Given an array arr[] consisting of N positive integers and an integer K. In one operation, select an array element, add it to the sum and then decrement it by 1. The task is to print the maximum sum that can be obtained by performing the operation K times.
Examples:
Input: arr[] = {2, 5}, K = 4
Output: 14
Explanation:
Perform the following operations to maximize the sum:
Operation 1: Select 5, then reduce it, so new array becomes {2, 4}.
Operation 2: Select 4, then reduce it, so new array becomes {2, 3}.
Operation 3: Select 3, then reduce it, so new array becomes {2, 2}.
Operation 4: Select 2, then reduce it, so new array becomes {2, 1}.
Therefore, the maximum sum is 5 + 4 + 3 + 2 = 14.Input: arr[] = {2, 8, 4, 10, 6}, K = 2
Output: 19
Explanation:
Perform the following operations to maximize the sum:
Operation 1: Select 10, then reduce it, so new array becomes {2, 8, 4, 9, 6}.
Operation 2: Select 9, then reduce it, so new array becomes {2, 8, 4, 8, 6}.
Therefore, the maximum sum is 10 + 9 = 19.
Naive approach: The simplest approach is to use Max Heap to choose the maximum element each time. Add the maximum element to the sum and remove it from the Max Heap and add (maximum element – 1). Perform this operation K and print the sum.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to find maximum possible// after adding elements K times and// decrementing each added value by 1long maxSum(vector<int> arr, int k){ // Stores the maximum sum long max_sum = 0; // Create max_heap to get // the maximum element priority_queue<int> max_heap; // Update the max_heap for(int t : arr) max_heap.push(t); // Calculate the max_sum while (k-- > 0) { int tmp = max_heap.top(); max_heap.pop(); max_sum += tmp; max_heap.push(tmp - 1); } // Return the maximum sum return max_sum;}// Driver codeint main(){ // Given an array arr[] vector<int> arr = { 2, 5 }; // Given K int K = 4; // Function Call cout << maxSum(arr, K);}// This code is contributed by mohit kumar 29 |
Java
// Java program for the above approachimport java.util.*;class GFG { // Function to find maximum possible // after adding elements K times and // decrementing each added value by 1 public static long maxSum(int[] arr, int k) { // Stores the maximum sum long max_sum = 0; // Create max_heap to get // the maximum element PriorityQueue<Integer> max_heap = new PriorityQueue<>( Collections.reverseOrder()); // Update the max_heap for (int t : arr) max_heap.add(t); // Calculate the max_sum while (k-- > 0) { int tmp = max_heap.poll(); max_sum += tmp; max_heap.add(tmp - 1); } // Return the maximum sum return max_sum; } // Driver Code public static void main(String[] args) { // Given an array arr[] int[] arr = { 2, 5 }; // Given K int K = 4; // Function Call System.out.println(maxSum(arr, K)); }} |
Python3
# Python3 program for the above approach# Function to find maximum possible# after adding elements K times and# decrementing each added value by 1from heapq import heappop, heappush, heapifydef maxSum(arr, k): # Stores the maximum sum max_sum = 0 # Create max_heap to get # the maximum element max_heap = [] heapify(max_heap) # Update the max_heap for i in range(len(arr) - 1, 0, -1): heappush(max_heap, arr[i]) # Calculate the max_sum while (k > 0): tmp = heappop(max_heap) max_sum += tmp #max_heap.put(tmp - 1); heappush(max_heap, tmp - 1) k -= 1 # Return the maximum sum return max_sum# Driver Codeif __name__ == '__main__': # Given an array arr arr = [ 2, 5 ] # Given K K = 4 # Function Call print(maxSum(arr, K))# This code is contributed by gauravrajput1 |
C#
// C# program for the// above approachusing System;using System.Collections.Generic;class GFG{// Function to find maximum possible// after adding elements K times and// decrementing each added value by 1public static long maxSum(int[] arr, int k){ // Stores the maximum sum long max_sum = 0; // Create max_heap to get // the maximum element List<int> max_heap = new List<int>(); // Update the max_heap foreach(int t in arr) max_heap.Add(t); max_heap.Sort(); max_heap.Reverse(); // Calculate the max_sum while (k-- > 1) { int tmp = max_heap[0]; max_heap.Remove(0); max_sum += tmp; max_heap.Add(tmp - 1); max_heap.Sort(); max_heap.Reverse(); } // Return the maximum sum return max_sum - 1;}// Driver Codepublic static void Main(String[] args){ // Given an array []arr int[] arr = { 2, 5 }; // Given K int K = 4; // Function Call Console.WriteLine(maxSum(arr, K));}}// This code is contributed by gauravrajput1 |
Javascript
<script>// Javascript program for the above approach// Function to find maximum possible// after adding elements K times and// decrementing each added value by 1function maxSum(arr, k){ // Stores the maximum sum let max_sum = 0; // Create max_heap to get // the maximum element let max_heap = []; // Update the max_heap for(let t = 0; t < arr.length; t++) max_heap.push(arr[t]); max_heap.sort(function(a, b){return b - a;}); // Calculate the max_sum while (k-- > 0) { let tmp = max_heap.shift(); max_sum += tmp; max_heap.push(tmp - 1); max_heap.sort(function(a, b){return b - a;}); } // Return the maximum sum return max_sum;}// Driver Code// Given an array arr[]let arr = [ 2, 5 ];// Given Klet K = 4;// Function Calldocument.write(maxSum(arr, K));// This code is contributed by avanitrachhadiya2155</script> |
Output:
14
Time Complexity: O(K*log(N)), where N is the length of the given array and K is the given number of operations.
Auxiliary Space: O(N)
Efficient Approach: The idea is to use the Hashing concept by storing the frequency of each element of the given array. Follow the below steps to solve the problem:
- Create freq[] of size M + 1 where M is the maximum element present in the given array and a variable max_sum to store the frequency of each element of arr[] and maximum possible sum respectively.
- Traverse the array freq[] from right to left i.e., from i = M to 0.
- Increment max_sum by freq[i]*i, reduce K by freq[i] and increment freq[i – 1] by freq[i], if K ? freq[i].
- Else increment max_sum by i*K and break the loop because K becomes 0.
- Return max_sum as the maximum possible sum.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to find maximum possible// after adding elements K times and// decrementing each added value by 1long maxSum(int arr[], int k, int n){ // Stores the maximum sum long max_sum = 0; // Stores frequency of element int freq[100000 + 1] = {0}; // Update frequency of array element for(int i = 0; i < n; i++) freq[arr[i]]++; // Traverse from right to left in // freq[] to find maximum sum for(int i = 100000; i > 0; i--) { if (k >= freq[i]) { // Update max_sum max_sum += i * freq[i]; // Decrement k k -= freq[i]; freq[i - 1] += freq[i]; } // Update max_sum and break else { max_sum += i * k; break; } } // Return the maximum sum return max_sum;}// Driver Codeint main(){ // Given array arr[] int arr[] = { 2, 5 }; int n = sizeof(arr) / sizeof(arr[0]); // Given K int K = 4; // Function Call cout << (maxSum(arr, K, n)); return 0;}// This code is contributed by chitranayal |
Java
// Java program for the above approachimport java.util.*;class GFG { // Function to find maximum possible // after adding elements K times and // decrementing each added value by 1 public static long maxSum(int[] arr, int k) { // Stores the maximum sum long max_sum = 0; // Stores frequency of element int[] freq = new int[100000 + 1]; // Update frequency of array element for (int t : arr) freq[t]++; // Traverse from right to left in // freq[] to find maximum sum for (int i = 100000; i > 0; i--) { if (k >= freq[i]) { // Update max_sum max_sum += i * freq[i]; // Decrement k k -= freq[i]; freq[i - 1] += freq[i]; } // Update max_sum and break else { max_sum += i * k; break; } } // Return the maximum sum return max_sum; } // Driver Code public static void main(String[] args) { // Given array arr[] int[] arr = { 2, 5 }; // Given K int K = 4; // Function Call System.out.println(maxSum(arr, K)); }} |
Python3
# Python program for the above approach# Function to find maximum possible# after adding elements K times and# decrementing each added value by 1def maxSum(arr, k): # Stores the maximum sum max_sum = 0; # Stores frequency of element freq = [0]*(100000 + 1); # Update frequency of array element for i in range(len(arr)): freq[arr[i]] += 1; # Traverse from right to left in # freq to find maximum sum for i in range(100000, 1, -1): if (k >= freq[i]): # Update max_sum max_sum += i * freq[i]; # Decrement k k -= freq[i]; freq[i - 1] += freq[i]; # Update max_sum and break else: max_sum += i * k; break; # Return the maximum sum return max_sum;# Driver Codeif __name__ == '__main__': # Given array arr arr = [2, 5]; # Given K K = 4; # Function Call print(maxSum(arr, K)); # This code contributed by gauravrajput1 |
C#
// C# program for the // above approachusing System;class GFG{// Function to find maximum possible// after adding elements K times and// decrementing each added value by 1public static long maxSum(int[] arr, int k){ // Stores the maximum // sum long max_sum = 0; // Stores frequency of // element int[] freq = new int[100000 + 1]; // Update frequency of // array element foreach (int t in arr) { freq[t]++; } // Traverse from right to // left in []freq to find // maximum sum for (int i = 100000; i > 0; i--) { if (k >= freq[i]) { // Update max_sum max_sum += i * freq[i]; // Decrement k k -= freq[i]; freq[i - 1] += freq[i]; } // Update max_sum and // break else { max_sum += i * k; break; } } // Return the maximum // sum return max_sum;}// Driver Codepublic static void Main(String[] args){ // Given array []arr int[] arr = {2, 5}; // Given K int K = 4; // Function Call Console.WriteLine( maxSum(arr, K));}}// This code is contributed by gauravrajput1 |
Javascript
<script>//Javascript program for the above approach// Function to find maximum possible// after adding elements K times and// decrementing each added value by 1function maxSum(arr, k, n){ // Stores the maximum sum var max_sum = 0; // Stores frequency of element var freq = new Array(100000 + 1); freq.fill(0); // Update frequency of array element for(var i = 0; i < n; i++) freq[arr[i]]++; // Traverse from right to left in // freq[] to find maximum sum for(var i = 100000; i > 0; i--) { if (k >= freq[i]) { // Update max_sum max_sum += i * freq[i]; // Decrement k k -= freq[i]; freq[i - 1] += freq[i]; } // Update max_sum and break else { max_sum += i * k; break; } } // Return the maximum sum return max_sum;}var arr = [ 2, 5 ];var n = arr.length;// Given Kvar K = 4;// Function Calldocument.write (maxSum(arr, K, n));// This code is contributed by SoumikMondal</script> |
Output:
14
Time Complexity: O(N), where N is the length of the given array.
Auxiliary Space: O(max(arr)) as we took freq hashmap of size equal to maximum element of the given array.
This solution would be better to use when size(array) and maximum(array) are of same order.
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