要删除的最小子数组的长度，以便对剩余的数组进行排序

给定一个由N 个整数组成的数组arr[] ，任务是打印要从arr[] 中删除的最小子数组的长度，以便对剩余的数组进行排序。

例子：

Input: arr[] = {1, 2, 3, 10, 4, 2, 3, 5}
Output: 3
Explanation:
The smallest subarray to be remove is {10, 4, 2} of length 3. The remaining array is {1, 2, 3, 3, 5} which are sorted.
Another correct solution is to remove the subarray [3, 10, 4].

Input: arr[] = {5, 4, 3, 2, 1}
Output: 4
Explanation:
Since the array is strictly decreasing, only a single element need to be kept in the array.
Therefore, required answer is 4.

处理方法：可以通过以下三种方法来去除子数组来解决问题：

1. Remove the subarray from the left i.e left suffix.
2. Remove the subarray from the right ie, prefix.
3. Remove the subarray from the middle and merge the left and right part of the array.

1. 从左到右遍历arr[] ，找到第一个左边的索引arr[left] > arr[left + 1] 。所以为了使数组排序而要删除的子数组长度是N-left-1。
2. 如果left == N – 1 ，则该数组已排序，因此返回 0。
3. 从右到左遍历arr[] ，找到第一个索引right ，即arr[right] < arr[right – 1] 。所以为了使数组排序而需要去除的子数组长度是正确的。
4. 现在初始化一个变量mincount并取N-left-1和right 中的最小值。
5. 现在计算要从数组中间删除的子数组的最小长度：
   1. 让i = 0, j = right 。并检查是否可以通过比较arr[i]和arr[j]删除i和j (不包括)之间的元素。
   2. 如果arr[j] >= arr[i] ，尝试使用j – i – 1更新答案并增加i以收紧窗口。
   3. 如果arr[j] < arr[i] ，则不能删除其间的元素，因此增加j以松开窗口。
   4. 遍历循环直到i > left或j == N 。并更新答案。
6. 返回mincount 。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include
using namespace std;
 
// Find the length of the shortest subarray
int findLengthOfShortestSubarray(int arr[], int N)
{
     
    // To store the result
    int minlength = INT_MAX;
 
    int left = 0;
    int right = N - 1;
 
    // Calculate the possible length of
    // the sorted subarray from left
    while (left < right &&
       arr[left + 1] >= arr[left])
    {
        left++;
    }
 
    // Array is sorted
    if (left == N - 1)
        return 0;
 
    // Calculate the possible length of
    // the sorted subarray from left
    while (right > left &&
       arr[right - 1] <= arr[right])
    {
        right--;
    }
 
    // Update the result
    minlength = min(N - left - 1, right);
 
    // Calculate the possible length
    // in the middle we can delete
    // and update the result
    int j = right;
    for(int i = 0; i < left + 1; i++)
    {
        if (arr[i] <= arr[j])
        {
             
            // Update the result
            minlength = min(minlength, j - i - 1);
        }
        else if (j < N - 1)
        {
            j++;
        }
        else
        {
            break;
        }
    }
 
    // Return the result
    return minlength;
}
 
// Driver Code
int main()
{
    int arr[] = { 6, 3, 10, 11, 15,
                  20, 13, 3, 18, 12 };
    int N = sizeof(arr) / sizeof(arr[0]);
     
    // Function call
    cout << (findLengthOfShortestSubarray(arr, N));
}
 
// This code is contributed by divyeshrabadiya07

Java

// Java program for the
// above approach
import java.util.*;
class GFG {
 
// Find the length of the shortest subarray
static int findLengthOfShortestSubarray(int arr[],
                                        int N)
{
  // To store the result
  int minlength = Integer.MAX_VALUE;
 
  int left = 0;
  int right = N - 1;
 
  // Calculate the possible length of
  // the sorted subarray from left
  while (left < right &&
         arr[left + 1] >= arr[left])
  {
    left++;
  }
 
  // Array is sorted
  if (left == N - 1)
    return 0;
 
  // Calculate the possible length of
  // the sorted subarray from left
  while (right > left &&
         arr[right - 1] <= arr[right])
  {
    right--;
  }
 
  // Update the result
  minlength = Math.min(N - left - 1,
                       right);
 
  // Calculate the possible length
  // in the middle we can delete
  // and update the result
  int j = right;
   
  for (int i = 0; i < left + 1; i++)
  {
    if (arr[i] <= arr[j])
    {
      // Update the result
      minlength = Math.min(minlength,
                           j - i - 1);
    }
    else if (j < N - 1)
    {
      j++;
    }
    else
    {
      break;
    }
  }
 
  // Return the result
  return minlength;
}
 
// Driver Code
public static void main(String[] args)
{
  int arr[] = {6, 3, 10, 11, 15,
               20, 13, 3, 18, 12};
  int N = arr.length;
 
  // Function call
  System.out.print(
         (findLengthOfShortestSubarray(arr, N)));
}
}
 
// This code is contributed by Chitranayal

Python3

# Python3 program for the above approach
import sys
 
# Find the length of the shortest subarray
def findLengthOfShortestSubarray(arr):
 
    # To store the result
    minlength = sys.maxsize
 
    left = 0
    right = len(arr) - 1
 
    # Calculate the possible length of
    # the sorted subarray from left
    while left < right and arr[left + 1] >= arr[left]:
        left += 1
 
    # Array is sorted
    if left == len(arr) - 1:
        return 0
 
    # Calculate the possible length of
    # the sorted subarray from left
    while right > left and arr[right-1] <= arr[right]:
        right -= 1
 
    # Update the result
    minlength = min(len(arr) - left - 1, right)
 
    # Calculate the possible length
    # in the middle we can delete
    # and update the result
    j = right
    for i in range(left + 1):
 
        if arr[i] <= arr[j]:
 
            # Update the result
            minlength = min(minlength, j - i - 1)
 
        elif j < len(arr) - 1:
 
            j += 1
 
        else:
 
            break
 
    # Return the result
    return minlength
 
 
# Driver Code
arr = [6, 3, 10, 11, 15, 20, 13, 3, 18, 12]
 
# Function Call
print(findLengthOfShortestSubarray(arr))

C#

// C# program for the
// above approach
using System;
 
class GFG {
 
// Find the length of the shortest subarray
static int findLengthOfShortestSubarray(int []arr,
                                        int N)
{
   
  // To store the result
  int minlength = int.MaxValue;
 
  int left = 0;
  int right = N - 1;
 
  // Calculate the possible length of
  // the sorted subarray from left
  while (left < right &&
         arr[left + 1] >= arr[left])
  {
    left++;
  }
 
  // Array is sorted
  if (left == N - 1)
    return 0;
 
  // Calculate the possible length of
  // the sorted subarray from left
  while (right > left &&
         arr[right - 1] <= arr[right])
  {
    right--;
  }
 
  // Update the result
  minlength = Math.Min(N - left - 1,
                       right);
 
  // Calculate the possible length
  // in the middle we can delete
  // and update the result
  int j = right;
   
  for(int i = 0; i < left + 1; i++)
  {
    if (arr[i] <= arr[j])
    {
       
      // Update the result
      minlength = Math.Min(minlength,
                           j - i - 1);
    }
    else if (j < N - 1)
    {
      j++;
    }
    else
    {
      break;
    }
  }
 
  // Return the result
  return minlength;
}
 
// Driver Code
public static void Main(String[] args)
{
  int []arr = { 6, 3, 10, 11, 15,
                20, 13, 3, 18, 12 };
  int N = arr.Length;
 
  // Function call
  Console.Write(findLengthOfShortestSubarray(
                 arr, N));
}
}
 
// This code is contributed by Amit Katiyar

Javascript

 
// Javascript program for the above approach
 
// Find the length of the shortest subarray
function findLengthOfShortestSubarray(arr, N)
{
     
    // To store the result
    let minlength = Number.MAX_VALUE;
     
    let left = 0;
    let right = N - 1;
     
    // Calculate the possible length of
    // the sorted subarray from left
    while (left < right &&
           arr[left + 1] >= arr[left])
    {
        left++;
    }
     
    // Array is sorted
    if (left == N - 1)
        return 0;
     
    // Calculate the possible length of
    // the sorted subarray from left
    while (right > left &&
           arr[right - 1] <= arr[right])
    {
        right--;
    }
     
    // Update the result
    minlength = Math.min(N - left - 1,
                         right);
     
    // Calculate the possible length
    // in the middle we can delete
    // and update the result
    let j = right;
     
    for(let i = 0; i < left + 1; i++)
    {
        if (arr[i] <= arr[j])
        {
             
            // Update the result
            minlength = Math.min(minlength,
                                 j - i - 1);
        }
        else if (j < N - 1)
        {
            j++;
        }
        else
        {
            break;
        }
    }
     
    // Return the result
    return minlength;
}
 
// Driver Code
let arr = [  6, 3, 10, 11, 15,
            20, 13, 3, 18, 12];
let N = arr.length;
 
// Function call
document.write(
    (findLengthOfShortestSubarray(arr, N)));
     
// This code is contributed by souravghosh0416
 

输出：

8

时间复杂度： O(N)
辅助空间： O(1)

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