Maximize sum of chosen Array elements with value at most M

Last Updated : 13 Jul, 2022

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# 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>

Output

Time Complexity: O(N * logN)
Auxiliary Space: O(1), since no extra space has been added.

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