# Maximum sum combination from two arrays

Given two arrays arr1[] and arr2[] each of size N. The task is to choose some elements from both the arrays such that no two elements have the same index and no two consecutive numbers can be selected from a single array. Find the maximum sum possible of above-chosen numbers.

Examples:

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.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach :
This problem is based on dynamic programming.

• Let dp(i, 1) be the maximum sum of the newly selected elements if the last element was taken from the position(i-1, 1).
• dp(i, 2) is the same but the last element taken has the position (i-1, 2)
• dp(i, 3) the same but we didn’t take any element from position i-1
• .

Recursion realtions are :

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) ).

We don’t actually need dp( i, 3), if we update dp(i, 1) as max(dp(i, 1), dp(i-1, 1)) and dp(i, 2) as max(dp(i, 2), dp(i-1, 2)).

Thus, dp(i, j) is the maximum total sum of the elements that are selected if the last element was taken from the position (i-1, 1) or less. The same with dp(i, 2). Therefore the answer to the above problem is max(dp(n, 1), dp(n, 2)).

Below is the implementation of the above approach :

## 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]; ` `     `  `    ``// For loop to calculate the value of dp ` `    ``for` `(``int` `i = 0; i < n; i++)  ` `    ``{ ` `        ``if``(i==0) ` `        ``{ ` `            ``dp[i] = arr1[i]; ` `            ``dp[i] = arr2[i]; ` `            ``continue``; ` `        ``} ` `         `  `       ``dp[i] = max(dp[i - 1], dp[i - 1] + arr1[i]); ` `       ``dp[i] = max(dp[i - 1], dp[i - 1] + arr2[i]); ` `    ``} ` `     `  `    ``// Return the required answer ` `    ``return` `max(dp[n-1], dp[n-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); ` `     `  `    ``// 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.. `

Output:

```29
```

Time Complexity: O(N)

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