Skip to content
Related Articles

Related Articles

Reduce given Array to 0 by maximising sum of chosen elements

View Discussion
Improve Article
Save Article
  • Difficulty Level : Medium
  • Last Updated : 25 Jul, 2022

Given an array arr[] containing N positive integers, the task is to maximize the array sum when after every sum operation all the remaining array elements decrease by 1.

Note: The value of an array element does not go below 0.

Examples:

nput: arr[] = {6, 2, 4, 5} 
Output: 12
Explanation: Add 6 initially to the final sum.
The final sum becomes 6 and the remaining array elements {1, 3, 4}.
Add 3 with the sum. The sum becomes 9 and the remaining elements {0, 3}
Add 3 with the sum. The sum becomes 12 and only 0 remains in the array.
Add 0 with the sum. The sum remains unchanged.
Input: arr[] = {5, 6, 4}
Output: 12
 

Naive approach:  Find the maximum element from the array. Add it to the sum and decrease all other elements by 1 and change the current element to 0 so it is not repeated again in the loop. Do this process until all the elements become 0.

C++




// C++ code to implement the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Find maximum possible sum
int maxSum(int arr[], int N)
{
    // Initialize ans with 0
    int ans = 0;
 
    // loop till atleast one element is greater than 0
    while (1) {
 
        // maximum element of array
        int maxValueIndex = max_element(arr, arr + N) - arr;
 
        // breaking condition when all elements become <=0
        if (arr[maxValueIndex] <= 0)
            break;
 
        // adding value to answer
        ans += arr[maxValueIndex];
        arr[maxValueIndex] = 0;
 
        // Iterate array
        for (int i = 0; i < N; i++) {
 
            arr[i]--;
        }
    }
 
    return ans;
}
 
// Driver code
int main()
{
    // Given array of values
    int arr[] = { 6, 2, 4, 5 };
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function call
    cout << maxSum(arr, N);
    return 0;
}

Java




// Java code to implement the above approach
import java.util.*;
class GFG
{
 
  // Find maximum possible sum
  static int maxSum(int arr[], int N)
  {
 
    // Initialize ans with 0
    int ans = 0;
 
    // loop till atleast one element is greater than 0
    while (true) {
 
      // maximum element of array
      int maxValue = Arrays.stream(arr).max().getAsInt();
      ;
      int maxValueIndex = 0;
      for (int i = 0; i < arr.length; i++) {
        if (arr[i] == maxValue) {
          maxValueIndex = i;
          break;
        }
      }
      // breaking condition when all elements become <=0
      if (arr[maxValueIndex] <= 0)
        break;
 
      // adding value to answer
      ans += arr[maxValueIndex];
      arr[maxValueIndex] = 0;
 
      // Iterate array
      for (int i = 0; i < N; i++) {
 
        arr[i]--;
      }
    }
 
    return ans;
  }
 
  // Driver code
  public static void main(String[] args)
  {
     
    // Given array of values
    int arr[] = { 6, 2, 4, 5 };
    int N = arr.length;
 
    // Function call
    System.out.print(maxSum(arr, N));
  }
}
 
// This code is contributed by gauravrajput1

Python3




# Python code to implement the above approach
 
# Find maximum possible sum
def maxSum(arr,  N):
   
    # Initialize ans with 0
    ans = 0
 
    # loop till atleast one element is greater than 0
    while (1):
 
        # maximum element of array
        maxValueIndex = arr.index(max(arr))
 
        # breaking condition when all elements become <=0
        if (arr[maxValueIndex] <= 0):
            break
 
        # adding value to answer
        ans += arr[maxValueIndex]
        arr[maxValueIndex] = 0
 
        # Iterate array
        for i in range(0, N):
            arr[i] -= 1
 
    return ans
 
# Driver code
 
# Given array of values
arr = [6, 2, 4, 5]
N = len(arr)
 
# Function call
print(maxSum(arr, N))
 
# This code is contributed by ninja_hattori.

C#




// C# code to implement the above approach
using System;
using System.Linq;
class GFG
{
 
  // Find maximum possible sum
  static int maxSum(int[] arr, int N)
  {
 
    // Initialize ans with 0
    int ans = 0;
 
    // loop till atleast one element is greater than 0
    while (true) {
 
      // maximum element of array
      int maxValue = arr.Max();
 
      int maxValueIndex = 0;
      for (int i = 0; i < arr.Length; i++) {
        if (arr[i] == maxValue) {
          maxValueIndex = i;
          break;
        }
      }
      // breaking condition when all elements become <=0
      if (arr[maxValueIndex] <= 0)
        break;
 
      // adding value to answer
      ans += arr[maxValueIndex];
      arr[maxValueIndex] = 0;
 
      // Iterate array
      for (int i = 0; i < N; i++) {
 
        arr[i]--;
      }
    }
 
    return ans;
  }
 
 
  // Driver code
  public static int Main()
  {
    // Given array of values
    int[] arr = { 6, 2, 4, 5 };
    int N = arr.Length;
 
    // Function call
    Console.Write(maxSum(arr, N));
    return 0;
  }
}
// This code is contributed by Pushpesh Raj

Javascript




<script>
 
// JavaScript code to implement the above approach
 
// Find maximum possible sum
function maxSum(arr, N)
{
    // Initialize ans with 0
    let ans = 0;
 
    // loop till atleast one element is greater than 0
    while (1) {
 
    // maximum element of array
        let maxValueIndex = arr.indexOf(Math.max(...arr));
 
    // breaking condition when all elements become <=0
        if (arr[maxValueIndex] <= 0)
            break;
 
    // adding value to answer
        ans += arr[maxValueIndex];
        arr[maxValueIndex] = 0;
 
    // Iterate array
        for (let i = 0; i < N; i++) {
 
            arr[i]--;
        }
    }
 
    return ans;
}
 
// Driver code
 
// Given array of values
let arr = [ 6, 2, 4, 5 ];
let N = arr.length;
 
// Function call
document.write(maxSum(arr, N));
 
// This code is contributed by shinjanpatra
</script>

Output

12

Time Complexity: O(N2)

Auxiliary Space: O(1),  since no extra space has been taken.

Efficient Approach: The solution of the problem is based on the concept of sorting. As the values decrease in each step, the higher values should be added first with the final sum. Follow the steps mentioned below to solve the problem:

  • Sort the array in descending order.
  • Run a loop from 1 to N.
    • For each index i the value added to the final sum will be (arr[i] – i) as it will be added after i decrement of the value.
    • As the sum value cannot go below 0, add max(arr[i]-i, 0) to the final sum.
  • Return the final sum.

Below is the implementation of the above approach.

C++




// C++ code to implement the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Find maximum possible sum
int maxSum(int arr[], int N)
{
    // Initialize ans with 0
    int ans = 0;
 
    // Sort array in descending order
    sort(arr, arr + N, greater<int>());
 
    // Iterate array
    for (int i = 0; i < N; i++) {
 
        // Starting value
        int value = arr[i];
 
        // Actual value when being added
        int current = max(0, value - i);
 
        // Add actual value with ans
        ans = ans + current;
    }
    return ans;
}
 
// Driver code
int main()
{
    // Given array of values
    int arr[] = { 6, 2, 4, 5 };
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function call
    cout << maxSum(arr, N);
    return 0;
}

Java




// Java code to implement the above approach
import java.util.*;
public class GFG
{
 
  // Utility program to reverse an array
  public static void reverse(int[] array)
  {
 
    // Length of the array
    int n = array.length;
 
    // Swaping the first half elements with last half
    // elements
    for (int i = 0; i < n / 2; i++) {
 
      // Storing the first half elements temporarily
      int temp = array[i];
 
      // Assigning the first half to the last half
      array[i] = array[n - i - 1];
 
      // Assigning the last half to the first half
      array[n - i - 1] = temp;
    }
  }
 
  // Find maximum possible sum
  static int maxSum(int[] arr, int N)
  {
 
    // Initialize ans with 0
    int ans = 0;
 
    // Sorting the array in ascending order
    Arrays.sort(arr);
 
    // Reversing the array
    reverse(arr);
 
    // Iterate array
    for (int i = 0; i < N; i++) {
 
      // Starting value
      int value = arr[i];
 
      // Actual value when being added
      int current = Math.max(0, value - i);
 
      // Add actual value with ans
      ans = ans + current;
    }
    return ans;
  }
 
  // Driver code
  public static void main(String args[])
  {
     
    // Given array of values
    int[] arr = { 6, 2, 4, 5 };
    int N = arr.length;
 
    // Function call
    System.out.println(maxSum(arr, N));
  }
}
 
// This code is contributed by Samim Hossain Mondal.

Python




# Python code to implement the above approach
 
# Find maximum possible sum
def maxSum(arr, N):
 
    # Initialize ans with 0
    ans = 0
 
    # Sort array in descending order
    arr.sort(reverse = True)
 
    # Iterate array
    for i in range(N):
 
        #Starting value
        value = arr[i]
 
        # Actual value when being added
        current = max(0, value - i)
 
        # Add actual value with ans
        ans = ans + current
     
    return ans
 
# Driver code
if __name__ == "__main__":
 
    # Given array of values
    arr = [ 6, 2, 4, 5 ]
    N = len(arr)
 
    # Function call
    print(maxSum(arr, N))
     
    # This code is contributed by hrithikgarg03188.

C#




// C# code to implement the above approach
using System;
class GFG
{
 
  // Find maximum possible sum
  static int maxSum(int[] arr, int N)
  {
 
    // Initialize ans with 0
    int ans = 0;
 
    // Sort array in descending order
 
    Array.Sort<int>(
      arr, delegate(int m, int n) { return n - m; });
 
    // Iterate array
    for (int i = 0; i < N; i++) {
 
      // Starting value
      int value = arr[i];
 
      // Actual value when being added
      int current = Math.Max(0, value - i);
 
      // Add actual value with ans
      ans = ans + current;
    }
    return ans;
  }
 
  // Driver code
  public static int Main()
  {
    // Given array of values
    int[] arr = { 6, 2, 4, 5 };
    int N = arr.Length;
 
    // Function call
    Console.Write(maxSum(arr, N));
    return 0;
  }
}
// This code is contributed by Taranpreet

Javascript




<script>
        // JavaScript code for the above approach
 
        // Find maximum possible sum
        function maxSum(arr, N)
        {
         
            // Initialize ans with 0
            let ans = 0;
 
            // Sort array in descending order
            arr.sort(function (a, b) { return b - a })
 
            // Iterate array
            for (let i = 0; i < N; i++) {
 
                // Starting value
                let value = arr[i];
 
                // Actual value when being added
                let current = Math.max(0, value - i);
 
                // Add actual value with ans
                ans = ans + current;
            }
            return ans;
        }
 
        // Driver code
 
        // Given array of values
        let arr = [6, 2, 4, 5];
        let N = arr.length;
 
        // Function call
        document.write(maxSum(arr, N));
 
       // This code is contributed by Potta Lokesh
    </script>

Output

12

Time Complexity: O(N log N)
Auxiliary Space: O(1) 


My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!