Find all differences in an array in O(n)

Question

Question: Given a sorted array A find all possible difference of elements from A.

My solution:

for (int i=0; i

        

Answer 1

First of all the array need to be sorted

lets think a sorted array ar = {1,2,3,4}

so what we were doing at the O(n^2)

for(int i=0; i

if we do the operations here elaborately then it will look like below

when i = 0 | sum = sum + {(2-1)+(3-1)+(4-1)}                   
when i = 1 | sum = sum + {(3-2)+(4-2)}               
when i = 2 | sum = sum + {(4-3)}

if we write them all

sum = ( -1-1-1) + (2+ -2-2) + (3+3 -3) + (4+4+4 ) 

we can see that

the number at index 0 is added to the sum for 0 time and substracted from the sum for 3 time.
the number at index 1 is added to the sum for 1 time and substracted from the sum for 2 time.
the number at index 2 is added to the sum for 2 time and substracted from the sum for 1 time.
the number at index 3 is added to the sum for 3 time and substracted from the sum for 0 time.

so for we can say that,

the number at index i will be added to the sum for i time 
and will be substracted from the sum for (n-i)-1 time

Then the generalized expression for each element will be

sum = sum + (i*a[i]) – ((n-i)-1)*a[i];