Sum of all Subarrays | Set 1

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 ;
    }

  

posted @ 2017-10-20 21:42  apanda009  阅读(307)  评论(0编辑  收藏  举报