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Maximum number of apples that can be eaten by a person
  • Difficulty Level : Hard
  • Last Updated : 23 Feb, 2021

Given two arrays apples[] and days[] representing the count of apples an apple tree produces and the number of days these apples are edible from the ith day respectively, the task is to find the maximum number of apples a person can eat if the person can eat at most one apple in a day.

Examples

Input: apples[] = { 1, 2, 3, 5, 2 }, days[] = { 3, 2, 1, 4, 2 } 
Output:
Explanation: 
On 1st day, person eats the apple produced by apple tree on the 1st day. 
On 2nd day, person eats the apple produced by apple tree on the 2nd day. 
On 3rd day, person eats the apple produced by apple tree on the 2nd day. 
On 4th to 7th day, person eats the apple produced by apple tree on the 4th day.

Input: apples[] = { 3, 0, 0, 0, 0, 2 }, days[] = { 3, 0, 0, 0, 0, 2 } 
Output:
 

Approach: The idea is to eat the apples having the closest expiration date. Follow the steps below to solve the problem:



Below is the implementation of the above approach:
 

C++




// C++ program of the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the maximum number of apples
// a person can eat such that the person eat
// at most one apple in a day.
int cntMaxApples(vector<int> apples, vector<int> days)
{
 
    // Stores count of apples and number
    // of days those apples are edible
    typedef pair<int, int> P;
 
    // Store count of apples and number
    // of days those apples are edible
    priority_queue<P, vector
 
<p>, greater
 
<p> > pq;
 
    // Stores indices of the array
    int i = 0;
 
    // Stores count of days
    int n = apples.size();
 
    // Stores maximum count of
    // edible apples
    int total_apples = 0;
 
    // Traverse both the arrays
    while (i < n || !pq.empty()) {
 
        // If top element of the apple
        // is not already expired
        if (i < n && apples[i] != 0) {
 
            // Insert count of apples and
            // their expiration date
            pq.push({ i + days[i] - 1, apples[i] });
        }
 
        // Remove outdated apples
        while (!pq.empty() && pq.top().first < i) {
            pq.pop();
        }
 
        // Insert all the apples produces by
        // tree on current day
        if (!pq.empty()) {
 
            // Stores top element of pq
            auto curr = pq.top();
 
            // Remove top element of pq
            pq.pop();
 
            // If count of apples in curr
            // is greater than 0
            if (curr.second > 1) {
 
                // Insert count of apples and
                // their expiration date
                pq.push({ curr.first,
                          curr.second - 1 });
            }
 
            // Update total_apples
            ++total_apples;
        }
 
        // Update index
        ++i;
    }
 
    return total_apples;
}
 
// Driver Code
int main()
{
    vector<int> apples = { 1, 2, 3, 5, 2 };
    vector<int> days = { 3, 2, 1, 4, 2 };
    cout << cntMaxApples(apples, days);
    return 0;
}

Python3




# Python3 program of the above approach
 
# Function to find the maximum number of apples
# a person can eat such that the person eat
# at most one apple in a day.
def cntMaxApples(apples, days) :
 
    # Store count of apples and number
    # of days those apples are edible
    pq = []
     
    # Stores indices of the array
    i = 0
     
    # Stores count of days
    n = len(apples)
     
    # Stores maximum count of
    # edible apples
    total_apples = 0
     
    # Traverse both the arrays
    while (i < n or len(pq) > 0) :
     
      # If top element of the apple
      # is not already expired
      if (i < n and apples[i] != 0) :
     
        # Insert count of apples and
        # their expiration date
        pq.append([i + days[i] - 1, apples[i]])
        pq.sort()
     
      # Remove outdated apples
      while (len(pq) > 0 and pq[0][0] < i) :
        pq.pop(0)
     
      # Insert all the apples produces by
      # tree on current day
      if (len(pq) > 0) :
     
        # Stores top element of pq
        curr = pq[0]
     
        # Remove top element of pq
        pq.pop(0)
     
        # If count of apples in curr
        # is greater than 0
        if (len(curr) > 1) :
     
          # Insert count of apples and
          # their expiration date
          pq.append([curr[0], curr[1] - 1])
          pq.sort()
     
        # Update total_apples
        total_apples += 1
     
      # Update index
      i += 1
     
    return total_apples
     
apples = [ 1, 2, 3, 5, 2 ]
days = [ 3, 2, 1, 4, 2 ]
 
print(cntMaxApples(apples, days))
 
# This code is contributed by divyesh072019

C#




// C# program of the above approach
using System;
using System.Collections.Generic;
class GFG {
 
  // Function to find the maximum number of apples
  // a person can eat such that the person eat
  // at most one apple in a day.
  static int cntMaxApples(int[] apples, int[] days)
  {
 
    // Store count of apples and number
    // of days those apples are edible
    List<Tuple<int,int>> pq = new List<Tuple<int,int>>();
 
    // Stores indices of the array
    int i = 0;
 
    // Stores count of days
    int n = apples.Length;
 
    // Stores maximum count of
    // edible apples
    int total_apples = 0;
 
    // Traverse both the arrays
    while (i < n || pq.Count > 0) {
 
      // If top element of the apple
      // is not already expired
      if (i < n && apples[i] != 0) {
 
        // Insert count of apples and
        // their expiration date
        pq.Add(new Tuple<int,int>(i + days[i] - 1, apples[i]));
        pq.Sort();
      }
 
      // Remove outdated apples
      while (pq.Count > 0 && pq[0].Item1 < i) {
        pq.RemoveAt(0);
      }
 
      // Insert all the apples produces by
      // tree on current day
      if (pq.Count > 0) {
 
        // Stores top element of pq
        Tuple<int,int> curr = pq[0];
 
        // Remove top element of pq
        pq.RemoveAt(0);
 
        // If count of apples in curr
        // is greater than 0
        if (curr.Item2 > 1) {
 
          // Insert count of apples and
          // their expiration date
          pq.Add(new Tuple<int,int>(curr.Item1, curr.Item2 - 1));
          pq.Sort();
        }
 
        // Update total_apples
        ++total_apples;
      }
 
      // Update index
      ++i;
    }
 
    return total_apples;
  }   
 
  // Driver code
  static void Main() {
    int[] apples = { 1, 2, 3, 5, 2 };
    int[] days = { 3, 2, 1, 4, 2 };
 
    Console.Write(cntMaxApples(apples, days));
  }
}
 
// This code is contributed by divyeshrabadiya07.
Output: 
7

 

Time Complexity: O(NlogN) 
Auxiliary Space: O(N)
 

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