通过重复从串联为 3 的倍数的对中删除元素来最大化数组的总和

给定一个由N 个正整数组成的数组arr[] ，任务是在重复删除串联是3的倍数的两个元素中的任何一个后，找到剩余数组元素的最大可能总和。

例子：

Input: arr[] = {23, 12, 43, 3, 56}
Output: 91
Explanation:
Initially the array is {23, 12, 43, 3, 56}. Following removal of array elements are performed:

• Pair {23, 43}: Concatenation = 2343, which is divisible by 3. Now, removing 43 modifies the array to {23, 12, 3, 56}.
• Pair {12, 3}: Concatenation = 123, which is divisible by 3. Now, removing 3 modifies the array to {23, 12, 56}.

After the above operations, sum of the array elements = 12 + 56 + 23 = 91.

Input: arr[] = {324, 32, 53, 67, 330}
Output: 415

方法：给定的问题可以通过使用以下事实来解决：一个数的余数除以3 时，等于其数字之和除以3时的余数。 请按照以下步骤解决问题：

• 初始化三个变量，比如maxRem0 、 maxRem1和maxRem2 ，分别存储余数为0 、 1和2的元素。
• 遍历给定的数组arr[]并执行以下步骤：
  • 初始化变量digitSum以存储数字总和。
  • 如果digitSum % 3 == 0 ，则将maxRem0的值更新为max(maxRem0, arr[i]) 。
  • 否则，如果余数是1或2 ，则
• 完成上述步骤后，打印maxRem0与maxRem1和maxRem2的最大值之和作为结果。

下面是上述方法的实现：

C++

// C++ approach for the above approach
#include 
using namespace std;
 
// Function to calculate sum
// of digits of an integer
int getSum(int n)
{
    int ans = 0;
     
    // char[] arr = (String.valueOf(n)).toCharArray();
    string arr = to_string(n);
    for(int i = 0; i < arr.length(); i++)
    {
        ans += int(arr[i]);
    }
    return ans;
}
 
// Function to calculate maximum sum
// of array after removing pairs whose
// concatenation is divisible by 3
void getMax(int arr[], int n)
{
 
    // Stores the sum of
    // digits of array element
    int maxRem0 = 0;
 
    int rem1 = 0;
    int rem2 = 0;
 
    for(int i = 0; i < n; i++)
    {
         
        // Find the sum of digits
        int digitSum = getSum(arr[i]);
 
        // If i is divisible by 3
        if (digitSum % 3 == 0)
            maxRem0 = max(maxRem0, arr[i]);
 
        // Otherwise, if i modulo 3 is 1
        else if (digitSum % 3 == 1)
            rem1 += arr[i];
 
        // Otherwise, if i modulo 3 is 2
        else
            rem2 += arr[i];
    }
 
    // Return the resultant
    // sum of array elements
    cout << (maxRem0 + max(rem1, rem2));
}
 
// Driver Code
int main()
{
    int arr[] = { 23, 12, 43, 3, 56 };
    int n = sizeof(arr) / sizeof(arr[0]);
     
    getMax(arr, n);
}
 
// This code is contributed by ukasp

Java

// Java approach for the above approach
class GFG{
 
// Function to calculate sum
// of digits of an integer
static int getSum(int n)
{
    int ans = 0;
    char[] arr = (String.valueOf(n)).toCharArray();
     
    for(char ch : arr)
    {
        ans += Character.getNumericValue(ch);
    }
    return ans;
}
 
// Function to calculate maximum sum
// of array after removing pairs whose
// concatenation is divisible by 3
static void getMax(int[] arr)
{
     
    // Stores the sum of
    // digits of array element
    int maxRem0 = 0;
 
    int rem1 = 0;
    int rem2 = 0;
 
    for(int i:arr)
    {
         
        // Find the sum of digits
        int digitSum = getSum(i);
 
        // If i is divisible by 3
        if (digitSum % 3 == 0)
            maxRem0 = Math.max(maxRem0, i);
 
        // Otherwise, if i modulo 3 is 1
        else if (digitSum % 3 == 1)
            rem1 += i;
 
        // Otherwise, if i modulo 3 is 2
        else
            rem2 += i;
    }
     
    // Return the resultant
    // sum of array elements
    System.out.print(maxRem0 +
                     Math.max(rem1, rem2));
}
 
// Driver Code
public static void main(String[] args)
{
    int[] arr = { 23, 12, 43, 3, 56 };
    getMax(arr);
}
}
 
// This code is contributed by abhinavjain194

Python3

# Python3 program for the above approach
 
# Function to calculate sum
# of digits of an integer
def getSum(n):
  ans = 0
  for i in str(n):
    ans += int(i)
     
  return ans
 
# Function to calculate maximum sum
# of array after removing pairs whose
# concatenation is divisible by 3
def getMax(arr):
       
    # Stores the sum of
    # digits of array element
    maxRem0 = 0
     
    rem1 = 0
    rem2 = 0
 
    for i in arr:
       
        # Find the sum of digits
        digitSum = getSum(i)
         
        # If i is divisible by 3
        if digitSum % 3 == 0:
            maxRem0 = max(maxRem0, i)
             
        # Otherwise, if i modulo 3 is 1
        elif digitSum % 3 == 1:
            rem1 += i
                         
        # Otherwise, if i modulo 3 is 2
        else:
            rem2 += i
             
    # Return the resultant
    # sum of array elements
    print( maxRem0 + max(rem1, rem2))
 
# Driver Code
 
# Given array
arr = [ 23, 12, 43, 3, 56 ]
getMax(arr)

C#

// C# program for the above approach
using System;
 
class GFG{
     
  
// Function to calculate sum
// of digits of an integer
static int getSum(int n)
{
    int ans = 0;
    string str = n.ToString();
    Char[] arr = str.ToCharArray();
      
    foreach(Char ch in arr)
    {
        ans += (int)Char.GetNumericValue(ch);
    }
    return ans;
}
  
// Function to calculate maximum sum
// of array after removing pairs whose
// concatenation is divisible by 3
static void getMax(int[] arr)
{
      
    // Stores the sum of
    // digits of array element
    int maxRem0 = 0;
  
    int rem1 = 0;
    int rem2 = 0;
  
    foreach(int i in arr)
    {
          
        // Find the sum of digits
        int digitSum = getSum(i);
  
        // If i is divisible by 3
        if (digitSum % 3 == 0)
            maxRem0 = Math.Max(maxRem0, i);
  
        // Otherwise, if i modulo 3 is 1
        else if (digitSum % 3 == 1)
            rem1 += i;
  
        // Otherwise, if i modulo 3 is 2
        else
            rem2 += i;
    }
      
    // Return the resultant
    // sum of array elements
    Console.WriteLine(maxRem0 +
                     Math.Max(rem1, rem2));
}
 
 
// Driver Code
static void Main()
{
    int[] arr = { 23, 12, 43, 3, 56 };
    getMax(arr);
}
}
 
// This code is contributed by souravghosh0416.

Javascript

 
        // Javascript program for the
        // above approach
 
        // Function to calculate sum
        // of digits of an integer
        function getSum(n) {
            let ans = 0;
 
            let arr = n.toString();
            for (let i = 0; i < arr.length; i++)
            {
                ans += arr[i];
            }
            return ans;
        }
 
        // Function to calculate maximum sum
        // of array after removing pairs whose
        // concatenation is divisible by 3
        function getMax(arr, n) {
 
            // Stores the sum of
            // digits of array element
            let maxRem0 = 0;
 
            let rem1 = 0;
            let rem2 = 0;
 
            for (let i = 0; i < n; i++) {
 
                // Find the sum of digits
                let digitSum = getSum(arr[i]);
 
                // If i is divisible by 3
                if (digitSum % 3 == 0)
                    maxRem0 = Math.max(maxRem0, arr[i]);
 
                // Otherwise, if i modulo 3 is 1
                else if (digitSum % 3 == 1)
                    rem1 += arr[i];
 
                // Otherwise, if i modulo 3 is 2
                else
                    rem2 += arr[i];
            }
 
            // Return the resultant
            // sum of array elements
            document.write(maxRem0 + Math.max(rem1, rem2))
        }
 
        // Driver Code
        let arr = [23, 12, 43, 3, 56];
        let n = arr.length;
 
        getMax(arr, n);
 
        // This code is contributed by Hritik
         
    

输出：

91

时间复杂度： O(N)
辅助空间： O(1)

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