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

• Last Updated : 24 Dec, 2021

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:
Explanation:
Convert the given array to {4, 10, 10, 10} by incrementing arr[1], arr[2] and arr[3] twice each.
Now

is 0 + 1 + 1 + 1 = 3.
Input: N = 3, X = 122, arr[] = {3, 11, 14}
Output: 15

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 of 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 minimum of increments and        // remaining/10 to the        // answer        ans += min(increments,                   remaining / 10);    }     // output the final answer    cout << ans;} // Driver Codeint 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 approachimport 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 minimum of increments and        // remaining/10 to the        // answer        ans += Math.min(increments,                        remaining / 10);    }         // Output the final answer    System.out.print(ans);} // Driver codepublic 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 problemdef 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 minimum of increments        # and remaining/10 to the        # answer        ans += min(increments,                   remaining // 10)     # Output the final answer    print(ans) # Driver Codeif __name__ =="__main__":     N = 4    X = 6    A = [ 4, 8, 8, 8 ]         maximizeval10(A, N, X) # This code is contributed by chitranayal

## C#

 // C# program for the above approachusing System;using System.Collections.Generic;class GFG{     public static void maximizeval10(int[] a,                                 int n,                                 int k){  // Initialize variables  int increments = 0;  int ans = 0;  List<int> v = new List<int>();   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  v.Sort();   int sum = 0;   for(int i = 0; i < v.Count; 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 minimum of increments and    // remaining/10 to the    // answer    ans += Math.Min(increments,                    remaining / 10);  }   // Output the readonly answer  Console.Write(ans);} // Driver codepublic 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 shikhasingrajput

## Javascript

 

Output:

3

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

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