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Maximum sum combination from two arrays

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  • Difficulty Level : Medium
  • Last Updated : 04 Sep, 2021

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 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 :  

C++




// 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




// 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..

Javascript




<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>

Output: 

29

 

Time Complexity: O(N)
 


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