查找之前至少有 K 个非递增元素和之后至少有 K 个非递减元素的索引

给定一个大小为N的数组arr[]和一个整数K ，任务是找到给定数组中的所有索引，这些索引之前至少有K个非递增元素，之后至少有K个非递减元素。

例子：

Input: arr[] = {1, 1, 1, 1, 1}, K = 0
Output: 0 1 2 3 4
Explanation: Since K equals 0, every index satisfies the condition.

Input: arr[] = {1, 2, 3, 4, 5, 6}, K = 2
Output: -1
Explanation: No index has 2 non-increasing before it and 2 non-decreasing elements after it.

方法：可以通过使用前缀和后缀数组的概念找到解决方案。请按照以下步骤操作：

1. 形成prefix[]数组，其中prefix[i]表示i之前的元素个数，服从非递增顺序。
2. 形成suffix[]数组，其中suffix[i]表示i之后的元素个数，且服从非递减顺序。
3. 现在只有索引应该包含在 answer 中， prefix[i]和suffix[i]都大于或等于K 。

下面是上述方法的实现。

C++

// C++ code for the above approach
#include 
using namespace std;
 
// Function to find all the indices
vector findIndices(int arr[], int K,
                        int N)
{
    vector prefix(N), suffix(N);
    vector ans;
 
    prefix[0] = 0;
    for (int i = 1; i < N; i++) {
        if (arr[i] <= arr[i - 1])
            prefix[i] = prefix[i - 1] + 1;
        else
            prefix[i] = 0;
    }
 
    suffix[N - 1] = 0;
    for (int i = N - 2; i >= 0; i--) {
        if (arr[i] <= arr[i + 1])
            suffix[i] = suffix[i + 1] + 1;
        else
            suffix[i] = 0;
    }
 
    for (int i = 0; i < N; i++) {
        if (prefix[i] >= K && suffix[i] >= K)
            ans.push_back(i);
    }
 
    return ans;
}
 
// Driver code
int main()
{
    int arr[] = { 1, 1, 1, 1, 1 };
    int K = 0;
    int N = sizeof(arr) / sizeof(arr[0]);
    vector ans = findIndices(arr, K, N);
    for (int i = 0; i < ans.size(); i++) {
        cout << ans[i] << " ";
    }
    if (ans.size() == 0)
        cout << "-1";
    return 0;
}

Java

// Java code for the above approach
import java.util.ArrayList;
class GFG {
 
  // Function to find all the indices
  static ArrayList findIndices(int[] arr, int K, int N)
  {
    int[] prefix = new int[N];
 
    int[] suffix = new int[N];
    ArrayList ans = new ArrayList();
 
    prefix[0] = 0;
    for (int i = 1; i < N; i++) {
      if (arr[i] <= arr[i - 1])
        prefix[i] = prefix[i - 1] + 1;
      else
        prefix[i] = 0;
    }
 
    suffix[N - 1] = 0;
    for (int i = N - 2; i >= 0; i--) {
      if (arr[i] <= arr[i + 1])
        suffix[i] = suffix[i + 1] + 1;
      else
        suffix[i] = 0;
    }
 
    for (int i = 0; i < N; i++) {
      if (prefix[i] >= K && suffix[i] >= K)
        ans.add(i);
    }
 
    return ans;
  }
 
  // Driver code
  public static void main(String args[])
  {
    int[] arr = { 1, 1, 1, 1, 1 };
    int K = 0;
    int N = arr.length;
    ArrayList ans = findIndices(arr, K, N);
    for (int i = 0; i < ans.size(); i++) {
      System.out.print(ans.get(i) + " ");
    }
    if (ans.size() == 0)
      System.out.println("-1");
  }
}
 
// This code is contributed by gfgking

Python3

# Python code for the above approach
 
# Function to find all the indices
def findIndices (arr, K, N):
    prefix = [0] * N
    suffix = [0] * N
    ans = [];
 
    prefix[0] = 0;
    for i in range(1, N):
        if (arr[i] <= arr[i - 1]):
            prefix[i] = prefix[i - 1] + 1;
        else:
            prefix[i] = 0;
 
 
    suffix[N - 1] = 0;
    for i in range(N - 2, 1, -1):
        if (arr[i] <= arr[i + 1]):
            suffix[i] = suffix[i + 1] + 1;
        else:
            suffix[i] = 0;
 
 
    for i in range(N):
        if (prefix[i] >= K and suffix[i] >= K):
            ans.append(i);
 
    return ans;
 
# Driver code
arr = [1, 1, 1, 1, 1];
K = 0;
N = len(arr)
ans = findIndices(arr, K, N);
for i in range(len(ans)):
    print(ans[i], end=" ");
if (len(ans) == 0):
    print("-1");
 
# This code is contributed by Saurabh Jaiswal

C#

// C# code for the above approach
using System;
using System.Collections.Generic;
 
class GFG {
 
  // Function to find all the indices
  static List findIndices(int[] arr, int K, int N)
  {
    int[] prefix = new int[N];
 
    int[] suffix = new int[N];
    List ans = new List();
 
    prefix[0] = 0;
    for (int i = 1; i < N; i++) {
      if (arr[i] <= arr[i - 1])
        prefix[i] = prefix[i - 1] + 1;
      else
        prefix[i] = 0;
    }
 
    suffix[N - 1] = 0;
    for (int i = N - 2; i >= 0; i--) {
      if (arr[i] <= arr[i + 1])
        suffix[i] = suffix[i + 1] + 1;
      else
        suffix[i] = 0;
    }
 
    for (int i = 0; i < N; i++) {
      if (prefix[i] >= K && suffix[i] >= K)
        ans.Add(i);
    }
 
    return ans;
  }
 
  // Driver code
  public static void Main()
  {
    int[] arr = { 1, 1, 1, 1, 1 };
    int K = 0;
    int N = arr.Length;
    List ans = findIndices(arr, K, N);
    for (int i = 0; i < ans.Count; i++) {
      Console.Write(ans[i] + " ");
    }
    if (ans.Count == 0)
      Console.Write("-1");
  }
}
 
// This code is contributed by ukasp

Javascript

    // JavaScript code for the above approach
 
    // Function to find all the indices
    const findIndices = (arr, K, N) => {
        let prefix = new Array(N).fill(0);
        let suffix = new Array(N).fill(0);
        let ans = [];
 
        prefix[0] = 0;
        for (let i = 1; i < N; i++) {
            if (arr[i] <= arr[i - 1])
                prefix[i] = prefix[i - 1] + 1;
            else
                prefix[i] = 0;
        }
 
        suffix[N - 1] = 0;
        for (let i = N - 2; i >= 0; i--) {
            if (arr[i] <= arr[i + 1])
                suffix[i] = suffix[i + 1] + 1;
            else
                suffix[i] = 0;
        }
 
        for (let i = 0; i < N; i++) {
            if (prefix[i] >= K && suffix[i] >= K)
                ans.push(i);
        }
 
        return ans;
    }
 
    // Driver code
    let arr = [1, 1, 1, 1, 1];
    let K = 0;
    let N = arr.length;
    let ans = findIndices(arr, K, N);
    for (let i = 0; i < ans.length; i++) {
        document.write(`${ans[i]} `);
    }
    if (ans.length == 0)
        cout << "-1";
 
// This code is contributed by rakeshsahni
 

输出

0 1 2 3 4 

时间复杂度： 在）
辅助空间： O(N)