# Maximize array sum by X increments when each element is divided by 10

Given an array arr[] consisting of N non-negative elements and an integer X, the task is to make X increments such that the value of array sum when each element is divided by 10, i.e. is maximized. Print the maximum value of possible.

Note: The value of any element can’t be increased beyond 1000.
Examples:

Input: N = 4, X = 6, arr[] = {4, 8, 8, 8}
Output: 3
Explanation:
Convert the given array to {4, 10, 10, 10} by incrementing arr, arr and arr twice each.
Now is 0 + 1 + 1 + 1 = 3.

Input: N = 3, X = 122, arr[] = {3, 11, 14}
Output: 15

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:

1. For all the elements, calculate the number of increments required to increase the number to the next multiple of 10 and store these values in an array, say V.
2. Calculate the maximum number of times that an element can be incremented by 10 and keep its value <= 1000 and add this value to a variable, say increments which is initialized to 0.
3. Sort the array V to make it non-decreasing.
4. Then for each value in V, perform the required moves and increase some element to the next multiple of 10, this increases the answer by 1.
5. Do this, while the total moves performed, do not exceed X.
6. After going through all elements of V if still some moves are remaining then add to the answer minimum between increments and (remaining moves)/10 .

Below is the implementation to the above approach:

## C++

 // C++ program for the above problem     #include  using namespace std;     void maximizeval10(int a[],                     int n, int k)  {      // initialize variables      int increments = 0;      int ans = 0;      vector<int> v;         for (int i = 0; i < n; i++) {             // add the current          // contribution of the          // element to the answer          ans += (a[i] / 10);             // if the value is          // already maximum          // then we can't change it          if (a[i] == 1000)              continue;             else {              // moves required to move              // to the next multiple              // of 10              v.push_back(10 - a[i] % 10);                 // no of times we can              // add 10 to this value              // so that its value              // does not exceed 1000.              increments += (100                             - ((a[i]) / 10)                             - 1);          }      }         // sort the array      sort(v.begin(), v.end());         int sum = 0;         for (int i = 0; i < v.size();           i++) {             // adding the values to          // increase the numbers          // to the next multiple of 10          sum += v[i];          if (sum <= k) {                 // if the total moves              // are less than X then              // increase the answer              ans++;          }          else                // if the moves exceed              // X then we cannot              // increase numbers              break;      }         // if there still remain      // some moves      if (sum < k) {             // remaining moves          int remaining = k - sum;             // add minimim of increments and          // remaining/10 to the          // answer          ans += min(increments,                     remaining / 10);      }         // output the final answer      cout << ans;  }     // Driver Code  int main()  {      int N = 4;      int X = 6;         int A[N] = { 4, 8, 8, 8 };      maximizeval10(A, N, X);         return 0;  }

## Java

 // Java program for the above approach  import java.util.*;      class GFG{         public static void maximizeval10(int[] a, int n,                                   int k)   {              // Initialize variables       int increments = 0;       int ans = 0;       Vector v = new Vector<>();         for(int i = 0; i < n; i++)       {              // Add the current           // contribution of the           // element to the answer           ans += (a[i] / 10);              // If the value is           // already maximum           // then we can't change it           if (a[i] == 1000)               continue;              else          {                             // Moves required to move               // to the next multiple               // of 10               v.add(10 - a[i] % 10);                  // No of times we can               // add 10 to this value               // so that its value               // does not exceed 1000.               increments += (100 - ((a[i]) /                              10) - 1);           }       }              // Sort the array       Collections.sort(v);          int sum = 0;          for(int i = 0; i < v.size(); i++)      {                      // Adding the values to           // increase the numbers           // to the next multiple of 10           sum += v.get(i);           if (sum <= k)           {                              // If the total moves               // are less than X then               // increase the answer               ans++;           }           else                // If the moves exceed               // X then we cannot               // increase numbers               break;       }          // If there still remain       // some moves       if (sum < k)       {                      // Remaining moves           int remaining = k - sum;              // Add minimim of increments and           // remaining/10 to the           // answer           ans += Math.min(increments,                           remaining / 10);       }              // Output the final answer       System.out.print(ans);  }      // Driver code  public static void main(String[] args)  {      int N = 4;       int X = 6;       int A[] = { 4, 8, 8, 8 };              maximizeval10(A, N, X);   }  }     // This code is contributed by divyeshrabadiya07

## Python3

 # Python3 program for the above problem  def maximizeval10(a, n, k):             # Initialize variables      increments = 0     ans = 0     v = []         for i in range (n):             # Add the current          # contribution of the          # element to the answer          ans += (a[i] // 10)             # If the value is already           # maximum then we can't          # change it          if (a[i] == 1000):              continue         else:                             # Moves required to move              # to the next multiple              # of 10              v.append(10 - a[i] % 10)                 # No of times we can              # add 10 to this value              # so that its value              # does not exceed 1000.              increments += (100 - ((a[i]) //                                      10) - 1);         # Sort the array      v.sort()         sum = 0     for i in range(len(v)):             # Adding the values to          # increase the numbers          # to the next multiple of 10          sum += v[i]          if (sum <= k):                 # If the total moves              # are less than X then              # increase the answer              ans += 1                    else:                 # If the moves exceed              # X then we cannot              # increase numbers              break        # If there still remain      # some moves      if (sum < k):             # Remaining moves          remaining = k - sum            # Add minimim of increments           # and remaining/10 to the          # answer          ans += min(increments,                     remaining // 10)         # Output the final answer      print(ans)     # Driver Code  if __name__ =="__main__":         N = 4     X = 6     A = [ 4, 8, 8, 8 ]             maximizeval10(A, N, X)     # This code is contributed by chitranayal

Output:

3


Time Complexity: O(N * log(N))
Auxillary Space complexity: O(N)

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