++)
        {
            Pair x = new Pair(Array[i],
                              count_Arr[i + 1]);
            element.add(x);
        }
 
        int start = 1;
        for (int i = 0; i < N; i++)
        {
            int end = start + element.elementAt(0).getSecond() - 1;
 
            // If Kth element is in range of element[i]
            // return element[i]
            if (K >= start && K <= end)
                return element.elementAt(i).getFirst();
 
            start += element.elementAt(i).getSecond();
        }
 
        // If K is out of bound
        return -1;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int[] arr = { 2, 4, 5, 1 };
        int N = arr.length;
        int K = 13;
        System.out.println(solve(arr, N, K));
    }
}   
 
// This code is contiributed by
// sanjeev2552

Python3

# Python3 implementation of the approach
 
# Function to return the Kth element in B[]
def solve(Array, N, K) :
 
    # Initialize the count Array
    count_Arr = [0]*(N + 2) ;
    factor = 1;
    size = N;
 
    # Reduce N repeatedly to half its value
    while (size) :
        start = 1;
        end = size;
 
        # Add count to start
        count_Arr[1] += factor * N;
 
        # Subtract same count after end index
        count_Arr[end + 1] -= factor * N;
        factor += 1;
        size //= 2;
 
    for i in range(2, N + 1) :
        count_Arr[i] += count_Arr[i - 1];
 
    # Store each element of Array[] with their count
    element = [];
     
    for i in range(N) :
        element.append(( Array[i], count_Arr[i + 1] ));
 
    # Sort the elements wrt value
    element.sort();
 
    start = 1;
    for i in range(N) :
        end = start + element[i][1] - 1;
 
        # If Kth element is in range of element[i]
        # return element[i]
        if (K >= start and K <= end) :
            return element[i][0];
 
        start += element[i][1];
 
    # If K is out of bound
    return -1;
 
 
# Driver code
if __name__ == "__main__" :
 
    arr = [ 2, 4, 5, 1 ];
    N = len(arr);
    K = 13;
 
    print(solve(arr, N, K));
 
    # This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;
using System.Collections.Generic;
     
class GFG
{
 
    // Pair class implementation to use Pair
    public class Pair
    {
        public int first;
        public int second;
 
        public Pair(int first, int second)
        {
            this.first = first;
            this.second = second;
        }
 
        public int getFirst()
        {
            return first;
        }
 
        public int getSecond()
        {
            return second;
        }
    }
 
    // Function to return the Kth element in B[]
    static int solve(int[] Array, int N, int K)
    {
 
        // Initialize the count Array
        int[] count_Arr = new int[N + 2];
        int factor = 1;
        int size = N;
 
        // Reduce N repeatedly to half its value
        while (size > 0)
        {
            int end = size;
 
            // Add count to start
            count_Arr[1] += factor * N;
 
            // Subtract same count after end index
            count_Arr[end + 1] -= factor * N;
            factor++;
            size /= 2;
        }
 
        for (int i = 2; i <= N; i++)
            count_Arr[i] += count_Arr[i - 1];
 
        // Store each element of Array[]
        // with their count
        List element = new List();
        for (int i = 0; i < N; i++)
        {
            Pair x = new Pair(Array[i],
                              count_Arr[i + 1]);
            element.Add(x);
        }
 
        int start = 1;
        for (int i = 0; i < N; i++)
        {
            int end = start + element[0].getSecond() - 1;
 
            // If Kth element is in range of element[i]
            // return element[i]
            if (K >= start && K <= end)
                return element[i].getFirst();
 
            start += element[i].getSecond();
        }
 
        // If K is out of bound
        return -1;
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        int[] arr = { 2, 4, 5, 1 };
        int N = arr.Length;
        int K = 13;
        Console.WriteLine(solve(arr, N, K));
    }
}
 
// This code is contributed by Rajput-Ji

Javascript

 
// JavaScript implementation of the approach
 
// Function to return the Kth element in B[]
function solve(arr, N, K) {
 
    // Initialize the count Array
    let count_Arr = new Array(N + 1).fill(0);
    let factor = 1;
    let size = N;
 
    // Reduce N repeatedly to half its value
    while (size) {
        let start = 1;
        let end = size;
 
        // Add count to start
        count_Arr[1] += factor * N;
 
        // Subtract same count after end index
        count_Arr[end + 1] -= factor * N;
        factor++;
        size = Math.floor(size / 2);
    }
 
    for (let i = 2; i <= N; i++)
        count_Arr[i] += count_Arr[i - 1];
 
    // Store each element of Array[] with their count
    let element = [];
    for (let i = 0; i < N; i++) {
        element.push([arr[i], count_Arr[i + 1]]);
    }
 
    // Sort the elements wrt value
    element.sort((a, b) => a - b);
 
    let start = 1;
    for (let i = 0; i < N; i++) {
        let end = start + element[i][1] - 1;
 
        // If Kth element is in range of element[i]
        // return element[i]
        if (K >= start && K <= end) {
            return element[i][0];
        }
 
        start += element[i][1];
    }
 
    // If K is out of bound
    return -1;
}
 
// Driver code
 
let arr = [2, 4, 5, 1];
let N = arr.length;
let K = 13;
 
document.write(solve(arr, N, K));
 
 
// This code is contributed by gfgking
 

输出：

2

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