Find original array from encrypted array (An array of sums of other elements) - GeeksforGeeks

Find original array from encrypted array (An array of sums of other elements) - GeeksforGeeks
Find original array from a given encrypted array of size n. Encrypted array is obtained by replacing each element of the original array by the sum of the remaining array elements.

Let n = 4, and
the original array be ori[] = {a, b, c, d}
encrypted array is given as:
arr[] = {b+c+d, a+c+d, a+b+d, a+b+c}

Elements of encrypted array are :
arr[0] = (b+c+d), arr[1] = (a+c+d), 
arr[2] = (a+b+d), arr[3] = (a+b+c)
add up all the elements
sum =  arr[0] + arr[1] + arr[2] + arr[3]
       = (b+c+d) + (a+c+d) + (a+b+d) + (a+b+c)
       = 3(a+b+c+d) 
Sum of elements of ori[] = sum / n-1
                        = sum/3 
                        = (a+b+c+d)
Thus, for a given encrypted array arr[] of size n, the sum of 
the elements of the original array ori[] can be calculated as:
sum =  (arr[0]+arr[1]+....+arr[n-1]) / (n-1)

Then, elements of ori[] are calculated as:
ori[0] = sum - arr[0]
ori[1] = sum - arr[1] 
        .
        .
ori[n-1] = sum - arr[n-1]   

void findAndPrintOriginalArray(int arr[], int n)

{

    // total sum of elements

    // of encrypted array

    int arr_sum = 0;

    for (int i=0; i<n; i++)

        arr_sum += arr[i];

 

    // total sum of elements

    // of original array

    arr_sum = arr_sum/(n-1);

 

    // calculating and displaying

    // elements of original array

    for (int i=0; i<n; i++)

        cout << (arr_sum - arr[i]) << " ";

}

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