Given pairwise sum of n numbers, find the numbers - GeeksforGeeks

Given pairwise sum of n numbers, find the numbers - GeeksforGeeks
Pairwise sum of n (where n >= 3) numbers are given in a specified order, find the numbers. The order has pair sum of first and second, then first and third, first and fourth, … second and third, second and fourth, .. and so on. Consider an example: n = 4, let the numbers be {a, b, c, d}, their pairwise sum is given in the order arr[] = {a+b, a+c, a+d, b+c, b+d, c+d}.

n = 3, {a+b, a+c, b+c}
We can find b-a = arr[2] - arr[1] 
                = (b+c) - (a+c)
We can find b = (arr[0] + (b-a))/2
              = (a + b + (b - a))/2  
              = b
We can find a = arr[0] - b
              = a  

n = 4, {a+b, a+c, a+d, b+c, b+d, c+d}
We can find b-a = arr[3] - arr[1]
                = (b+c) - (a+c)  
We can find b = (arr[0] + (b-a)) / 2
              = ((a+b) + (b-a)) / 2
            a = arr[0] - b
              = (a+b) - b
            c = arr[1] - a
              = (a+c) - a
            d = arr[2] - a
              = (a+d) - a

Observation : 
b_minus_a = b - a = arr[n-1] - arr[1]
b = (arr[0] + b_minus_a)/2
a = (arr[0] - b)
c = arr[1] - a
d = arr[2] - a
..........

n = 5, {a+b, a+c, a+d, a+e, b+c, 
        b+d, b+e, c+d, c+e, d+e}

Then calculate b-a = arr[n-1] - arr[1]
                   = (b+c) - (a+c)
Then b = (arr[0] + (b-a)) / 2
       = ((a+b) + (b-a)) / 2
     a = arr[0] - b
       = (a+b) - b
Then for i=1 to n-2, 
remaining numbers are calculated as
arr[i] - a, like
       c = arr[1] - a
         = (a+c) - a
       d = arr[2] - a
         = (a+c) - a      and so on,
          .
          .
          .
          .
last number = arr[n-2] - a 

void findNumbers(int arr[], int n)

{

    int num[n];

 

    // b-a is calculated here

    int b_minus_a = arr[n-1] - arr[1];

 

    // b is calculated here

    num[1] = (arr[0] + b_minus_a) / 2;

 

    // a is calculated here

    num[0] = arr[0] - num[1];

 

    // to calculate all the other numbers

    for (int i=1; i<=(n-2); i++)

        num[i+1] = arr[i] - num[0];

 

    // display the numbers

    cout << "Numbers are: ";

    for (int i=0; i<n; i++)

        cout << num[i] << " ";

}

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