Maximize the Product of Sum by changing Array parts into given two varieties

Given an array of optimistic integer arr[] of size N and an integer Z, (Z > arr[i] for all 0 ≤ i ≤ N – 1). Every integer of the array may be transformed into the next two varieties:

• Preserve it unchanged
• Change it to Z – arr[i].

The duty is to maximise the product of the sum of those two kinds of parts.

Observe: There ought to be current at the least one ingredient of every kind. 

Examples:

Enter: N = 5, arr[] = {500, 100, 400, 560, 876}, Z = 1000
Output: 290400
Clarification: arr[] = {500, 100, 400, 560, 876}
Convert parts current at indices 0, 3 and 4 to first kind =  (500, 560, 876)
Convert parts current at indices 1 and a pair of to second kind 
= (Z-arr[1],  Z-arr[2]) = (1000 – 100, 1000 – 400) = (900, 600)
Sum of all first kind parts = 500+560+876 = 1936
Sum of all second kind parts = 900 + 600 = 1500  
Product of every kind sum = 1936*1500 = 290400. 
Which is most potential for this case.

Enter: N = 4, arr[] = {1, 4, 6, 3}, Z = 7
Output: 100
Clarification: Change the first and final ingredient to 2nd kind, i.e.,
{7-1, 7-3} = {6, 4}. The sum is (6 + 4) = 10.
Preserve the 2nd and third ingredient as it’s. Their sum = (4 + 6) = 10 .
Product is 10*10 = 100. That is the utmost product potential.

Method: The issue may be solved utilizing Sorting based mostly on the next thought:

The thought is to kind the arr[] in reducing order, Calculate product of all potential mixtures of the categories. Get hold of most product amongst all mixtures. 

Illustration:

Enter: N = 4, arr[] = {1, 4, 6, 3}, Z = 7

After sorting arr[] in reducing order = {6, 4, 3, 1}

Now now we have 3 potential mixtures for selecting all parts as first or second kind:

• 1 of first kind, 3 of second kind 
• 2 of first kind, 2 of second kind 
• 3 of first kind, 1 of second kind 

Let’s see the product and sum at every mixture for reducing ordered arr[]:

Selecting first ingredient as first kind and subsequent 3 parts as second kind:

• Sum of first kind parts = 6
• Sum of second kind parts = ((7 – 4)+(7 – 3)+(7 – 1))= 13
• Product of first and second = 6 * 13 = 78

Selecting first two parts as first kind and final 2 parts as second kind:

• Sum of first kind parts = 6 + 4 = 10
• Sum of second kind parts = (7 – 3)+(7 – 1))= 10
• Product of first and second varieties = 10 * 10 = 100

Selecting first three parts as first kind and final ingredient as second kind:

• Sum of first kind parts = 6 + 4 + 3 = 13
• Sum of second kind parts = (7 – 1)) = 6
• Product of first and second varieties = 13 * 6 = 78

As we are able to clearly see that 2nd mixture has most worth of product.Due to this fact, output for this case is : 

Most Product: 100

Comply with the steps to unravel the issue:

• Type the enter array arr[].
• Traverse from the top of the array to calculate the product for all potential mixtures:
  • Think about all of the ingredient until index i as first kind, and the suffix parts as second kind.
  • Calculate the product of this mix.
  • Replace the utmost product accordingly.
• Print the utmost product obtained.

Beneath is the implementation of the above strategy.

Time Complexity: O(N²)
Auxiliary House: O(1)