Maximum value obtained by performing given operations in an Array

Given an array arr[], the task is to find out the maximum obtainable value. The user is allowed to add or multiply the two consecutive elements. However, there has to be at least one addition operation between two multiplication operations (i.e), two consecutive multiplication operations are not allowed.
Let the array elements be 1, 2, 3, 4 then 1 * 2 + 3 + 4 is a valid operation whereas 1 + 2 * 3 * 4 is not a a valid operation as there are consecutive multiplication operations.
Examples: 
 

Input : 5 -1 -5 -3 2 9 -4
Output : 33
Explanation:
The maximum value obtained by following the above conditions is 33. 
The sequence of operations are given as:
5 + (-1) + (-5) * (-3) + 2 * 9 + (-4) = 33
Input : 5 -3 -5 2 3 9 4
Output : 62

Approach:
This problem can be solved by using dynamic programming. 
 

1. Assuming 2D array dp[][] of dimensions n * 2.
2. dp[i][0] represents the maximum value of the array up to ith position if the last operation is addition.
3. dp[i][1] represents the maximum value of the array up to ith position if the last operation is multiplication.

Now, since consecutive multiplication operation is not allowed, the recurrence relation can be considered as : 
 

dp[i][0] = max(dp[ i - 1][0], dp[ i - 1][1]) + a[ i + 1];
dp[i][1] = dp[i - 1][0] - a[i] + a[i] * a[i + 1];

The base cases are: 
 

dp[0][0] = a[0] + a[1];
dp[0][1] = a[0] * a[1];

Below is the implementation of the above approach: 
 

// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;

// A function to calculate the maximum value
void findMax(int a[], int n)
{
    int dp[n][2];
    memset(dp, 0, sizeof(dp));
    
    // basecases
    dp[0][0] = a[0] + a[1];
    dp[0][1] = a[0] * a[1];
   
    //Loop to iterate and add the max value in the dp array
    for (int i = 1; i <= n - 2; i++) {
        dp[i][0] = max(dp[i - 1][0], dp[i - 1][1]) + a[i + 1];
        dp[i][1] = dp[i - 1][0] - a[i] + a[i] * a[i + 1];
    }

    cout << max(dp[n - 2][0], dp[n - 2][1]);
}

// Driver Code
int main()
{
    int arr[] = { 5, -1, -5, -3, 2, 9, -4 };
    findMax(arr, 7);
}

// Java implementation of the above approach 
import java.util.*;
class GFG 
{
    
    // A function to calculate the maximum value 
    static void findMax(int []a, int n) 
    { 
        int dp[][] = new int[n][2]; 
        int i, j;
        
        for (i = 0; i < n ; i++)
            for(j = 0; j < 2; j++)
                dp[i][j] = 0;

        // basecases 
        dp[0][0] = a[0] + a[1]; 
        dp[0][1] = a[0] * a[1]; 
        
        // Loop to iterate and add the 
        // max value in the dp array 
        for (i = 1; i <= n - 2; i++)
        { 
            dp[i][0] = Math.max(dp[i - 1][0], 
                                dp[i - 1][1]) + a[i + 1]; 
            dp[i][1] = dp[i - 1][0] - a[i] + 
                        a[i] * a[i + 1]; 
        } 
    
        System.out.println(Math.max(dp[n - 2][0], 
                                    dp[n - 2][1])); 
    } 
    
    // Driver Code 
    public static void main (String[] args) 
    {
        int arr[] = { 5, -1, -5, -3, 2, 9, -4 }; 
        findMax(arr, 7); 
    } 
}

// This code is contributed by AnkitRai01

# Python3 implementation of the above approach 
import numpy as np

# A function to calculate the maximum value 
def findMax(a, n) : 

    dp = np.zeros((n, 2));
    
    # basecases 
    dp[0][0] = a[0] + a[1]; 
    dp[0][1] = a[0] * a[1]; 
    
    # Loop to iterate and add the max value in the dp array 
    for i in range(1, n - 1) :
        dp[i][0] = max(dp[i - 1][0], dp[i - 1][1]) + a[i + 1]; 
        dp[i][1] = dp[i - 1][0] - a[i] + a[i] * a[i + 1]; 

    print(max(dp[n - 2][0], dp[n - 2][1]), end =""); 

# Driver Code 
if __name__ == "__main__" : 

    arr = [ 5, -1, -5, -3, 2, 9, -4 ]; 
    findMax(arr, 7); 
    
# This code is contributed by AnkitRai01

// C# implementation of the above approach 
using System;

class GFG 
{
    
    // A function to calculate the maximum value 
    static void findMax(int []a, int n) 
    { 
        int [,]dp = new int[n, 2]; 
        int i, j;
        
        for (i = 0; i < n ; i++)
            for(j = 0; j < 2; j++)
                dp[i, j] = 0;

        // basecases 
        dp[0, 0] = a[0] + a[1]; 
        dp[0, 1] = a[0] * a[1]; 
        
        // Loop to iterate and add the 
        // max value in the dp array 
        for (i = 1; i <= n - 2; i++)
        { 
            dp[i, 0] = Math.Max(dp[i - 1, 0], dp[i - 1, 1]) + a[i + 1];
                                
            dp[i, 1] = dp[i - 1, 0] - a[i] + a[i] * a[i + 1]; 
        } 
    
        Console.WriteLine(Math.Max(dp[n - 2, 0], dp[n - 2, 1])); 
    } 
    
    // Driver Code 
    public static void Main() 
    {
        int []arr = { 5, -1, -5, -3, 2, 9, -4 }; 
        findMax(arr, 7); 
    } 
}

// This code is contributed by AnkitRai01

<script>
// javascript implementation of the above approach 
// A function to calculate the maximum value
    function findMax(a , n) {
        var dp = Array(n).fill().map(()=>Array(2).fill(0));
        var i, j;

        for (i = 0; i < n; i++)
            for (j = 0; j < 2; j++)
                dp[i][j] = 0;

        // basecases
        dp[0][0] = a[0] + a[1];
        dp[0][1] = a[0] * a[1];

        // Loop to iterate and add the
        // max value in the dp array
        for (i = 1; i <= n - 2; i++) {
            dp[i][0] = Math.max(dp[i - 1][0], dp[i - 1][1]) + a[i + 1];
            dp[i][1] = dp[i - 1][0] - a[i] + a[i] * a[i + 1];
        }

        document.write(Math.max(dp[n - 2][0], dp[n - 2][1]));
    }

    // Driver Code
    
        var arr = [ 5, -1, -5, -3, 2, 9, -4 ];
        findMax(arr, 7);

// This code contributed by Rajput-Ji 
</script>

33

Time complexity: O(n) where n is size of given array
Auxiliary space: O(n) because it is using extra space for array "dp"

Efficient approach : Space optimization O(1)

In previous approach the current value ( dp[i] ) is depend upon the ( dp[i-1] ) so which is just the previous computation of DP array. So to optimize the space complexity we only require the computation of previous index.

Implementation Steps:

• Create variables prev_sum and prev_prod to get the computation of just previous index of DP.
• Initialize them with a[0] + a[1] and a[0] * a[1] resp.
• Now create variables curr_sum and  curr_prod to get the current value from previous computation.
• Create variable max_val to store the final result.
• Now iterate over subproblems and get the current value from prev_sum and prev_prod.
• After every iteration assign values of curr_sum to prev_sum and curr_prod to prev_prod to iterate further.
• At last print answer stored in max_val.

Implementation:

// C++ program for above approach

#include <bits/stdc++.h>
using namespace std;

// A function to calculate the maximum value
void findMax(int a[], int n)
{
    // initialize variables for previous sum and product
    int prev_sum = a[0] + a[1];
    int prev_prod = a[0] * a[1];

    // initialize variables for current sum and product
    int curr_sum, curr_prod;

    // to store the final result
    int max_val;

    // Loop to iterate and update the max value
    for (int i = 1; i <= n - 2; i++) {
        // get current value from previous computation
        // store in previous sum and previous product
        curr_sum = max(prev_sum, prev_prod) + a[i + 1];
        curr_prod = prev_sum - a[i] + a[i] * a[i + 1];

        // update answer
        max_val = max(curr_sum, curr_prod);

        // assigning values of currsum and currproduct
        // to prev sum and product to iterate further
        prev_sum = curr_sum;
        prev_prod = curr_prod;
    }

    // print the final result
    cout << max_val;
}

// Driver Code
int main()
{
    int arr[] = { 5, -1, -5, -3, 2, 9, -4 };

    // function call
    findMax(arr, 7);
}

import java.util.*;

class Main {
    // A function to calculate the maximum value
    static void findMax(int a[], int n)
    {
        // initialize variables for previous sum and product
        int prev_sum = a[0] + a[1];
        int prev_prod = a[0] * a[1];

        // initialize variables for current sum and product
        int curr_sum, curr_prod;

        // to store the final result
        int max_val = 0;

        // Loop to iterate and update the max value
        for (int i = 1; i <= n - 2; i++) {
            // get current value from previous computation
            // store in previous sum and previous product
            curr_sum
                = Math.max(prev_sum, prev_prod) + a[i + 1];
            curr_prod = prev_sum - a[i] + a[i] * a[i + 1];

            // update answer
            max_val = Math.max(curr_sum, curr_prod);

            // assigning values of currsum and currproduct
            // to prev sum and product to iterate further
            prev_sum = curr_sum;
            prev_prod = curr_prod;
        }

        // print the final result
        System.out.println(max_val);
    }

    // Driver Code
    public static void main(String[] args)
    {
        int arr[] = { 5, -1, -5, -3, 2, 9, -4 };

        // function call
        findMax(arr, 7);
    }
}

# Python program for above approach

# A function to calculate the maximum value


def findMax(a, n):
    # initialize variables for previous sum and product
    prev_sum = a[0] + a[1]
    prev_prod = a[0] * a[1]

    # initialize variables for current sum and product
    curr_sum, curr_prod = 0, 0

    # to store the final result
    max_val = 0

    # Loop to iterate and update the max value
    for i in range(1, n - 1):
        # get current value from previous computation
        # store in previous sum and previous product
        curr_sum = max(prev_sum, prev_prod) + a[i + 1]
        curr_prod = prev_sum - a[i] + a[i] * a[i + 1]

        # update answer
        max_val = max(curr_sum, curr_prod)

        # assigning values of currsum and currproduct
        # to prev sum and product to iterate further
        prev_sum = curr_sum
        prev_prod = curr_prod

    # print the final result
    print(max_val)


# Driver Code
arr = [5, -1, -5, -3, 2, 9, -4]

# function call
findMax(arr, 7)

using System;

public class Program
{
    public static void Main()
    {
        int[] arr = { 5, -1, -5, -3, 2, 9, -4 };
        findMax(arr, 7);
    }

    // A function to calculate the maximum value
    static void findMax(int[] a, int n)
    {
        // initialize variables for previous sum and product
        int prev_sum = a[0] + a[1];
        int prev_prod = a[0] * a[1];

        // initialize variables for current sum and product
        int curr_sum, curr_prod;

        // to store the final result
        int max_val = 0;

        // Loop to iterate and update the max value
        for (int i = 1; i <= n - 2; i++)
        {
            // get current value from previous computation
            // store in previous sum and previous product
            curr_sum = Math.Max(prev_sum, prev_prod) + a[i + 1];
            curr_prod = prev_sum - a[i] + a[i] * a[i + 1];

            // update answer
            max_val = Math.Max(curr_sum, curr_prod);

            // assigning values of currsum and currproduct
            // to prev sum and product to iterate further
            prev_sum = curr_sum;
            prev_prod = curr_prod;
        }

        // print the final result
        Console.WriteLine(max_val);
    }
}

function findMax(arr) {
  // Initialize variables for previous sum and product
  let prev_sum = arr[0] + arr[1];
  let prev_prod = arr[0] * arr[1];

  // Initialize variables for current sum and product
  let curr_sum, curr_prod;

  // Initialize a variable to store the final result
  let max_val;

  // Loop to iterate and update the max value
  for (let i = 1; i < arr.length - 1; i++) {
    // Get the current value from previous computation
    // Store it in current sum and current product
    curr_sum = Math.max(prev_sum, prev_prod) + arr[i + 1];
    curr_prod = prev_sum - arr[i] + arr[i] * arr[i + 1];

    // Update the answer
    max_val = Math.max(curr_sum, curr_prod);

    // Assign values of curr_sum and curr_prod
    // to prev_sum and prev_prod to iterate further
    prev_sum = curr_sum;
    prev_prod = curr_prod;
  }

  // Print the final result
  console.log(max_val);
}

// Driver Code
const arr = [5, -1, -5, -3, 2, 9, -4];

// Function call
findMax(arr);

Output

33

Time complexity: O(n) where n is size of given array
Auxiliary space: O(1) because no extra space is used.