Sum of all Subarrays

Given an integer array arr[], compute the sum of all possible sub-arrays of the array. A sub-array is a contiguous part of the array.

Examples:

Input: arr[] = [1, 4, 5, 3, 2]
Output: 116
Explanation: Sum of all possible subarrays of the array [1, 4, 5, 3, 2] is 116.

Input: arr[] = [1, 2, 3, 4]
Output: 50
Explanation: Sum of all possible subarrays of the array [1, 2, 3, 4] is 50.

[Naive Approach] Using Nested Loop - O(n²) Time and O(1) Space

The main idea of this approach is to iterate over all possible subarrays of the given array using two nested loops. The outer loop selects the starting index of the subarray, while the inner loop extends the subarray to include all possible ending points. During this process, a running sum (temp) is maintained for each subarray and added to the total result.

116

[Expected Approach] Element Contribution Method - O(n) Time and O(1) Space

If we take a close look then we observe a pattern. 

Let's take an example: arr[] = [1, 4, 5, 3, 2]
All subarrays :  [1], [1, 4], [1, 4, 5], [1, 4, 5, 3], [1, 4, 5, 3, 2], [4], [4, 5], [4, 5, 3], [4, 5, 3, 2], [5], [5, 3], [5, 3, 2], [3], [3, 2], [2]

👁 subbaray-sum

Third element arr[2] appears 3 time when subarray start with arr[0]: [1, 4, 5], [1, 4, 5, 3], [1, 4, 5, 3, 2]
Third element arr[2] appears 3 time when subarray start with arr[1]: [4, 5], [4, 5, 3], [4, 5, 3, 2]
Third element arr[2] appears 3 time when subarray start with arr[2]: [5], [5, 3], [5, 3, 2]

So, We can clearly see that, For any element arr[i] in an array of size n, it appears in exactly (i + 1) * (n - i) subarrays.

For arr[] = [1, 2, 3], sum of subarrays is: arr[0] * ( 0 + 1 ) * ( 3 - 0 ) +  arr[1] * ( 1 + 1 ) * ( 3 - 1 ) + arr[2] * ( 2 + 1 ) * ( 3 - 2 ) = 1*3*3 + 2*2*2 + 3*3*1  = 20

Step by step Implementation:

• Declare an integer answer equal to zero
• Run a for loop for i from [0, n-1]: Add arr[i] * (i+1) * (n-i) into the answer at each iteration
• Return answer

116