需要删除的最小子数组的长度以使剩余元素连续

给定一个由N 个整数组成的数组arr[] ，任务是找到需要移除的最小子数组的长度，以使剩余的数组元素连续。

例子：

Input: arr[] = {1, 2, 3, 7, 5, 4, 5}
Output: 2
Explanation:
Removing the subarray {7, 5} from the array arr[] modifies the array to {1, 2, 3, 4, 5}, which makes all array elements consecutive. Therefore, the length of the subarray removed is 2, which is minimum.

Input: arr[] = {4, 5, 6, 8, 9, 10}
Output: 3

朴素的方法：解决给定问题的最简单方法是删除生成数组arr[] 的所有可能子数组，并检查它们中的每一个，检查它们的删除是否使剩余的数组元素连续。检查所有子数组后，打印得到满足条件的最小子数组的长度。

时间复杂度： O(N ³ )
辅助空间： O(1)

高效的方法：上述方法可以通过存储连续元素的最长前缀和后缀的长度，然后找到需要删除的子数组的最小长度进行优化，使得前缀和后缀的串联形成一个连续元素的序列.
请按照以下步骤解决问题：

• 初始化两个变量，假设L为0 ， R为(N – 1)分别存储连续元素最长前缀的结束索引和最长后缀的起始索引。
• 将L的值更新为第一个索引，其中arr[i] + 1不等于arr[i + 1] ，使得arr[0, …, L]是一个连续的前缀数组。
• 将R的值更新到从末尾开始的第一个索引，其中arr[i]不等于arr[i – 1] + 1 ，使得arr[R, …, N – 1]是一个连续的后缀数组。
• 初始化一个变量，比如ans，以存储(N – L – 1)和R的最小值 存储所需的结果。
• 如果arr[R] ≤ arr[L] + 1 的值，则存储右索引， R1为arr[0, …, L, R1, …, N – 1]是一个连续数组。
• 如果(R1 – L – 1)的值小于ANS的值，然后更新ANS至(R1 – L – 1)的值。
• 完成以上步骤后，打印ans的值作为结果。

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 length of the
// smallest subarray to be removed to
// make remaining array elements consecutive
void shortestSubarray(int* A, int N)
{
    int i;
 
    // Store the ending index of the
    // longest prefix consecutive array
    int left_index;
 
    // Traverse the array to find the
    // longest prefix consecutive sequence
    for (i = 0; i < N - 1; i++) {
        if (A[i] + 1 != A[i + 1])
            break;
    }
 
    // A[0...left_index] is the
    // prefix consecutive sequence
    left_index = i;
 
    // Store the starting index of the
    // longest suffix consecutive sequence
    int right_index;
 
    // Traverse the array to find the
    // longest suffix consecutive sequence
    for (i = N - 1; i >= 1; i--) {
        if (A[i] != A[i - 1] + 1)
            break;
    }
 
    // A[right_index...N-1] is
    // the consecutive sequence
    right_index = i;
 
    int updated_right;
 
    // Store the smallest subarray
    // required to be removed
    int minLength = min(N - left_index - 1,
                        right_index);
 
    // Check if subarray from the
    // middle can be removed
    if (A[right_index]
        <= A[left_index] + 1) {
 
        // Update the right index s.t.
        // A[0, N-1] is consecutive
        updated_right = right_index
                        + A[left_index]
                        - A[right_index] + 1;
 
        // If updated_right < N, then
        // update the minimumLength
        if (updated_right < N)
            minLength = min(minLength,
                            updated_right
                                - left_index - 1);
    }
 
    // Print the required result
    cout << minLength;
}
 
// Driver Code
int main()
{
    int arr[] = { 1, 2, 3, 7, 4, 3, 5 };
    int N = sizeof(arr) / sizeof(arr[0]);
    shortestSubarray(arr, N);
 
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
 
class GFG{
 
// Function to find the length of the
// smallest subarray to be removed to
// make remaining array elements consecutive
static void shortestSubarray(int A[], int N)
{
    int i;
 
    // Store the ending index of the
    // longest prefix consecutive array
    int left_index;
 
    // Traverse the array to find the
    // longest prefix consecutive sequence
    for(i = 0; i < N - 1; i++)
    {
        if (A[i] + 1 != A[i + 1])
            break;
    }
 
    // A[0...left_index] is the
    // prefix consecutive sequence
    left_index = i;
 
    // Store the starting index of the
    // longest suffix consecutive sequence
    int right_index;
 
    // Traverse the array to find the
    // longest suffix consecutive sequence
    for(i = N - 1; i >= 1; i--)
    {
        if (A[i] != A[i - 1] + 1)
            break;
    }
 
    // A[right_index...N-1] is
    // the consecutive sequence
    right_index = i;
 
    int updated_right;
 
    // Store the smallest subarray
    // required to be removed
    int minLength = Math.min(N - left_index - 1,
                             right_index);
 
    // Check if subarray from the
    // middle can be removed
    if (A[right_index] <= A[left_index] + 1)
    {
         
        // Update the right index s.t.
        // A[0, N-1] is consecutive
        updated_right = right_index + A[left_index] -
                     A[right_index] + 1;
 
        // If updated_right < N, then
        // update the minimumLength
        if (updated_right < N)
            minLength = Math.min(minLength,
                                 updated_right -
                                 left_index - 1);
    }
 
    // Print the required result
    System.out.println(minLength);
}
 
// Driver Code
public static void main(String[] args)
{
    int arr[] = { 1, 2, 3, 7, 4, 3, 5 };
    int N = arr.length;
     
    shortestSubarray(arr, N);
}
}
 
// This code is contributed by Kingash

Python3

# Python3 program for the above approach
 
# Function to find the length of the
# smallest subarray to be removed to
# make remaining array elements consecutive
def shortestSubarray(A, N):
     
    i = 0
 
    # Store the ending index of the
    # longest prefix consecutive array
    left_index = 0
 
    # Traverse the array to find the
    # longest prefix consecutive sequence
    for i in range(N - 1):
        if (A[i] + 1 != A[i + 1]):
            break
 
    # A[0...left_index] is the
    # prefix consecutive sequence
    left_index = i
 
    # Store the starting index of the
    # longest suffix consecutive sequence
    right_index = 0
 
    # Traverse the array to find the
    # longest suffix consecutive sequence
    i = N - 1
     
    while (i >= 1):
        if (A[i] != A[i - 1] + 1):
            break
         
        i -= 1
 
    # A[right_index...N-1] is
    # the consecutive sequence
    right_index = i
 
    updated_right = 0
 
    # Store the smallest subarray
    # required to be removed
    minLength = min(N - left_index - 1, right_index)
 
    # Check if subarray from the
    # middle can be removed
    if (A[right_index] <= A[left_index] + 1):
         
        # Update the right index s.t.
        # A[0, N-1] is consecutive
        updated_right = (right_index + A[left_index] -
                                       A[right_index] + 1)
 
        # If updated_right < N, then
        # update the minimumLength
        if (updated_right < N):
            minLength = min(minLength, updated_right -
                                       left_index - 1)
 
    # Print the required result
    print(minLength)
 
# Driver Code
if __name__ == '__main__':
     
    arr = [ 1, 2, 3, 7, 4, 3, 5 ]
    N = len(arr)
     
    shortestSubarray(arr, N)
 
# This code is contributed by SURENDRA_GANGWAR

C#

// C# program for the above approach
using System;
 
class GFG{
     
// Function to find the length of the
// smallest subarray to be removed to
// make remaining array elements consecutive
static void shortestSubarray(int[] A, int N)
{
    int i;
 
    // Store the ending index of the
    // longest prefix consecutive array
    int left_index;
 
    // Traverse the array to find the
    // longest prefix consecutive sequence
    for(i = 0; i < N - 1; i++)
    {
        if (A[i] + 1 != A[i + 1])
            break;
    }
 
    // A[0...left_index] is the
    // prefix consecutive sequence
    left_index = i;
 
    // Store the starting index of the
    // longest suffix consecutive sequence
    int right_index;
 
    // Traverse the array to find the
    // longest suffix consecutive sequence
    for(i = N - 1; i >= 1; i--)
    {
        if (A[i] != A[i - 1] + 1)
            break;
    }
 
    // A[right_index...N-1] is
    // the consecutive sequence
    right_index = i;
 
    int updated_right;
 
    // Store the smallest subarray
    // required to be removed
    int minLength = Math.Min(N - left_index - 1,
                             right_index);
 
    // Check if subarray from the
    // middle can be removed
    if (A[right_index] <= A[left_index] + 1)
    {
         
        // Update the right index s.t.
        // A[0, N-1] is consecutive
        updated_right = right_index + A[left_index] -
                     A[right_index] + 1;
 
        // If updated_right < N, then
        // update the minimumLength
        if (updated_right < N)
            minLength = Math.Min(minLength,
                                 updated_right -
                                 left_index - 1);
    }
 
    // Print the required result
    Console.WriteLine(minLength);
}
 
// Driver code
static public void Main()
{
    int[] arr = { 1, 2, 3, 7, 4, 3, 5 };
    int N = arr.Length;
     
    shortestSubarray(arr, N);
}
}
 
// This code is contributed by offbeat

Javascript

 
// JavaScript program to implement
// the above approach
 
// Function to find the length of the
// smallest subarray to be removed to
// make remaining array elements consecutive
function shortestSubarray(A, N)
{
    let i;
  
    // Store the ending index of the
    // longest prefix consecutive array
    let left_index;
  
    // Traverse the array to find the
    // longest prefix consecutive sequence
    for(i = 0; i < N - 1; i++)
    {
        if (A[i] + 1 != A[i + 1])
            break;
    }
  
    // A[0...left_index] is the
    // prefix consecutive sequence
    left_index = i;
  
    // Store the starting index of the
    // longest suffix consecutive sequence
    let right_index;
  
    // Traverse the array to find the
    // longest suffix consecutive sequence
    for(i = N - 1; i >= 1; i--)
    {
        if (A[i] != A[i - 1] + 1)
            break;
    }
  
    // A[right_index...N-1] is
    // the consecutive sequence
    right_index = i;
  
    let updated_right;
  
    // Store the smallest subarray
    // required to be removed
    let minLength = Math.min(N - left_index - 1,
                             right_index);
  
    // Check if subarray from the
    // middle can be removed
    if (A[right_index] <= A[left_index] + 1)
    {
          
        // Update the right index s.t.
        // A[0, N-1] is consecutive
        updated_right = right_index + A[left_index] -
                     A[right_index] + 1;
  
        // If updated_right < N, then
        // update the minimumLength
        if (updated_right < N)
            minLength = Math.min(minLength,
                                 updated_right -
                                 left_index - 1);
    }
  
    // Prlet the required result
    document.write(minLength);
}
 
// Driver code
    let arr = [ 1, 2, 3, 7, 4, 3, 5 ];
    let N = arr.length;
      
    shortestSubarray(arr, N);
 
// This code is contributed by susmitakundugoaldanga.

输出：

4

时间复杂度： O(N)
辅助空间： O(1)

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