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 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 the above-chosen numbers. 

Examples: 

Input : arr1[] = {9, 3, 5, 7, 3}, arr2[] = {5, 8, 1, 4, 5} 
Output : 29 
Select first, third and fifth 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. 

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 relations 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 :  

// CPP program to maximum sum 
// combination from two arrays
#include <bits/stdc++.h>
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 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 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# 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..

<script>
    // Javascript program to maximum sum combination from two arrays
    
    // Function to maximum sum 
    // combination from two arrays
    function Max_Sum(arr1, arr2, n)
    {
        // To store dp value
        let dp = new Array(n);
        for (let i = 0; i < n; i++)
        {
            dp[i] = new Array(2);
            for (let j = 0; j < 2; j++)
            {
                dp[i][j] = 0;    
            }
        }

        // For loop to calculate the value of dp
        for (let 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]);
    }
    
    let arr1 = [9, 3, 5, 7, 3];
    let arr2 = [5, 8, 1, 4, 5];
  
    let n = arr1.length;
      
    // Function call
    document.write(Max_Sum(arr1, arr2, n));

</script>

29

Time Complexity: O(N), where N is the length of the given arrays.
Auxiliary Space: O(N)

Efficient approach : Space optimization O(1)

To optimize the space complexity since we only need to access the values of dp[i] and dp[i-1], we can just use variables to store these values instead of an entire array. This way, the space complexity will be reduced from O(N) to O(1)

Implementation Steps:

• Initialize prev1 and prev2 with the first elements of arr1 and arr2 respectively.
• Create two variables curr1 and curr2.
• Use a loop to iterate over the arrays from index 1 to n-1.
• Update prev1 and prev2 to curr1 and curr2 respectively for further iterations.
• Return the maximum of prev1 and prev2 as the maximum sum combination from the two arrays.

Implementation :

#include <bits/stdc++.h>
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 prev1 = arr1[0], prev2 = arr2[0];
    int curr1, curr2;

    // For loop to calculate the value of dp
    for (int i = 1; i < n; i++)
    {
        curr1 = max(prev1, prev2 + arr1[i]);
        curr2 = max(prev2, prev1 + arr2[i]);
      
          // assigning values for further iteration
        prev1 = curr1;
        prev2 = curr2;
    }

    // Return the required answer
    return max(prev1, prev2);
}

// 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;
}

import java.util.*;

public class Main 
{
    // Function to maximum sum combination from two arrays
    static int Max_Sum(int[] arr1, int[] arr2, int n)
    {
        // To store dp value
        int prev1 = arr1[0], prev2 = arr2[0];
        int curr1, curr2;

        // For loop to calculate the value of dp
        for (int i = 1; i < n; i++) {
            curr1 = Math.max(prev1, prev2 + arr1[i]);
            curr2 = Math.max(prev2, prev1 + arr2[i]);

            // assigning values for further iteration
            prev1 = curr1;
            prev2 = curr2;
        }

        // Return the required answer
        return Math.max(prev1, prev2);
    }

    // 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));
    }
}

def Max_Sum(arr1, arr2, n):
    # To store dp value
    prev1 = arr1[0]
    prev2 = arr2[0]
    curr1 = 0
    curr2 = 0

    # For loop to calculate the value of dp
    for i in range(1, n):
        curr1 = max(prev1, prev2 + arr1[i])
        curr2 = max(prev2, prev1 + arr2[i])

        # assigning values for further iteration
        prev1 = curr1
        prev2 = curr2

    # Return the required answer
    return max(prev1, prev2)


# Driver code
arr1 = [9, 3, 5, 7, 3]
arr2 = [5, 8, 1, 4, 5]
n = len(arr1)

# Function call
print(Max_Sum(arr1, arr2, n))

using System;

class MainClass {
    // Function to maximum sum combination from two arrays
    public static int Max_Sum(int[] arr1, int[] arr2, int n)
    {
        // To store dp value
        int prev1 = arr1[0], prev2 = arr2[0];
        int curr1, curr2;

        // For loop to calculate the value of dp
        for (int i = 1; i < n; i++) {
            curr1 = Math.Max(prev1, prev2 + arr1[i]);
            curr2 = Math.Max(prev2, prev1 + arr2[i]);

            // assigning values for further iteration
            prev1 = curr1;
            prev2 = curr2;
        }

        // Return the required answer
        return Math.Max(prev1, prev2);
    }

    // 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));
    }
}

// Function to maximum sum combination from two arrays
function Max_Sum(arr1, arr2, n) {
    // To store dp value
    let prev1 = arr1[0],
        prev2 = arr2[0];
    let curr1, curr2;
    // For loop to calculate the value of dp
    for (let i = 1; i < n; i++) {
        curr1 = Math.max(prev1, prev2 + arr1[i]);
        curr2 = Math.max(prev2, prev1 + arr2[i]);

        // assigning values for further iteration
        prev1 = curr1;
        prev2 = curr2;
    }

    // Return the required answer
    return Math.max(prev1, prev2);
}

// Driver code
let arr1 = [9, 3, 5, 7, 3];
let arr2 = [5, 8, 1, 4, 5];
let n = arr1.length;

// Function call
console.log(Max_Sum(arr1, arr2, n));
//This code is contributed by sarojmcy2e

Output

29

Time Complexity: O(N), where N is the length of the given arrays.
Auxiliary Space: O(1)