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
Method 1 (Simple Solution)
A simple solution is to generate all sub-array and compute their sum.
Below is the implementation of above idea.
C++
#include<bits/stdc++.h>
using namespace std;
long int SubArraySum(int arr[], int n)
{
long int result = 0,temp=0;
for (int i=0; i <n; i++)
{
temp=0;
for (int j=i; j<n; j++)
{
temp+=arr[j];
result += temp ;
}
}
return result ;
}
int main()
{
int arr[] = {1, 2, 3} ;
int n = sizeof(arr)/sizeof(arr[0]);
cout << "Sum of SubArray : "
<< SubArraySum(arr, n) << endl;
return 0;
}
|
Java
class GFG {
public static long SubArraySum(int arr[], int n)
{
long result = 0,temp=0;
for (int i = 0; i < n; i ++)
{
temp=0;
for (int j = i; j < n; j ++)
{
temp+=arr[j];
result += temp ;
}
}
return result ;
}
public static void main(String[] args)
{
int arr[] = {1, 2, 3} ;
int n = arr.length;
System.out.println("Sum of SubArray : " +
SubArraySum(arr, n));
}
}
|
Python 3
def SubArraySum(arr, n):
temp,result = 0,0
for i in range(0, n):
temp=0;
for j in range(i, n):
temp+=arr[j]
result += temp
return result
arr = [1, 2, 3]
n = len(arr)
print ("Sum of SubArray :"
,SubArraySum(arr, n))
|
C#
using System;
class GFG {
public static long SubArraySum(int []arr,
int n)
{
long result = 0,temp=0;
for (int i = 0; i < n; i ++)
{
temp=0;
for (int j = i; j < n; j ++)
{
temp+=arr[j];
result += temp ;
}
}
return result ;
}
public static void Main()
{
int []arr = {1, 2, 3} ;
int n = arr.Length;
Console.Write("Sum of SubArray : " +
SubArraySum(arr, n));
}
}
|
PHP
<?php
function SubArraySum($arr, $n)
{
$result = 0;
$temp=0;
for ($i = 0; $i < $n; $i++)
{
$temp=0;
for ($j = $i; $j < $n; $j++)
{
$temp+=$arr[$j]
$result += $temp ;
}
}
return $result ;
}
$arr = array(1, 2, 3) ;
$n = sizeof($arr);
echo "Sum of SubArray : ",
SubArraySum($arr, $n),"\n";
?>
|
Javascript
<script>
function SubArraySum(arr, n)
{
let result = 0,temp=0;
for (let i=0; i <n; i++)
{
temp=0;
for (let j=i; j<n; j++)
{
temp+=arr[j];
result += temp ;
}
}
return result ;
}
let arr = [1, 2, 3] ;
let n = arr.length;
document.write("Sum of SubArray : "
+ SubArraySum(arr, n) + "<br>");
</script>
|
Output:
Sum of SubArray : 20
Time Complexity : O(n2)
Method 2 (efficient solution)
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 subarrays 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
= 20Below is the implementation of above idea.
C++
#include<bits/stdc++.h>
using namespace std;
long int SubArraySum( int arr[] , int n )
{
long int result = 0;
for (int i=0; i<n; i++)
result += (arr[i] * (i+1) * (n-i));
return result ;
}
int main()
{
int arr[] = {1, 2, 3} ;
int n = sizeof(arr)/sizeof(arr[0]);
cout << "Sum of SubArray : "
<< SubArraySum(arr, n) << endl;
return 0;
}
|
Java
class GFG {
public static long SubArraySum( int arr[] , int n )
{
long result = 0;
for (int i=0; i<n; i++)
result += (arr[i] * (i+1) * (n-i));
return result ;
}
public static void main(String[] args)
{
int arr[] = {1, 2, 3} ;
int n = arr.length;
System.out.println("Sum of SubArray " +
SubArraySum(arr, n));
}
}
|
Python 3
def SubArraySum(arr, n ):
result = 0
for i in range(0, n):
result += (arr[i] * (i+1) * (n-i))
return result
arr = [1, 2, 3]
n = len(arr)
print ("Sum of SubArray : ",
SubArraySum(arr, n))
|
C#
using System;
class GFG
{
public static long SubArraySum(int []arr ,
int n )
{
long result = 0;
for (int i = 0; i < n; i++)
result += (arr[i] *
(i + 1) * (n - i));
return result ;
}
static public void Main ()
{
int []arr = {1, 2, 3} ;
int n = arr.Length;
Console.WriteLine("Sum of SubArray: " +
SubArraySum(arr, n));
}
}
|
PHP
<?php
function SubArraySum($arr , $n)
{
$result = 0;
for ($i = 0; $i < $n; $i++)
$result += ($arr[$i] *
($i + 1) *
($n - $i));
return $result ;
}
$arr = array(1, 2, 3) ;
$n = sizeof($arr);
echo "Sum of SubArray : ",
SubArraySum($arr, $n) ,"\n";
#This code is contributed by aj_36
?>
|
Output :
Sum of SubArray : 20
Time complexity : O(n)
Reference :
https://www.quora.com/Can-we-find-the-sum-of-all-sub-arrays-of-an-integer-array-in-O-n-time
Sum of all subsequences of an array
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