给定操作可以从给定数组的Kth索引达到的最远索引

给定由N个整数和三个整数X ， Y和K组成的数组arr [] ，任务是找到可以通过以下操作达到的最远索引：

• 如果arr [i]≥arr [i + 1]：从索引i移至i + 1 。
• 如果ARR [I] 由1要么减小量X如果X的值> 0或递减Y(ARR第[i + 1] – ARR [I])，如果Y的值>(改编[i + 1] – arr [i]) 。

例子：

Input: arr[] = {4, 2, 7, 6, 9, 14, 12}, X = 1, Y = 5, K = 0
Output: 4
Explanation: 
Initially, K = 0.
arr[0] > arr[1]: Therefore, move to index 1.
arr[1] < arr[2]: Decrement X by 1 and move to index 2. Now X = 0.
arr[2] > arr[3]: Move to index 3.
arr[3] < arr[4]: Decrement Y by 3 and move to index 4. Now Y = 2
arr[4] < arr[5]: Neither X > 0 nor Y > 5. Hence, it is not possible to move to the next index.
Therefore, the maximum index that can be reached is 4.

Input: arr[] = {14, 3, 19, 3}, X = 17, Y = 0, K = 1
Output: 3

方法：想法是使用X表示索引之间的最大差异，使用Y表示其余差异。请按照以下步骤解决此问题：

• 声明优先级队列。
• 遍历给定的数组arr []并执行以下操作：
  • 如果当前元素( arr [i] )大于下一个元素( arr [i + 1] )，则移至下一个索引。
  • 否则，将(arr [i + 1] – arr [i])之差推入优先级队列。
  • 如果优先级队列的大小大于Y ，则将X减少优先级队列的顶部元素并弹出该元素。
  • 如果X小于0 ，则可以达到的最远索引是i 。
• 完成上述步骤后，如果X的值至少为0 ，则可以达到的最远索引为(N – 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 farthest index
// that can be reached
void farthestHill(int arr[], int X,
                  int Y, int N, int K)
{
    int i, diff;
 
    // Declare a priority queue
    priority_queue pq;
 
    // Iterate the array
    for (i = K; i < N - 1; i++) {
 
        // If current element is
        // greater than the next element
        if (arr[i] >= arr[i + 1])
            continue;
 
        // Otherwise, store their difference
        diff = arr[i + 1] - arr[i];
 
        // Push diff into pq
        pq.push(diff);
 
        // If size of pq exceeds Y
        if (pq.size() > Y) {
 
            // Decrease X by the
            // top element of pq
            X -= pq.top();
 
            // Remove top of pq
            pq.pop();
        }
 
        // If X is exhausted
        if (X < 0) {
 
            // Current index is the
            // farthest possible
            cout << i;
            return;
        }
    }
 
    // Print N-1 as farthest index
    cout << N - 1;
}
 
// Driver Code
int main()
{
    int arr[] = { 4, 2, 7, 6, 9, 14, 12 };
    int X = 5, Y = 1;
    int K = 0;
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function Call
    farthestHill(arr, X, Y, N, K);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
 
class GFG{
     
// Function to find the farthest index
// that can be reached
public static void farthestHill(int arr[], int X,
                                int Y, int N, int K)
{
    int i, diff;
     
    // Declare a priority queue
    PriorityQueue pq = new PriorityQueue();
   
    // Iterate the array
    for(i = K; i < N - 1; i++)
    {
         
        // If current element is
        // greater than the next element
        if (arr[i] >= arr[i + 1])
            continue;
   
        // Otherwise, store their difference
        diff = arr[i + 1] - arr[i];
   
        // Push diff into pq
        pq.add(diff);
   
        // If size of pq exceeds Y
        if (pq.size() > Y)
        {
             
            // Decrease X by the
            // top element of pq
            X -= pq.peek();
   
            // Remove top of pq
            pq.poll();
        }
   
        // If X is exhausted
        if (X < 0)
        {
             
            // Current index is the
            // farthest possible
            System.out.print(i);
            return;
        }
    }
   
    // Print N-1 as farthest index
    System.out.print(N - 1);
}
 
// Driver code
public static void main(String[] args)
{
    int arr[] = { 4, 2, 7, 6, 9, 14, 12 };
    int X = 5, Y = 1;
    int K = 0;
    int N = arr.length;
   
    // Function Call
    farthestHill(arr, X, Y, N, K);
}
}
 
// This code is contributed by divyeshrabadiya07

Python3

# Python3 program for the above approach
  
# Function to find the farthest index
# that can be reached
def farthestHill(arr, X, Y, N, K):
     
    # Declare a priority queue
    pq = []
  
    # Iterate the array
    for i in range(K, N - 1, 1):
  
        # If current element is
        # greater than the next element
        if (arr[i] >= arr[i + 1]):
            continue
  
        # Otherwise, store their difference
        diff = arr[i + 1] - arr[i]
  
        # Push diff into pq
        pq.append(diff)
  
        # If size of pq exceeds Y
        if (len(pq) > Y):
  
            # Decrease X by the
            # top element of pq
            X -= pq[-1]
  
            # Remove top of pq
            pq[-1]
         
        # If X is exhausted
        if (X < 0):
             
            # Current index is the
            # farthest possible
            print(i)
            return
         
    # Print N-1 as farthest index
    print(N - 1)
 
# Driver Code
arr = [ 4, 2, 7, 6, 9, 14, 12 ]
X = 5
Y = 1
K = 0
N = len(arr)
  
# Function Call
farthestHill(arr, X, Y, N, K)
 
# This code is contributed by code_hunt

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to find the farthest index
// that can be reached
public static void farthestHill(int[] arr, int X,
                                int Y, int N, int K)
{
    int i, diff;
     
    // Declare a priority queue
    List pq = new List();
     
    // Iterate the array
    for(i = K; i < N - 1; i++)
    {
         
        // If current element is
        // greater than the next element
        if (arr[i] >= arr[i + 1])
            continue;
             
        // Otherwise, store their difference
        diff = arr[i + 1] - arr[i];
         
        // Push diff into pq
        pq.Add(diff);
        pq.Sort();
        pq.Reverse();
         
        // If size of pq exceeds Y
        if (pq.Count > Y)
        {
             
            // Decrease X by the
            // top element of pq
            X -= pq[0];
             
            // Remove top of pq
            pq.RemoveAt(0);
        }
 
        // If X is exhausted
        if (X < 0)
        {
             
            // Current index is the
            // farthest possible
            Console.Write(i);
            return;
        }
    }
     
    // Print N-1 as farthest index
    Console.Write(N - 1);
}
 
// Driver code
public static void Main(String[] args)
{
    int[] arr = { 4, 2, 7, 6, 9, 14, 12 };
    int X = 5, Y = 1;
    int K = 0;
    int N = arr.Length;
     
    // Function Call
    farthestHill(arr, X, Y, N, K);
}
}
 
// This code is contributed by gauravrajput1

输出：

4

时间复杂度： O(N * log(E))，其中E是优先级队列中元素的最大数量。
辅助空间： O(E)