Skip to content
Related Articles

Related Articles

Maximize sum of subsets from two arrays having no consecutive values
  • Difficulty Level : Hard
  • Last Updated : 04 Feb, 2021

Given two arrays arr1[] and arr2[] of equal length, the task is to find the maximum sum of any subset possible by selecting elements from both the arrays such that no two elements in the subset should be consecutive.

Examples:

Input: arr1[] = {-1, -2, 4, -4, 5}, arr2[] = {-1, -2, -3, 4, 10}
Output: 14
Explanation:
Required subset {4, 10}. Therefore, sum = 4 + 10 = 14.

Input: arr1[] = {2, 5, 4, 2000}, arr2[] = {-2000, 100, 23, 40}
Output: 2100

Naive Approach: The simplest approach is to generate all possible subsets from both the given arrays such that no two adjacent elements are consecutive and calculate sum of each subset. Finally, print the maximum sum possible. 
Time Complexity: O(N*2N)
Auxiliary Space: O(2N)

Efficient Approach: The above approach can be optimized using Dynamic Programming. Follow the steps below to solve the problem:



  • Initialize an auxiliary array dp[] of size N.
  • Here, dp[i] stores the maximum possible sum of a subset from both the arrays such that no two elements are consecutive.
  • Declare a function maximumSubsetSum():
    • Base Cases:
      • dp[1] = max(arr1[1], arr2[1]).
      • dp[2] = max(max(arr1[1], arr2[1]), max(arr1[2], arr2[2])).
    • For all other cases, following three conditions arise:
      • dp[i] = max(arr1[i], arr2[i], arr1[i] + dp[i – 2], arr2[i] + dp[i – 2], dp[i – 1]).
  • Finally, print dp[N] as the required answer.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to calculate maximum subset sum
void maximumSubsetSum(int arr1[],int arr2[], int length)
{
 
    // Initialize array to store dp states
    int dp[length+1];
 
    // Base Cases
    if (length == 1)
    {
        cout << (max(arr1[0], arr2[0]));
        return;
    }
 
    if (length == 2)
    {
        cout << (max(max(arr1[1], arr2[1]), max(arr1[0], arr2[0])));
        return;
    }
    else
    {
 
        // Pre initializing for dp[0] & dp[1]
        dp[0] = max(arr1[0], arr2[0]);
        dp[1] = max(max(arr1[1], arr2[1]), max(arr1[0], arr2[0]));
 
        int index = 2;
        while (index < length)
        {
 
            // Calculating dp[index] based on
            // above formula
            dp[index] = max(max(arr1[index], arr2[index]),
                max(max(arr1[index] + dp[index - 2],
                        arr2[index] + dp[index - 2]),
                    dp[index - 1]));
            ++index;
        }
 
        // Print maximum subset sum
        cout<<(dp[length - 1]);
    }
}
 
// Driver Code
int main()
{
   
  // Given arrays
  int arr1[] = { -1, -2, 4, -4, 5 };
  int arr2[] = { -1, -2, -3, 4, 10 };
 
  // Length of the array
  int length = 5;
  maximumSubsetSum(arr1, arr2, length);
  return 0;
}
 
// This code is contributed by mohit kumar 29

Java




// Java program for the above approach
 
import java.io.*;
import java.util.*;
class GFG {
 
    // Function to calculate maximum subset sum
    static void maximumSubsetSum(int arr1[],
                                 int arr2[],
                                 int length)
    {
 
        // Initialize array to store dp states
        int dp[] = new int[length + 1];
 
        // Base Cases
        if (length == 1) {
            System.out.print(
                Math.max(arr1[0], arr2[0]));
            return;
        }
 
        if (length == 2) {
            System.out.print(
                Math.max(
                    Math.max(arr1[1], arr2[1]),
                    Math.max(arr1[0], arr2[0])));
            return;
        }
        else {
 
            // Pre initializing for dp[0] & dp[1]
            dp[0] = Math.max(arr1[0], arr2[0]);
            dp[1] = Math.max(
                Math.max(arr1[1], arr2[1]),
                Math.max(arr1[0], arr2[0]));
 
            int index = 2;
            while (index < length) {
 
                // Calculating dp[index] based on
                // above formula
                dp[index] = Math.max(
                    Math.max(arr1[index], arr2[index]),
                    Math.max(
                        Math.max(
                            arr1[index] + dp[index - 2],
                            arr2[index] + dp[index - 2]),
                        dp[index - 1]));
                ++index;
            }
 
            // Print maximum subset sum
            System.out.print(dp[length - 1]);
        }
    }
 
    // Driver Code
    public static void main(String[] args)
    {
 
        // Given arrays
        int arr1[] = { -1, -2, 4, -4, 5 };
        int arr2[] = { -1, -2, -3, 4, 10 };
 
        // Length of the array
        int length = arr1.length;
 
        maximumSubsetSum(arr1, arr2, length);
    }
}

Python3




# Python program of the above approach
 
# Function to calculate maximum subset sum
def maximumSubsetSum(arr1, arr2, length) :
 
    # Initialize array to store dp states
    dp = [0] * (length+1)
 
    # Base Cases
    if (length == 1) :     
        print(max(arr1[0], arr2[0]))
        return
    if (length == 2) :
        print(max(max(arr1[1], arr2[1]), max(arr1[0], arr2[0])))
        return 
    else  :
 
        # Pre initializing for dp[0] & dp[1]
        dp[0] = max(arr1[0], arr2[0])
        dp[1] = max(max(arr1[1], arr2[1]), max(arr1[0], arr2[0]))
        index = 2
        while (index < length) :
 
            # Calculating dp[index] based on
            # above formula
            dp[index] = max(max(arr1[index], arr2[index]),
                max(max(arr1[index] + dp[index - 2],
                        arr2[index] + dp[index - 2]),
                    dp[index - 1]))
            index += 1
         
        # Prmaximum subset sum
        print(dp[length - 1])
     
# Driver Code
 
# Given arrays
arr1 = [ -1, -2, 4, -4, 5 ]
arr2 = [ -1, -2, -3, 4, 10 ]
 
# Length of the array
length = 5
maximumSubsetSum(arr1, arr2, length)
 
# This code is contributed by susmitakundugoaldanga.

C#




// C# program for the above approach
using System;
class GFG
{
 
  // Function to calculate maximum subset sum
  static void maximumSubsetSum(int[] arr1,
                               int[] arr2,
                               int length)
  {
 
    // Initialize array to store dp states
    int[] dp = new int[length + 1];
 
    // Base Cases
    if (length == 1) {
      Console.WriteLine(Math.Max(arr1[0], arr2[0]));
      return;
    }
 
    if (length == 2)
    {
      Console.WriteLine(Math.Max(
        Math.Max(arr1[1], arr2[1]),
        Math.Max(arr1[0], arr2[0])));
      return;
    }
    else
    {
 
      // Pre initializing for dp[0] & dp[1]
      dp[0] = Math.Max(arr1[0], arr2[0]);
      dp[1] = Math.Max(Math.Max(arr1[1], arr2[1]),
                       Math.Max(arr1[0], arr2[0]));
 
      int index = 2;
      while (index < length) {
 
        // Calculating dp[index] based on
        // above formula
        dp[index] = Math.Max(Math.Max(arr1[index], arr2[index]),
                             Math.Max(Math.Max(arr1[index] +
                                               dp[index - 2],
                                               arr2[index] +
                                               dp[index - 2]),
                                      dp[index - 1]));
        ++index;
      }
 
      // Print maximum subset sum
      Console.WriteLine(dp[length - 1]);
    }
  }
 
  // Driver Code
  static public void Main()
  {
 
    // Given arrays
    int[] arr1 = { -1, -2, 4, -4, 5 };
    int[] arr2 = { -1, -2, -3, 4, 10 };
 
    // Length of the array
    int length = arr1.Length;
 
    maximumSubsetSum(arr1, arr2, length);
  }
}
 
// This code is contributed by code_hunt.

 
 

Output: 
14

 

 

Time Complexity: O(N)
Auxiliary Space: O(N)

 

Attention reader! Don’t stop learning now. Get hold of all the important mathematical concepts for competitive programming with the Essential Maths for CP Course at a student-friendly price. To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

My Personal Notes arrow_drop_up
Recommended Articles
Page :