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，以便对其余数组进行排序。
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.
方法：该问题可以通过以下三种删除子数组的方法来解决：
- Remove the subarray from the left i.e left suffix.
- Remove the subarray from the right ie, prefix.
- Remove the subarray from the middle and merge the left and right part of the array.
- 从左到右遍历arr [] ，找到arr [left]> arr [left + 1]的第一个左索引。因此要删除以使数组排序的子数组长度为N-left-1。
- 如果left == N – 1 ，则此数组已经排序，因此返回0。
- 遍历ARR []由右至左，发现的常用3 [右]
- 现在初始化一个变量mincount并取N-left-1和right的最小值。
- 现在也计算要从数组中间删除的子数组的最小长度：
- 令i = 0，j = right 。并检查是否可以通过比较arr [i]和arr [j]删除介于i和j (不包括)之间的元素。
- 如果arr [j]> = arr [i] ，请尝试使用j – i – 1更新答案，并递增i以收紧窗口。
- 如果arr [j]
，则元素之间不能删除，因此增加j可以松开窗口。 - 遍历循环，直到i> left或j == N。并更新答案。
- 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 ChitranayalPython3
# 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输出：
8
时间复杂度： O(N)
辅助空间： O(1)