通过根据给定条件选择元素，可能的最大修改数组总和

给定一个大小为N的数组arr[] ，任务是找到数组元素的最大可能总和，以便可以根据以下条件选择元素：

• 对于每个索引i元素的值是arr[i] ，然后我们可以将从1 到 min(N, arr[i]) 的任何元素添加到总和中，否则离开该元素。
• 如果某个元素已经添加到总和中，那么我们不能将相同的元素再次添加到总和中。

例子：

Input: arr[] = {1, 2, 2, 4}
Output: 7
Explanation: 
Initially, the total sum is 0. 
Sum after adding element at index 1 to sum = 1. Elements added to sum = {1} 
Sum after adding element at index 2 to sum = 3. Elements added to sum = {1, 2} 
Now, coming at index 3 we can not add any element from (1 to 2) to the sum because we have already added the elements {1, 2} into the sum. So, leave the element at index 3.
Then add the element at index 4 to sum since 4 was not added to sum before therefore the maximum sum we can get is 3+4=7.

Input: arr[] = {4, 4, 1, 5}
Output: 13

方法：这个问题可以使用贪心技术来解决。以下是步骤：

1. 按升序对数组的元素进行排序。
2. 保持可以添加到总和的最大可能数量（比如 maxPossible） 。
3. 使用数组的最大元素初始化maxPossible并将其添加到总和中。
4. 从头到尾遍历数组，检查给定的元素值之前是否已添加到总和中。
5. 如果之前已添加元素值，则我们将(element – 1)添加到总和，如果元素值小于maxPossible，则我们将添加该数字并将maxPossible更改为元素值。
6. 完成上述步骤后打印总和的值。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
#define ll long long
using namespace std;
 
// Function that finds the maximum sum
// of the array elements according to
// the given condition
ll findMaxValue(int arr[], int n)
{
    // Sort the array
    sort(arr, arr + n);
 
    // Take the max value from the array
    ll ans = arr[n - 1];
    ll maxPossible = arr[n - 1];
 
    // Iterate from the end of the array
    for (int i = n - 2; i >= 0; --i) {
 
        if (maxPossible > 0) {
 
            // Check for the number had
            // come before or not
            if (arr[i] >= maxPossible) {
 
                // Take the value which is
                // not occured already
                ans += (maxPossible - 1);
                maxPossible = maxPossible - 1;
            }
            else {
 
                // Change max to new value
                maxPossible = arr[i];
                ans += maxPossible;
            }
        }
    }
 
    // Return the maximum sum
    return ans;
}
 
// Driver Code
int main()
{
 
    // Given array arr[]
    int arr[] = { 4, 4, 1, 5 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    // Function Call
    cout << findMaxValue(arr, n);
}

Java

// Java program for the above approach
import java.util.*;
class GFG{
 
// Function that finds the maximum sum
// of the array elements according to
// the given condition
static int findMaxValue(int arr[], int n)
{
    // Sort the array
    Arrays.sort(arr);
 
    // Take the max value from the array
    int ans = arr[n - 1];
    int maxPossible = arr[n - 1];
 
    // Iterate from the end of the array
    for (int i = n - 2; i >= 0; --i)
    {
        if (maxPossible > 0)
        {
 
            // Check for the number had
            // come before or not
            if (arr[i] >= maxPossible)
            {
 
                // Take the value which is
                // not occured already
                ans += (maxPossible - 1);
                maxPossible = maxPossible - 1;
            }
            else
            {
 
                // Change max to new value
                maxPossible = arr[i];
                ans += maxPossible;
            }
        }
    }
 
    // Return the maximum sum
    return ans;
}
 
// Driver Code
public static void main(String[] args)
{
 
    // Given array arr[]
    int arr[] = { 4, 4, 1, 5 };
    int n = arr.length;
 
    // Function Call
    System.out.print(findMaxValue(arr, n));
}
}
 
// This code is contributed by sapnasingh4991

Python3

# Python3 program for the above approach
 
# Function that finds the maximum sum
# of the array elements according to
# the given condition
def findMaxValue(arr, n):
     
    # Sort the array
    arr.sort()
 
    # Take the max value from the array
    ans = arr[n - 1]
    maxPossible = arr[n - 1]
     
    # Iterate from the end of the array
    for i in range(n - 2, -1, -1):
        if (maxPossible > 0):
             
            # Check for the number had
            # come before or not
            if (arr[i] >= maxPossible):
                 
                # Take the value which is
                # not occured already
                ans += (maxPossible - 1)
                maxPossible = maxPossible - 1
            else:
                 
                # Change max to new value
                maxPossible = arr[i]
                ans += maxPossible
                 
    # Return the maximum sum
    return ans
 
# Driver code
if __name__=='__main__':
     
    # Given array arr[]
    arr = [ 4, 4, 1, 5 ]
    n = len(arr)
 
    # Function call
    print(findMaxValue(arr,n))
 
# This code is contributed by rutvik_56

C#

// C# program for the above approach
using System;
class GFG{
 
// Function that finds the maximum sum
// of the array elements according to
// the given condition
static int findMaxValue(int []arr, int n)
{
    // Sort the array
    Array.Sort(arr);
 
    // Take the max value from the array
    int ans = arr[n - 1];
    int maxPossible = arr[n - 1];
 
    // Iterate from the end of the array
    for (int i = n - 2; i >= 0; --i)
    {
        if (maxPossible > 0)
        {
 
            // Check for the number had
            // come before or not
            if (arr[i] >= maxPossible)
            {
 
                // Take the value which is
                // not occured already
                ans += (maxPossible - 1);
                maxPossible = maxPossible - 1;
            }
            else
            {
 
                // Change max to new value
                maxPossible = arr[i];
                ans += maxPossible;
            }
        }
    }
 
    // Return the maximum sum
    return ans;
}
 
// Driver Code
public static void Main(String[] args)
{
 
    // Given array []arr
    int []arr = { 4, 4, 1, 5 };
    int n = arr.Length;
 
    // Function Call
    Console.Write(findMaxValue(arr, n));
}
}
 
// This code is contributed by sapnasingh4991

Javascript

 
// JavaScript program for the above approach
 
// Function that finds the maximum sum
// of the array elements according to
// the given condition
function findMaxValue(arr, n)
{
    // Sort the array
    arr.sort((a, b) => a - b);
 
    // Take the max value from the array
    let ans = arr[n - 1];
    let maxPossible = arr[n - 1];
 
    // Iterate from the end of the array
    for (let i = n - 2; i >= 0; --i) {
 
        if (maxPossible > 0) {
 
            // Check for the number had
            // come before or not
            if (arr[i] >= maxPossible) {
 
                // Take the value which is
                // not occured already
                ans += (maxPossible - 1);
                maxPossible = maxPossible - 1;
            }
            else {
 
                // Change max to new value
                maxPossible = arr[i];
                ans += maxPossible;
            }
        }
    }
 
    // Return the maximum sum
    return ans;
}
 
// Driver Code
 
    // Given array arr[]
    let arr = [ 4, 4, 1, 5 ];
    let n = arr.length;
 
    // Function Call
    document.write(findMaxValue(arr, n));
 
 
// This code is contributed by Surbhi Tyagi.
 

输出：

13

时间复杂度： O(N)
辅助空间： O(1)

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