当无法修改数组时，数组中的第K个最小元素使用恒定空间

给定大小为N且整数K的数组arr [] ，任务是在常量额外空间中从数组中找到第K^个最小元素，并且该数组无法修改。

例子：

Input: arr[] = {7, 10, 4, 3, 20, 15}, K = 3
Output: 7
Given array in sorted is {3, 4, 7, 10, 15, 20}
where 7 is the third smallest element.

Input: arr[] = {12, 3, 5, 7, 19}, K = 2
Output: 5

方法：首先，我们从数组中找到最小和最大元素。然后，我们设置低=最小值，高=最大值和中=(低+高)/ 2 。
现在，执行修改后的二进制搜索，对于每个中位数，我们计算小于mid且等于mid的元素数。如果countLess 并且countLess + countEqual≥k，那么mid是我们的答案，否则我们必须修改我们的低位和高位。

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 smallest
// element from the array
int kthSmallest(int* arr, int k, int n)
{
  
    // Minimum and maximum element from the array
    int low = *min_element(arr, arr + n);
    int high = *max_element(arr, arr + n);
  
    // Modified binary search
    while (low <= high) {
  
        int mid = low + (high - low) / 2;
  
        // To store the count of elements from the array
        // which are less than mid and
        // the elements which are equal to mid
        int countless = 0, countequal = 0;
        for (int i = 0; i < n; ++i) {
            if (arr[i] < mid)
                ++countless;
            else if (arr[i] == mid)
                ++countequal;
        }
  
        // If mid is the kth smallest
        if (countless < k
            && (countless + countequal) >= k) {
            return mid;
        }
  
        // If the required element is less than mid
        else if (countless >= k) {
            high = mid - 1;
        }
  
        // If the required element is greater than mid
        else if (countless < k
                 && countless + countequal < k) {
            low = mid + 1;
        }
    }
}
  
// Driver code
int main()
{
    int arr[] = { 7, 10, 4, 3, 20, 15 };
    int n = sizeof(arr) / sizeof(int);
    int k = 3;
  
    cout << kthSmallest(arr, k, n);
  
    return 0;
}

Java

// Java implementation of the approach
import java.util.*;
  
class GFG 
{
  
// Function to return the kth smallest
// element from the array
static int kthSmallest(int[] arr, int k, int n)
{
  
    // Minimum and maximum element from the array
    int low = Arrays.stream(arr).min().getAsInt();
    int high = Arrays.stream(arr).max().getAsInt();
  
    // Modified binary search
    while (low <= high)
    {
  
        int mid = low + (high - low) / 2;
  
        // To store the count of elements from the array
        // which are less than mid and
        // the elements which are equal to mid
        int countless = 0, countequal = 0;
        for (int i = 0; i < n; ++i) 
        {
            if (arr[i] < mid)
                ++countless;
            else if (arr[i] == mid)
                ++countequal;
        }
  
        // If mid is the kth smallest
        if (countless < k
            && (countless + countequal) >= k) 
        {
            return mid;
        }
  
        // If the required element is less than mid
        else if (countless >= k) 
        {
            high = mid - 1;
        }
  
        // If the required element is greater than mid
        else if (countless < k
                && countless + countequal < k)
        {
            low = mid + 1;
        }
    }
    return Integer.MIN_VALUE;
}
  
// Driver code
public static void main(String[] args) 
{
    int arr[] = { 7, 10, 4, 3, 20, 15 };
    int n = arr.length;
    int k = 3;
  
    System.out.println(kthSmallest(arr, k, n));
}
}
  
// This code is contributed by 29AjayKumar

Python3

# Python3 implementation of the approach 
  
# Function to return the kth smallest 
# element from the array 
def kthSmallest(arr, k, n) : 
  
    # Minimum and maximum element from the array 
    low = min(arr); 
    high = max(arr); 
  
    # Modified binary search 
    while (low <= high) :
  
        mid = low + (high - low) // 2; 
  
        # To store the count of elements from the array 
        # which are less than mid and 
        # the elements which are equal to mid 
        countless = 0; countequal = 0; 
          
        for i in range(n) :
              
            if (arr[i] < mid) :
                countless += 1; 
                  
            elif (arr[i] == mid) :
                countequal += 1; 
  
  
        # If mid is the kth smallest 
        if (countless < k and (countless + countequal) >= k) :
            return mid; 
          
  
        # If the required element is less than mid 
        elif (countless >= k) :
            high = mid - 1; 
  
        # If the required element is greater than mid 
        elif (countless < k and countless + countequal < k) :
            low = mid + 1; 
      
# Driver code 
if __name__ == "__main__" : 
      
    arr = [ 7, 10, 4, 3, 20, 15 ]; 
    n = len(arr); 
    k = 3; 
  
    print(kthSmallest(arr, k, n)); 
  
# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;
using System.Linq; 
  
class GFG 
{
  
// Function to return the kth smallest
// element from the array
static int kthSmallest(int[] arr, int k, int n)
{
  
    // Minimum and maximum element from the array
    int low = arr.Min();
    int high = arr.Max();
  
    // Modified binary search
    while (low <= high)
    {
  
        int mid = low + (high - low) / 2;
  
        // To store the count of elements from the array
        // which are less than mid and
        // the elements which are equal to mid
        int countless = 0, countequal = 0;
        for (int i = 0; i < n; ++i) 
        {
            if (arr[i] < mid)
                ++countless;
            else if (arr[i] == mid)
                ++countequal;
        }
  
        // If mid is the kth smallest
        if (countless < k
            && (countless + countequal) >= k) 
        {
            return mid;
        }
  
        // If the required element is less than mid
        else if (countless >= k) 
        {
            high = mid - 1;
        }
  
        // If the required element is greater than mid
        else if (countless < k
                && countless + countequal < k)
        {
            low = mid + 1;
        }
    }
    return int.MinValue;
}
  
// Driver code
public static void Main(String[] args) 
{
    int []arr = { 7, 10, 4, 3, 20, 15 };
    int n = arr.Length;
    int k = 3;
  
    Console.WriteLine(kthSmallest(arr, k, n));
}
}
  
// This code is contributed by Rajput-Ji

输出：

7

时间复杂度： O(N log(Max – Min))，其中Max和Min分别是数组中的最大元素和最小元素，N是数组的大小。