两个数组的最大和组合

给定两个数组arr1 []和arr2 []，每个数组的大小均为N。任务是从两个数组中选择一些元素，以使没有两个元素具有相同的索引，并且不能从单个数组中选择两个连续的数字。找到上述选择的数字的最大和。

例子：

Input : arr1[] = {9, 3, 5, 7, 3}, arr2[] = {5, 8, 1, 4, 5}
Output : 29
Select first, third and fivth element from the first array.
Select the second and fourth element from the second array.

Input : arr1[] = {1, 2, 9}, arr2[] = {10, 1, 1}
Output : 19
Select last element from the first array and first element from the second array.

方法 ：
此问题基于动态编程。

如果最后一个元素是从位置(i-1，1 )获得的，则dp(i，1)是新选择的元素的最大和。

dp(i，2)相同，但最后一个元素的位置为(i-1，2)

dp(i，3)相同，但是我们没有从位置i-1取任何元素

。

递归实现为：

dp(i, 1)=max(dp (i – 1, 2) + arr(i, 1), dp(i – 1, 3) + arr(i, 1), arr(i, 1) );
dp(i, 2)=max(dp(i – 1, 1) + arr(i, 2 ), dp(i – 1, 3) + arr (i, 2), arr(i, 2));
dp(i, 3)=max(dp(i- 1, 1), dp( i-1, 2) ).

如果我们将dp(i，1)更新为max(dp(i，1)，dp(i-1，1))和dp(i，2)作为max ，我们实际上并不需要dp(i，3) (dp(i，2)，dp(i-1，2)) 。

因此， dp(i，j)是如果最后一个元素是从位置(i-1，1 )或更小的位置处选择的元素的最大总和。与dp(i，2)相同。因此，上述问题的答案是max(dp(n，1)，dp(n，2)) 。

下面是上述方法的实现：

C++

// CPP program to maximum sum 
// combination from two arrays
#include 
using namespace std;
  
  
// Function to maximum sum 
// combination from two arrays
int Max_Sum(int arr1[], int arr2[], int n)
{
    // To store dp value
    int dp[n][2];
      
    // For loop to calculate the value of dp
    for (int i = 0; i < n; i++) 
    {
        if(i==0)
        {
            dp[i][0] = arr1[i];
            dp[i][1] = arr2[i];
            continue;
        }
          
       dp[i][0] = max(dp[i - 1][0], dp[i - 1][1] + arr1[i]);
       dp[i][1] = max(dp[i - 1][1], dp[i - 1][0] + arr2[i]);
    }
      
    // Return the required answer
    return max(dp[n-1][0], dp[n-1][1]);
}
  
// Driver code
int main()
{
    int arr1[] = {9, 3, 5, 7, 3};
    int arr2[] = {5, 8, 1, 4, 5};
  
    int n = sizeof(arr1) / sizeof(arr1[0]);
      
    // Function call
    cout << Max_Sum(arr1, arr2, n);
  
    return 0;
}

Java

// Java program to maximum sum 
// combination from two arrays
class GFG
{
  
// Function to maximum sum 
// combination from two arrays
static int Max_Sum(int arr1[], 
                   int arr2[], int n)
{
    // To store dp value
    int [][]dp = new int[n][2];
      
    // For loop to calculate the value of dp
    for (int i = 0; i < n; i++) 
    {
        if(i == 0)
        {
            dp[i][0] = arr1[i];
            dp[i][1] = arr2[i];
            continue;
        }
          
        dp[i][0] = Math.max(dp[i - 1][0], 
                            dp[i - 1][1] + arr1[i]);
        dp[i][1] = Math.max(dp[i - 1][1],   
                            dp[i - 1][0] + arr2[i]);
    }
      
    // Return the required answer
    return Math.max(dp[n - 1][0],
                    dp[n - 1][1]);
}
  
// Driver code
public static void main(String[] args) 
{
    int arr1[] = {9, 3, 5, 7, 3};
    int arr2[] = {5, 8, 1, 4, 5};
  
    int n = arr1.length;
      
    // Function call
    System.out.println(Max_Sum(arr1, arr2, n));
}
}
  
// This code is contributed
// by PrinciRaj1992

Python3

# Python3 program to maximum sum 
# combination from two arrays
  
# Function to maximum sum 
# combination from two arrays
def Max_Sum(arr1, arr2, n):
      
    # To store dp value
    dp = [[0 for i in range(2)] 
             for j in range(n)]
      
    # For loop to calculate the value of dp
    for i in range(n):
        if(i == 0):
            dp[i][0] = arr1[i]
            dp[i][1] = arr2[i]
            continue
        else:
            dp[i][0] = max(dp[i - 1][0], 
                           dp[i - 1][1] + arr1[i])
            dp[i][1] = max(dp[i - 1][1], 
                           dp[i - 1][0] + arr2[i])
      
    # Return the required answer
    return max(dp[n - 1][0], 
               dp[n - 1][1])
  
# Driver code
if __name__ == '__main__':
    arr1 = [9, 3, 5, 7, 3]
    arr2 = [5, 8, 1, 4, 5]
  
    n = len(arr1)
      
    # Function call
    print(Max_Sum(arr1, arr2, n))
  
# This code is contributed by
# Surendra_Gangwar

C#

// C# program to maximum sum 
// combination from two arrays
using System;
  
class GFG
{
  
// Function to maximum sum 
// combination from two arrays
static int Max_Sum(int []arr1, 
                   int []arr2, int n)
{
    // To store dp value
    int [,]dp = new int[n, 2];
      
    // For loop to calculate the value of dp
    for (int i = 0; i < n; i++) 
    {
        if(i == 0)
        {
            dp[i, 0] = arr1[i];
            dp[i, 1] = arr2[i];
            continue;
        }
          
        dp[i, 0] = Math.Max(dp[i - 1, 0], 
                            dp[i - 1, 1] + arr1[i]);
        dp[i, 1] = Math.Max(dp[i - 1, 1], 
                            dp[i - 1, 0] + arr2[i]);
    }
      
    // Return the required answer
    return Math.Max(dp[n - 1, 0],
                    dp[n - 1, 1]);
} 
  
// Driver code
public static void Main() 
{
    int []arr1 = {9, 3, 5, 7, 3};
    int []arr2 = {5, 8, 1, 4, 5};
  
    int n = arr1.Length;
      
    // Function call
    Console.WriteLine(Max_Sum(arr1, arr2, n));
}
}
  
// This code is contributed
// by anuj_67..

输出：

29

时间复杂度： O(N)