Given an integer array ‘arr[]’ of size n, find sum of all sub-arrays of given array.
Examples:
Input : arr[] = {1, 2, 3}
Output : 20
Explanation : {1} + {2} + {3} + {2 + 3} +
{1 + 2} + {1 + 2 + 3} = 20
Input : arr[] = {1, 2, 3, 4}
Output : 50
// Simple Java program to compute sum of
// subarray elements
class GFG {
// Computes sum all sub-array
public static long SubArraySum(int arr[], int n)
{
long result = 0;
// Pick starting point
for (int i = 0; i < n; i ++)
{
// Pick ending point
for (int j = i; j < n; j ++)
{
// sum subarray between current
// starting and ending points
for (int k = i; k <= j; k++)
result += arr[k] ;
}
}
return result ;
}
If we take a close look then we observe a pattern. Let take an example
arr[] = [1, 2, 3], n = 3
All subarrays : [1], [1, 2], [1, 2, 3],
[2], [2, 3], [3]
here first element 'arr[0]' appears 3 times
second element 'arr[1]' appears 4 times
third element 'arr[2]' appears 3 times
Every element arr[i] appears in two types of subsets:
i) In sybarrays beginning with arr[i]. There are
(n-i) such subsets. For example [2] appears
in [2] and [2, 3].
ii) In (n-i)*i subarrays where this element is not
first element. For example [2] appears in
[1, 2] and [1, 2, 3].
Total of above (i) and (ii) = (n-i) + (n-i)*i
= (n-i)(i+1)
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 + 2*4 + 3*3
= 20
public static long SubArraySum( int arr[] , int n )
{
long result = 0;
// computing sum of subarray using formula
for (int i=0; i<n; i++)
result += (arr[i] * (i+1) * (n-i));
// return all subarray sum
return result ;
}