Given an array of positive integer arr[], and a number K. the task is to maximize the sum of the array by replacing any K elements of the array by taking modulus with any positive integer which is less than arr[i] i.e, (arr[i] = arr[i]%X where X ≤ arr[i]).
Examples:
Input: arr[] = {5, 7, 18, 12, 11, 3}, K = 4
Output: 41
Explanation: The replacement should be {5%3, 7%4, 18, 12, 11%6, 3%2}Input: arr[] = {8, 2, 28, 12, 7, 9}, K = 4
Output: 55
Explanation: The replacement should be {8%5, 2%2, 28, 12, 7%4, 9}
Approach: For every element arr[i] in the array arr[], module it with (arr[i]/2 +1) which will give the highest possible value of arr[i] after the operation. Following are the steps to solve the problem
- Sort the array arr[].
- Iterate over the range [0, K) using the variable i and perform the following tasks:
- For every element arr[i], module it with (arr[i]/2 +1) and update the result.
- Find the sum of the updated array and output it.
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 the maximum possible sumint find(int arr[], int K, int N){ // Sorting the array sort(arr, arr + N); int sum = 0; // Loop to take update K for (int i = 0; i < K; i++) { // Smallest number in array arr[i] %= (arr[i] / 2) + 1; } // Loop to find sum for (int i = 0; i < N; i++) { sum += arr[i]; } return sum;}// Driver Codeint main(){ int arr[] = { 5, 7, 18, 12, 11, 3 }; int K = 4; int N = sizeof(arr) / sizeof(arr[0]); cout << find(arr, K, N); return 0;} |
C
// C program for the above approach#include <stdio.h>#include <stdlib.h>#include <limits.h>#include <limits.h>void sort(int arr[], int n){ int i, j, temp; for (i = 0; i < n - 1; i++) { for (j = 0; j < n - i - 1; j++) { if (arr[j] > arr[j + 1]) { temp = arr[j]; arr[j] = arr[j + 1]; arr[j + 1] = temp; } } }}// Function to find the maximum possible sumint find(int arr[], int K, int N){ // Sorting the array sort(arr, N); int sum = 0; // Loop to take update K for (int i = 0; i < K; i++) { // Smallest number in array arr[i] %= (arr[i] / 2) + 1; } // Loop to find sum for (int i = 0; i < N; i++) { sum += arr[i]; } return sum;}int main(){ int arr[] = {5, 7, 18, 12, 11, 3}; int K = 4; int N = sizeof(arr) / sizeof(arr[0]); printf("%d", find(arr, K, N)); return 0;}// This code is contributed by abhinavprkash. |
Java
// Java program for the above approachimport java.util.Arrays;class GFG { // Function to find the maximum possible sum static int find(int arr[], int K, int N) { // Sorting the array Arrays.sort(arr); int sum = 0; // Loop to take update K for (int i = 0; i < K; i++) { // Smallest number in array arr[i] %= (arr[i] / 2) + 1; } // Loop to find sum for (int i = 0; i < N; i++) { sum += arr[i]; } return sum; } // Driver Code public static void main(String args[]) { int arr[] = { 5, 7, 18, 12, 11, 3 }; int K = 4; int N = arr.length; System.out.println(find(arr, K, N)); }}// This code is contributed by saurabh_jaiswal. |
Python3
# python3 program for the above approach# Function to find the maximum possible sumdef find(arr, K, N): # Sorting the array arr.sort() sum = 0 # Loop to take update K for i in range(0, K): # Smallest number in array arr[i] %= (arr[i] // 2) + 1 # Loop to find sum for i in range(0, N): sum += arr[i] return sum# Driver Codeif __name__ == "__main__": arr = [5, 7, 18, 12, 11, 3] K = 4 N = len(arr) print(find(arr, K, N))# This code is contributed by rakeshsahni |
C#
// C# program for the above approachusing System;class GFG { // Function to find the maximum possible sum static int find(int[] arr, int K, int N) { // Sorting the array Array.Sort(arr); int sum = 0; // Loop to take update K for (int i = 0; i < K; i++) { // Smallest number in array arr[i] %= (arr[i] / 2) + 1; } // Loop to find sum for (int i = 0; i < N; i++) { sum += arr[i]; } return sum; } // Driver Code public static void Main() { int[] arr = { 5, 7, 18, 12, 11, 3 }; int K = 4; int N = arr.Length; Console.Write(find(arr, K, N)); }}// This code is contributed by ukasp. |
Javascript
<script> // JavaScript code for the above approach // Function to find the maximum possible sum function find(arr, K, N) { // Sorting the array arr.sort(function (a, b) { return a - b }) let sum = 0; // Loop to take update K for (let i = 0; i < K; i++) { // Smallest number in array arr[i] %= Math.floor(arr[i] / 2) + 1; } // Loop to find sum for (let i = 0; i < N; i++) { sum += arr[i]; } return sum; } // Driver Code let arr = [5, 7, 18, 12, 11, 3]; let K = 4; let N = arr.length; document.write(find(arr, K, N));// This code is contributed by Potta Lokesh </script> |
Output
41
Time Complexity: O(N * logN) where N is the size of the array
Auxiliary Space: O(1)
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