Open In App

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

Last Updated : 24 Dec, 2021
Comments
Improve
Suggest changes
Like Article
Like
Report

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. 

\sum\limits^{N-1}_{i=0} \lfloor arr_i/10 \rfloor     
 

is maximized. Print the maximum value of 


\sum\limits^{N-1}_{i=0} \lfloor arr_i/10 \rfloor     
 

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[1], arr[2] and arr[3] twice each. 
Now 

\sum\limits^{N-1}_{i=0} \lfloor arr_i/10 \rfloor     

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 <bits/stdc++.h>
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 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<Integer> 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 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 minimum 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
C# 
// C# program for the above approach
using 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 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 shikhasingrajput
JavaScript 
<script>
// Javascript program for the above approach
    
function maximizeval10(a,n,k)
{
    // Initialize variables
    let increments = 0;
    let ans = 0;
    let v = [];
 
    for(let i = 0; i < n; i++)
    {
 
        // Add the current
        // contribution of the
        // element to the answer
        ans += Math.floor(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(10 - a[i] % 10);
 
            // No of times we can
            // add 10 to this value
            // so that its value
            // does not exceed 1000.
            increments += (100 - (Math.floor(a[i]) / 10) - 1);
        }
    }
     
    // Sort the array
    v.sort(function(a,b){return a-b;});
 
    let sum = 0;
 
    for(let i = 0; i < v.length; 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
        let remaining = k - sum;
 
        // Add minimum of increments and
        // remaining/10 to the
        // answer
        ans += Math.min(increments,
                        Math.floor(remaining / 10));
    }
     
    // Output the final answer
    document.write(ans);        
}

// Driver code
let N = 4;
let X = 6;
let A=[4, 8, 8, 8];
maximizeval10(A, N, X);
    

// This code is contributed by avanitrachhadiya2155
</script>

Output: 
3

 

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


N
nishitsharma1
Improve
Article Tags :
Practice Tags :

Similar Reads

We use cookies to ensure you have the best browsing experience on our website. By using our site, you acknowledge that you have read and understood our Cookie Policy & Privacy Policy
Lightbox