Given an integer K and an array of integers arr, the task is to find the maximum element from the array and after every retrieval the number will get decremented by 1. Repeat these steps exactly K number of times and print the sum of all the values retrieved in the end.
Examples:
Input: K = 3, arr[] = {2, 3, 5, 4}
Output: 13
For K = 1, current maximum is 5 (Sum = 5 and arr[] = {2, 3, 4, 4})
For K = 2, current maximum is 4 (Sum = 5 + 4 = 9 and arr[] = {2, 3, 3, 4})
For K = 3, current maximum is 4 (Sum = 9 + 4 = 13 and arr[] = {2, 3, 3, 3})
Hence, the result is 13Input: K = 4, arr[] = {1, 2, 4}
Output: 11
Approach: The main idea is to use a max heap which will have the maximum element at it’s root at any instance of time.
- Create a max heap of all the elements of the array.
- Get the root element of the heap and add it to the sum.
- Pop the root element and decrement it by 1 then insert it again into the heap.
- Repeat the above two steps exactly K number of times.
- Print the total sum in the end.
Below is the implementation of the above approach:
C++
// C++ implementation of above approach#include <bits/stdc++.h>using namespace std;#define ll long longll getSum(int arr[], int K, int n){ ll sum = 0; priority_queue<ll> maxHeap; for (ll i = 0; i < n; i++) { // put all array elements // in a max heap maxHeap.push(arr[i]); } while (K--) { // Get the current maximum element ll currentMax = maxHeap.top(); // Add it to the sum sum += currentMax; // Remove the current max from the heap maxHeap.pop(); // Add the current max back to the // heap after decrementing it by 1 maxHeap.push(currentMax - 1); } return sum;}// driver codeint main(){ int arr[] = { 2, 3, 5, 4 }, K = 3; int n = sizeof(arr) / sizeof(arr[0]); cout << getSum(arr, K, n) << endl;} |
Java
// Java implementation of above approach import java.util.*;class Solution{static int getSum(int arr[], int K, int n) { int sum = 0; PriorityQueue<Integer> maxHeap = new PriorityQueue<Integer>(n,Collections.reverseOrder()); for (int i = 0; i < n; i++) { // put all array elements // in a max heap maxHeap.add(arr[i]); } while (K-->0) { // Get the current maximum element int currentMax = (int)maxHeap.peek(); // Add it to the sum sum += currentMax; // Remove the current max from the heap maxHeap.remove(); // Add the current max back to the // heap after decrementing it by 1 maxHeap.add(currentMax - 1); } return sum; } // driver code public static void main(String args[]) { int arr[] = { 2, 3, 5, 4 }, K = 3; int n =arr.length; System.out.println(getSum(arr, K, n)); } }//contributed by Arnab Kundu |
Python3
# Python3 implementation of above approach# importing the heapq moduleimport heapq# function to get maximum sumdef getSum(arr, K, n): Sum = 0 maxHeap = arr # creating a maxheap heapq._heapify_max(maxHeap) while (K > 0): # Get the current maximum element currentMax = maxHeap[0] # Add it to the sum Sum += currentMax # decrementing the root of the max heap by 1 maxHeap[0] -= 1 # maintaining the heap property # as we have reduced the value of root by 1 # so it may distort the heap property heapq._heapify_max(maxHeap) K -= 1 return Sum# Driver codearr = [2, 3, 5, 4]K = 3n = len(arr)print(getSum(arr, K, n))'''This code is contributed by Rajat Kumar''' |
C#
// C# implementation of above approach using System;using System.Collections.Generic;class GFG { static int getSum(int[] arr, int K, int n) { int sum = 0; List<int> maxHeap = new List<int>(); for (int i = 0; i < n; i++) { // put all array elements // in a max heap maxHeap.Add(arr[i]); } maxHeap.Sort(); maxHeap.Reverse(); while (K-- > 0) { // Get the current maximum element int currentMax = maxHeap[0]; // Add it to the sum sum += currentMax; // Remove the current max from the heap maxHeap.RemoveAt(0); // Add the current max back to the // heap after decrementing it by 1 maxHeap.Add(currentMax - 1); maxHeap.Sort(); maxHeap.Reverse(); } return sum; } // Driver code static void Main() { int[] arr = { 2, 3, 5, 4 }; int K = 3; int n = arr.Length; Console.Write(getSum(arr, K, n)); }}// This code is contributed by divyesh072019. |
Javascript
<script> // Javascript implementation of above approach function getSum(arr, K, n) { let sum = 0; let maxHeap = []; for (let i = 0; i < n; i++) { // put all array elements // in a max heap maxHeap.push(arr[i]); } maxHeap.sort(function(a, b){return a - b}); maxHeap.reverse(); while (K-- > 0) { // Get the current maximum element let currentMax = maxHeap[0]; // Add it to the sum sum += currentMax; // Remove the current max from the heap maxHeap.shift(); // Add the current max back to the // heap after decrementing it by 1 maxHeap.push(currentMax - 1); maxHeap.sort(function(a, b){return a - b}); maxHeap.reverse(); } return sum; } // Driver code let arr = [ 2, 3, 5, 4 ]; let K = 3; let n = arr.length; document.write(getSum(arr, K, n)); // This code is contributed by suresh07.</script> |
Output:
13
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