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Farthest index that can be reached from the Kth index of given array by given operations
  • Last Updated : 07 Dec, 2020

Given an array arr[] consisting of N integers and three integers X, Y, and K, the task is to find the farthest index that can be reached by the following operations:

  • If arr[i] ≥ arr[i + 1]: Move from index i to i + 1.
  • If arr[i] < arr[i+1]: Either decrement X by 1 if the value of X > 0 or decrement Y by (arr[i + 1] – arr[i]) if the value of Y > (arr[i+1] – arr[i]).

Examples:

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

Approach: The idea is to use X for the maximum difference between indexes and Y for the remaining difference. Follow the steps below to solve this problem:



  • Declare a priority queue.
  • Traverse the given array arr[] and perform the following operations:
    • If the current element (arr[i]) is greater than the next element (arr[i + 1]), then move to the next index.
    • Otherwise, push the difference of (arr[i + 1] – arr[i]) into the priority queue.
    • If the size of the priority queue is greater than Y, then decrement X by the top element of the priority queue and pop that element.
    • If X is less than 0, the farthest index that can be reached is i.
  • After completing the above steps, if the value of X is at least 0, then the farthest index that can be reached is (N – 1).

Below is the implementation of the above approach:

C++

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// C++ program for the above approach
 
#include <bits/stdc++.h>
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<int> 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;
}

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Java

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// 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<Integer> pq = new PriorityQueue<Integer>();
   
    // 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

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Python3

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# 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

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C#

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// 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<int> pq = new List<int>();
     
    // 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

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Output: 

4

 

Time Complexity: O(N*log(E)), where E is the maximum number of elements in the priority queue.
Auxiliary Space: O(E)

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