通过合并两个给定的数组可能获得的最大前缀和

给定分别由N和M个整数组成的两个数组A []和B [] ，任务是计算可以通过合并两个数组获得的最大前缀和。

例子：

Input : A[] = {2, -1, 4, -5}, B[]={4, -3, 12, 4, -3}
Output : 22
Explanation: Merging the two arrays to generate the sequence {2, 4, -1, -3, 4, 12, 4, -5, -3}. Maximum prefix sum = Sum of {arr[0], …, arr[6]} = 22.

Input: A[] = {2, 1, 13, 5, 14}, B={-1, 4, -13}
Output: 38
Explanation: Merging the two arrays to generate the sequence {2, 1, -1, 13, 5, 14, -13}. Maximum prefix sum = Sum of {arr[0], …, arr[6]} = 38.

天真的方法：最简单的方法是使用递归，可以使用“记忆化”对其进行优化。请按照以下步骤解决问题：

• 初始化map ，int> dp []以进行记忆。
• 定义一个递归函数，例如maxPresum(x，y)来找到最大前缀和：
  • 如果DP [{X，Y}]已经计算出，然后返回DP [{X，Y}]。
  • 否则，如果x == N || y == M，然后返回0 。
  • 将dp [{x，y}]更新为dp [{x，y}] = max(dp [{x，y}，a [x] + maxPresum(x + 1，y)，b [y] + maxPresum( x，y + 1)) ，然后返回dp [{x，y}]。
• 将最大前缀总和打印为maxPresum(0，0) 。

下面是上述方法的实现：

C++

// C++ implementation of above approach
#include 
 
#define int long long
using namespace std;
 
// Stores the dp states
map, int> dp;
 
// Recursive Function to calculate the maximum
// prefix sum ontained by merging two arrays
int maxPreSum(vector a, vector b,
              int x, int y)
{
    // If subproblem is already computed
    if (dp.find({ x, y }) != dp.end())
        return dp[{ x, y }];
 
    // If x >= N or y >= M
    if (x == a.size() && y == b.size())
        return 0;
 
    int curr = dp[{ x, y }];
 
    // If x < N
    if (x == a.size()) {
        curr = max(curr, b[y]
                             + maxPreSum(a, b, x, y + 1));
    }
    // If y A = { 2, 1, 13, 5, 14 };
    vector B = { -1, 4, -13 };
    cout << maxPreSum(A, B, 0, 0) << endl;
    return 0;
}

C#

// C# implementation of above approach
using System;
using System.Collections.Generic;
 
class GFG{
     
// Stores the dp states
static Dictionary, int> dp = new Dictionary, int>();
 
// Recursive Function to calculate the maximum
// prefix sum ontained by merging two arrays
static int maxPreSum(int[] a, int[] b, int x, int y)
{
     
    // If subproblem is already computed
    if (dp.ContainsKey(new Tuple(x, y)))
        return dp[new Tuple(x, y)];
     
    // If x >= N or y >= M
    if (x == a.Length && y == b.Length)
        return 0;
     
    int curr = 0;
    if (dp.ContainsKey(new Tuple(x, y)))
    {
        curr = dp[new Tuple(x, y)];
    }
     
    // If x < N
    if (x == a.Length)
    {
        curr = Math.Max(curr, b[y] + maxPreSum(
            a, b, x, y + 1));
    }
     
    // If y(x, y)] = curr;
    return dp[new Tuple(x, y)];
}
 
// Driver code
static void Main()
{
    int[] A = { 2, 1, 13, 5, 14 };
    int[] B = { -1, 4, -13 };
     
    Console.WriteLine(maxPreSum(A, B, 0, 0));
}
}
 
// This code is contributed by divyesh072019

输出：

38

时间复杂度： O(N·M)
辅助空间： O(N * M)

高效方法：可以基于以下观察条件来优化上述方法：最大前缀和等于数组A []和B []的最大前缀和。请按照以下步骤解决问题：

• 计算数组A []的最大前缀和，并将其存储在变量中，例如X。
• 计算数组B []的最大前缀和，并将其存储在变量中，例如Y。
• 打印X和Y的总和。

下面是上述方法的实现：

C++

// C++ Program to implement
// the above approach
#include 
using namespace std;
 
int maxPresum(vector a, vector b)
{
    // Stores the maximum prefix
    // sum of the array A[]
    int X = max(a[0], 0);
 
    // Traverse the array A[]
    for (int i = 1; i < a.size(); i++) {
        a[i] += a[i - 1];
        X = max(X, a[i]);
    }
 
    // Stores the maximum prefix
    // sum of the array B[]
    int Y = max(b[0], 0);
 
    // Traverse the array B[]
    for (int i = 1; i < b.size(); i++) {
        b[i] += b[i - 1];
        Y = max(Y, b[i]);
    }
    return X + Y;
}
 
// Driver code
int main()
{
    vector A = { 2, -1, 4, -5 };
    vector B = { 4, -3, 12, 4, -3 };
    cout << maxPresum(A, B) << endl;
}

Java

// Java Program to implement
// the above approach
import java.util.*;
class GFG {
 
  static int maxPresum(int [] a, int [] b)
  {
     
    // Stores the maximum prefix
    // sum of the array A[]
    int X = Math.max(a[0], 0);
 
    // Traverse the array A[]
    for (int i = 1; i < a.length; i++)
    {
      a[i] += a[i - 1];
      X = Math.max(X, a[i]);
    }
 
    // Stores the maximum prefix
    // sum of the array B[]
    int Y = Math.max(b[0], 0);
 
    // Traverse the array B[]
    for (int i = 1; i < b.length; i++) {
      b[i] += b[i - 1];
      Y = Math.max(Y, b[i]);
    }
    return X + Y;
  }
 
  // Driver code
  public static void main(String [] args)
  {
    int [] A = { 2, -1, 4, -5 };
    int [] B = { 4, -3, 12, 4, -3 };
    System.out.print(maxPresum(A, B));
  }
}
 
// This code is contributed by ukasp.

Python3

# Python3 implementation of the
# above approach
 
def maxPresum(a, b) :
     
    # Stores the maximum prefix
    # sum of the array A[]
    X = max(a[0], 0)
 
    # Traverse the array A[]
    for i in range(1, len(a)):
        a[i] += a[i - 1]
        X = max(X, a[i])
     
    # Stores the maximum prefix
    # sum of the array B[]
    Y = max(b[0], 0)
 
    # Traverse the array B[]
    for i in range(1, len(b)):
        b[i] += b[i - 1]
        Y = max(Y, b[i])
     
    return X + Y
 
# Driver code
 
A = [ 2, -1, 4, -5 ]
B = [ 4, -3, 12, 4, -3 ]
print(maxPresum(A, B))
 
# This code is contributed by code_hunt.

C＃

// C# Program to implement
// the above approach
using System;
using System.Collections.Generic;
class GFG {
 
  static int maxPresum(List a, List b)
  {
     
    // Stores the maximum prefix
    // sum of the array A[]
    int X = Math.Max(a[0], 0);
 
    // Traverse the array A[]
    for (int i = 1; i < a.Count; i++)
    {
      a[i] += a[i - 1];
      X = Math.Max(X, a[i]);
    }
 
    // Stores the maximum prefix
    // sum of the array B[]
    int Y = Math.Max(b[0], 0);
 
    // Traverse the array B[]
    for (int i = 1; i < b.Count; i++) {
      b[i] += b[i - 1];
      Y = Math.Max(Y, b[i]);
    }
    return X + Y;
  }
 
  // Driver code
  static void Main()
  {
    List A = new List(new int[]{ 2, -1, 4, -5 });
    List B = new List(new int[]{ 4, -3, 12, 4, -3 });
    Console.WriteLine(maxPresum(A, B));
  }
}
 
// This code is contributed by divyeshrabadiya07.

输出：

22

时间复杂度： O(M + N)
辅助空间： O(1)