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Discover the Most sum of the Array by performing the given operations

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Given an Array A[] of measurement N with repeated parts and all array parts are constructive, the duty is to search out the most sum by making use of the given operations:

  • Choose any 2 indexes and choose 2 integers(say x and y) such that the product of the weather(x and y) is the same as the product of the weather on the chosen index.
  • Ultimately, exchange the array component with x and y


Enter: N = 3, A[] = {2, 3, 2}
Output: 14

  • i = 1,  j = 2, x = 6,  y = 1 After making use of the next operation we get sum = 9(6 + 1 + 2) and array now grow to be {6, 1, 2} now we once more selected i and j and can attempt to discover i and j
  • i = 1, j = 3 array grow to be {12, 1, 1} sum grow to be 14, thus the utmost sum will stay 14 and we can not enhance additional.

Enter: N = 2, A = {1, 3}
Output: 4

  • i = 1, j = 2, x = 3, y = 1 After making use of the next operation we get sum = 4 array now grow to be {3, 1}. If we once more apply the operation then the array will stay the identical so the utmost sum comes out to be 4. 

Strategy: Implement the thought beneath to resolve the issue:

Discover the operations then we are able to observe that we’re multiplying 2 array parts so at every time we are going to choose one integer as product and different is 1. why? As a result of this may fulfill our situation for the operation that’s given within the query that’s product of x and y must be equal to the product of array component on the index we have now Chosen. So, we are going to comply with this operation n -1 occasions and after following the operation n -1 occasions we are going to get our remaining array which encompass n – 1 ones and one component which is the same as product of all of the array parts.

Observe the beneath steps to implement the above concept:

  • Calculate the product of the array.
  • Add n – 1 to the ultimate product of the array.
  • The ensuing variable would be the most sum attainable.

Beneath is the implementation for the above strategy:


#embrace <bits/stdc++.h>

utilizing namespace std;


void maximumsum(int A[], int n)



    int ok = 1;



    for (int i = 0; i < n; i++) {

        ok *= A[i];




    ok += n - 1;



    cout << ok << endl;



int essential()



    int n = 3;

    int A[] = { 2, 3, 2 };



    maximumsum(A, n);


    return 0;


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


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