Maximize sum of chosen Array elements with value at most M
Given an array arr[] of N positive numbers and an integer M. The task is to maximize the value of M by adding array elements when arr[i] ≤ M.
Note: Any array element can be added at most once.
Examples:
Input: arr[] = {3, 9, 19, 5, 21}, M = 10
Output: 67
Explanation: One way to getthe value is
M > 3; 3 is added to M and it becomes 10+3 = 13
M > 9; 9 is added to M and it becomes 13+9 = 22
M > 19; 19 is added to M and it becomes 22+19 = 41
M > 5; 5 is added to M and it becomes 41+5 = 46
M > 21; 21 is added to M and it becomes 46+21 = 67
Thus, M = 67 at the end.Input: arr[] = {2, 13, 4, 19}, M = 6
Output: 12
Explanation: One way to get the value is
M > 4; 4 is added to M and it becomes 6+4 = 10
M > 2; 2 is added to M and it becomes 10+2 = 12
No other value in the array is smaller or equal to M.
Thus, M is 12 at the end.
Approach: The solution is based on the concept of sorting. Follow the steps mentioned below:
- First, sort the array in increasing order.
- For every index i, from 0 to N-1, do the following:
- If M ≥ arr[i], add arr[i] with M.
- If M< arr[i], stop iteration.
- Return the final value of M as the answer.
Below is the implementation of the above approach.
C++
// C++ code to implement the approach #include <bits/stdc++.h> using namespace std; // Function to calculate // the maximum value of M // that can be obtained int IsArrayHungry( int M, vector< int >& arr) { // Sort the array in increasing order. sort(arr.begin(), arr.end()); long long sum = M; int N = arr.size(); for ( int i = 0; i < N; i++) { if (sum >= arr[i]) sum += arr[i]; else break ; } return sum; } // Driver code int main() { vector< int > arr{ 3, 9, 19, 5, 21 }; int M = 10; int res = IsArrayHungry(M, arr); cout << res; return 0; } |
Java
// Java program for the above approach import java.io.*; import java.lang.*; import java.util.*; class GFG { // Function to calculate // the maximum value of M // that can be obtained static int IsArrayHungry( int M, int arr[ ]) { // Sort the array in increasing order. Arrays.sort(arr); int sum = M; int N = arr.length; for ( int i = 0 ; i < N; i++) { if (sum >= arr[i]) sum += arr[i]; else break ; } return sum; } // Driver code public static void main (String[] args) { int arr[ ] = { 3 , 9 , 19 , 5 , 21 }; int M = 10 ; int res = IsArrayHungry(M, arr); System.out.print(res); } } // This code is contributed by hrithikgarg03188. |
Python3
# Python 3 code to implement the approach # Function to calculate # the maximum value of M # that can be obtained def IsArrayHungry(M, arr): # Sort the array in increasing order. arr.sort() sum = M N = len (arr) for i in range (N): if ( sum > = arr[i]): sum + = arr[i] else : break return sum # Driver code if __name__ = = "__main__" : arr = [ 3 , 9 , 19 , 5 , 21 ] M = 10 res = IsArrayHungry(M, arr) print (res) # This code is contributed by ukasp. |
C#
// C# code to implement above approach using System; class GFG { // Function to calculate // the maximum value of M // that can be obtained static int IsArrayHungry( int M, int []arr) { // Sort the array in increasing order. Array.Sort(arr); int sum = M; int N = arr.Length; for ( int i = 0; i < N; i++) { if (sum >= arr[i]) sum += arr[i]; else break ; } return sum; } // Driver Code: public static void Main() { int []arr = { 3, 9, 19, 5, 21 }; int M = 10; int res = IsArrayHungry(M, arr); Console.WriteLine(res); } } // This code is contributed by Samim Hossain Mondal. |
Javascript
<script> // JavaScript code for the above approach // Function to calculate // the maximum value of M // that can be obtained function IsArrayHungry(M, arr) { // Sort the array in increasing order. arr.sort( function (a, b) { return a - b }) let sum = M; let N = arr.length; for (let i = 0; i < N; i++) { if (sum >= arr[i]) sum += arr[i]; else break ; } return sum; } // Driver code let arr = [3, 9, 19, 5, 21]; let M = 10; let res = IsArrayHungry(M, arr); document.write(res); // This code is contributed by Potta Lokesh </script> |
67
Time Complexity: O(N * logN)
Auxiliary Space: O(1), since no extra space has been added.