通过从随机索引反转子数组形成的排序数组中的第 K 个最小元素

给定一个大小为N的排序数组arr[]和一个整数K ，任务是找到数组中存在的^第K^个最小元素。给定的阵列已经通过反转子阵列获得{ARR [0]，ARR [R]}和{ARR [R + 1]，ARR [N – 1]}在一些随机指数R.如果该键不存在于数组，打印-1 。

例子：

Input: arr[] = { 4, 3, 2, 1, 8, 7, 6, 5 }, K = 2
Output: 2
Explanation: Sorted form of the array arr[] is { 1, 2, 3, 4, 5, 6, 7, 8 }. Therefore, the 2nd smallest element in the array arr[] is 2.

Input: arr[] = { 10, 8, 6, 5, 2, 1, 13, 12 }, K = 3
Output: 5

朴素的方法：解决问题最简单的方法是将给定的数组arr[]按升序排序并打印数组中^第K^个最小的元素。

时间复杂度： O(N*log(N))
辅助空间： O(1)

上述解决方案的替代方法：我们可以在不使用任何排序技术的情况下对数组进行排序，这肯定会降低时间复杂度。我们可以使用二分搜索找到数组中的枢轴点P (旋转发生在它周围) ，然后使用std::reverse()函数反转两个子数组[0, P + 1]和[P + 1, N]在 C++ 中。

反转子数组：找到枢轴点P 后，只需找到两个子数组的反转如下：

std::reverse(arr, arr + P + 1);

std::reverse(arr + P + 1, arr + N);

因此我们得到了排序后的数组，我们可以将第 K 个最小元素打印为arr[K-1] 。

C++

// C++ program for the above approach.
#include 
using namespace std;
 
/* Function to get pivot. For array 4, 3, 2, 1, 6, 5
   it returns 3 (index of 1) */
int findPivot(int arr[], int low, int high)
{
    // base cases
    if (high < low)
        return -1;
    if (high == low)
        return low;
 
    int mid = (low + high) / 2; /*low + (high - low)/2;*/
    if (mid < high && arr[mid] < arr[mid + 1])
        return mid;
 
    if (mid > low && arr[mid] > arr[mid - 1])
        return (mid - 1);
 
    if (arr[low] <= arr[mid])
        return findPivot(arr, low, mid - 1);
 
    return findPivot(arr, mid + 1, high);
}
 
// Driver Code
int main()
{
 
    // Given Input
    int arr[] = { 10, 8, 6, 5, 2, 1, 13, 12 };
    int N = sizeof(arr) / sizeof(arr[0]);
    int K = 3;
 
    // Function Call
    int P = findPivot(arr, 0, N - 1);
 
    // reversing first subarray
    reverse(arr, arr + P + 1);
 
    // reversing second subarray
    reverse(arr + P + 1, arr + N);
    // printing output
    cout << arr[K - 1];
    return 0;
}
 
// This code is contributed by Pranjay Vats

Javascript

    // Javascript program for the above approach.
     
    /* Function to get pivot. For array 4, 3, 2, 1, 6, 5
   it returns 3 (index of 1) */
  function findPivot(arr, low, high)
  {
   
      // base cases
      if (high < low)
          return -1;
      if (high == low)
          return low;
 
      let mid = parseInt((low + high) / 2, 10); /*low + (high - low)/2;*/
      if (mid < high && arr[mid] < arr[mid + 1])
          return mid;
 
      if (mid > low && arr[mid] > arr[mid - 1])
          return (mid - 1);
 
      if (arr[low] <= arr[mid])
          return findPivot(arr, low, mid - 1);
 
      return findPivot(arr, mid + 1, high);
  }
   
  function reverse(i, j, arr)
  {
    let y = j - 1;
      for(let x = i; x < j; x++)
    {
        arr[x] = arr[y];
        y--;
    }
  }
   
  // Given Input
  let arr = [ 10, 8, 6, 5, 2, 1, 13, 12 ];
  let N = arr.length;
  let K = 3;
 
  // Function Call
  let P = findPivot(arr, 0, N - 1);
   
  // reversing first subarray
  reverse(0, P + 1, arr);
 
  // reversing second subarray
  reverse(P + 1, N, arr);
   
  // printing output
  document.write(arr[K - 1]);
   
  // This code is contributed by divyeshrabadiya07.

输出

5

时间复杂度： O(N)

辅助空间： O(1)

高效的方法：最优的想法是基于观察到第R 个元素是最小的元素，因为[1, R]范围内的元素是颠倒的。现在，如果随机索引为R ，则表示子数组[1, R]和[R + 1, N]按降序排序。因此，任务减少到找到可以使用二分搜索获得的R的值。最后，打印第K^个最小元素。

请按照以下步骤解决问题：

• 将l初始化为1 ，将h初始化为N，以存储二分搜索的搜索空间的边界元素索引。
• 循环而l+1 < h的值
  • 将中间元素存储在一个变量中， mid为(l+h)/2 。
  • 如果arr[l] ≥ arr[mid]。如果这是真的，然后通过更新l至中期检查中旬的右侧。
  • 否则，将r更新为mid 。
• 现在在找到R 之后，如果K ≤ R ，那么答案是arr[R-K+1]。否则， arr[N-(KR)+1] 。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find the Kth element in a
// sorted and rotated array at random point
int findkthElement(vector arr, int n, int K)
{
     
    // Set the boundaries for binary search
    int l = 0;
    int h = n - 1, r;
 
    // Apply binary search to find R
    while (l + 1 < h)
    {
         
        // Initialize the middle element
        int mid = (l + h) / 2;
         
        // Check in the right side of mid
        if (arr[l] >= arr[mid])
            l = mid;
             
        // Else check in the left side
        else
            h = mid;
    }
     
    // Random point either l or h
    if (arr[l] < arr[h])
        r = l;
    else
        r = h;
     
    // Return the kth smallest element
    if (K <= r + 1)
        return arr[r + 1 - K];
    else
        return arr[n - (K - (r + 1))];
}
 
// Driver Code
int main()
{
     
    // Given Input
    vector arr = { 10, 8, 6, 5, 2, 1, 13, 12 };
    int n = arr.size();
    int K = 3;
     
    // Function Call
    cout << findkthElement(arr, n, K);
}
 
// This code is contributed by mohit kumar 29

Java

// Java program for the above approach
class GFG{
     
// Function to find the Kth element in a
// sorted and rotated array at random point
public static int findkthElement(int arr[], int n, int K)
{
     
    // Set the boundaries for binary search
    int l = 0;
    int h = n - 1, r;
 
    // Apply binary search to find R
    while (l + 1 < h)
    {
         
        // Initialize the middle element
        int mid = (l + h) / 2;
         
        // Check in the right side of mid
        if (arr[l] >= arr[mid])
            l = mid;
             
        // Else check in the left side
        else
            h = mid;
    }
     
    // Random point either l or h
    if (arr[l] < arr[h])
        r = l;
    else
        r = h;
     
    // Return the kth smallest element
    if (K <= r + 1)
        return arr[r + 1 - K];
    else
        return arr[n - (K - (r + 1))];
}
 
// Driver Code
public static void main(String args[])
{
     
    // Given Input
    int []arr = { 10, 8, 6, 5, 2, 1, 13, 12 };
    int n = arr.length;
    int K = 3;
     
    // Function Call
    System.out.println(findkthElement(arr, n, K));
}
}
 
// This code is contributed by SoumikMondal

蟒蛇3

# Python program for the above approach
 
# Function to find the Kth element in a
# sorted and rotated array at random point
def findkthElement(arr, n, K):
   
      # Set the boundaries for binary search
    l = 0
    h = n-1
 
    # Apply binary search to find R
    while l+1 < h:
       
          # Initialize the middle element
        mid = (l+h)//2
 
        # Check in the right side of mid
        if arr[l] >= arr[mid]:
            l = mid
 
        # Else check in the left side
        else:
            h = mid
 
    # Random point either l or h
    if arr[l] < arr[h]:
        r = l
    else:
        r = h
 
    # Return the kth smallest element
    if K <= r+1:
        return arr[r+1-K]
    else:
        return arr[n-(K-(r+1))]
 
 
# Driver Code
if __name__ == "__main__":
   
      # Given Input
    arr = [10, 8, 6, 5, 2, 1, 13, 12]
    n = len(arr)
    K = 3
     
    # Function Call
    print(findkthElement(arr, n, K) )

C＃

using System.IO;
using System;
 
class GFG {
 
    // Function to find the Kth element in a
    // sorted and rotated array at random point
    public static int findkthElement(int[] arr, int n,
                                     int K)
    {
 
        // Set the boundaries for binary search
        int l = 0;
        int h = n - 1, r;
 
        // Apply binary search to find R
        while (l + 1 < h) {
 
            // Initialize the middle element
            int mid = (l + h) / 2;
 
            // Check in the right side of mid
            if (arr[l] >= arr[mid])
                l = mid;
 
            // Else check in the left side
            else
                h = mid;
        }
 
        // Random point either l or h
        if (arr[l] < arr[h])
            r = l;
        else
            r = h;
 
        // Return the kth smallest element
        if (K <= r + 1)
            return arr[r + 1 - K];
        else
            return arr[n - (K - (r + 1))];
    }
 
    static void Main()
    {
        // Given Input
        int[] arr = { 10, 8, 6, 5, 2, 1, 13, 12 };
        int n = arr.Length;
        int K = 3;
 
        // Function Call
        Console.WriteLine(findkthElement(arr, n, K));
    }
}
 
// This code is contributed by abhinavjain194.

Javascript

 
      // Function to find the Kth element in a
      // sorted and rotated array at random point
      function findkthElement(arr, n, K) {
        // Set the boundaries for binary search
        var l = 0;
        var h = n - 1,
          r;
 
        // Apply binary search to find R
        while (l + 1 < h) {
          // Initialize the middle element
          var mid = parseInt((l + h) / 2);
 
          // Check in the right side of mid
          if (arr[l] >= arr[mid]) l = mid;
          // Else check in the left side
          else h = mid;
        }
 
        // Random point either l or h
        if (arr[l] < arr[h]) r = l;
        else r = h;
 
        // Return the kth smallest element
        if (K <= r + 1) return arr[r + 1 - K];
        else return arr[n - (K - (r + 1))];
      }
 
      // Given Input
      var arr = [10, 8, 6, 5, 2, 1, 13, 12];
      var n = arr.length;
      var K = 3;
 
      // Function Call
      document.write(findkthElement(arr, n, K));
       

输出

5

时间复杂度： O(log(N))
辅助空间： O(1)

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