254. Sum of all sub arrays in O(n) Time

Objective: Given an array write an algorithm to find the sum of all the possible sub-arrays.

Example:

int [] a = {1, 2, 3};

Output: Possible subarrays –
{1}, {2}, {3}, {1, 2} , {2, 3}, {1, 2, 3}
So sum = 1+ 2+ 3 + 3 + 5 + 6 = 20

Approach:

As discussed in Print all sub arrays, find out all the sub-arrays, and then calculate the sum.

Time complexity: O(n^3)

Output:

Sum of elements of sub arrays is: 50

Better Approach:

Let’s observe the behavior for array = {1,2,3,4}

All sub arrays are:

[1] , [1 2], [1 2 3], [1 2 3 4],
[2], [2 3], [2 3 4],
[3], [3 4]
[4]

No of occurrences for each element
- 1 appears 4 times
- 2 appears 6 times
- 3 appears 6 times
- 4 appears 4 times
For each element at first place – If we observe closely, element at first position, the sub arrays are
For 1 = [1] , [1 2], [1 2 3], [1 2 3 4] and for 2 = [2], [2 3], [2 3 4], for 3 = [3], [3 4] so for element 1, no of occurrence at first position will be equal to n (n=4) here.
The next element which is ‘2’ the number of occurrences at the first position will be one less than n. means n – 1, and so on
So for i^th element in the array will have appearances at the first position in all the sub-arrays will be = (n-i).
So for the first position, occurrences are
- 1 appears 4 times.
- 2 appears 3 times.
- 3 appears 2 times.
- 4 appears 1 time.
From Step 1 if we subtract the number of occurrences in above step, the remaining occurrences are (i is the iteration index)
- 1 = 0, n = 4, i = 0
- 2 = 3, n = 4, i = 1
- 3 = 4, n = 4, i = 2
- 4 = 3, n = 4, i = 3
From the step above, the formula which will give this result will be = (n-i)*i
So Total number of occurrences for ith index element in array will be = (n-i) + (n-i)*I => (n-i)*(i+1)
So for array {1,2,3,4}
- 1*(4-0)*(0+1) +
- 2*(4-1)*(1+1) +
- 3*(4-2)*(2+1) +
- 4*(4-3)*(3+1) = 1*4 + 2*6 + 3*6 + 4*4 = 50

Time Complexity: O(n)

Output:

Sum of elements of sub arrays is: 50

Reference: Her e