Hello Everyone,
Given an unsorted array of nonnegative integers, find a continuous subarray which adds to a given number.
Examples :
Input: arr[] = {1, 4, 20, 3, 10, 5}, sum = 33 Ouptut: Sum found between indexes 2 and 4 Sum of elements between indices 2 and 4 is 20 + 3 + 10 = 33 Input: arr[] = {1, 4, 0, 0, 3, 10, 5}, sum = 7 Ouptut: Sum found between indexes 1 and 4 Sum of elements between indices 1 and 4 is 4 + 0 + 0 + 3 = 7 Input: arr[] = {1, 4}, sum = 0 Output: No subarray found There is no subarray with 0 sum
There may be more than one subarrays with sum as the given sum. The following solutions print first such subarray.
Simple Approach: A simple solution is to consider all subarrays one by one and check the sum of every subarray. Following program implements the simple solution. Run two loops: the outer loop picks a starting point I and the inner loop tries all subarrays starting from i.
Algorithm:
 Traverse the array from start to end.
 From every index start another loop from i to the end of array to get all subarray starting from i, keep a variable sum to calculate the sum.
 For every index in inner loop update sum = sum + array[j]
 If the sum is equal to the given sum then print the subarray.
/* A simple program to print subarray
with sum as given sum */
#include <bits/stdc++.h>
using
namespace
std;
/* Returns true if the there is a subarray
of arr[] with sum equal to 'sum' otherwise
returns false. Also, prints the result */
int
subArraySum(
int
arr[],
int
n,
int
sum)
{
int
curr_sum, i, j;
// Pick a starting point
for
(i = 0; i < n; i++) {
curr_sum = arr[i];
// try all subarrays starting with 'i'
for
(j = i + 1; j <= n; j++) {
if
(curr_sum == sum) {
cout <<
"Sum found between indexes "
<< i <<
" and "
<< j  1;
return
1;
}
if
(curr_sum > sum  j == n)
break
;
curr_sum = curr_sum + arr[j];
}
}
cout <<
"No subarray found"
;
return
0;
}
// Driver Code
int
main()
{
int
arr[] = { 15, 2, 4, 8, 9, 5, 10, 23 };
int
n =
sizeof
(arr) /
sizeof
(arr[0]);
int
sum = 23;
subArraySum(arr, n, sum);
return
0;
}
Output :
Sum found between indexes 1 and 4
Complexity Analysis:

Time Complexity: O(n^2) in worst case.
Nested loop is used to traverse the array so the time complexity is O(n^2) 
Space Complexity: O(1).
As constant extra space is required.
Efficient Approach: There is an idea if all the elements of the array are positive. If a subarray has sum greater than the given sum then there is no possibility that adding elements to the current subarray the sum will be x (given sum). Idea is to use a similar approach to a sliding window. Start with an empty subarray, add elements to the subarray until the sum is less than x. If the sum is greater than x, remove elements from the start of the current subarray.
Algorithm:
 Create three variables, l=0, sum = 0
 Traverse the array from start to end.
 Update the variable sum by adding current element, sum = sum + array[i]
 If the sum is greater than the given sum, update the variable sum as sum = sum – array[l], and update l as, l++.
 If the sum is equal to given sum, print the subarray and break the loop.
/* An efficient program to print
subarray with sum as given sum */
#include <iostream>
using
namespace
std;
/* Returns true if the there is a subarray of
arr[] with a sum equal to 'sum' otherwise
returns false. Also, prints the result */
int
subArraySum(
int
arr[],
int
n,
int
sum)
{
/* Initialize curr_sum as value of
first element and starting point as 0 */
int
curr_sum = arr[0], start = 0, i;
/* Add elements one by one to curr_sum and
if the curr_sum exceeds the sum,
then remove starting element */
for
(i = 1; i <= n; i++) {
// If curr_sum exceeds the sum,
// then remove the starting elements
while
(curr_sum > sum && start < i  1) {
curr_sum = curr_sum  arr[start];
start++;
}
// If curr_sum becomes equal to sum,
// then return true
if
(curr_sum == sum) {
cout <<
"Sum found between indexes "
<< start <<
" and "
<< i  1;
return
1;
}
// Add this element to curr_sum
if
(i < n)
curr_sum = curr_sum + arr[i];
}
// If we reach here, then no subarray
cout <<
"No subarray found"
;
return
0;
}
// Driver Code
int
main()
{
int
arr[] = { 15, 2, 4, 8, 9, 5, 10, 23 };
int
n =
sizeof
(arr) /
sizeof
(arr[0]);
int
sum = 23;
subArraySum(arr, n, sum);
return
0;
}
Output :
Sum found between indexes 1 and 4
Complexity Analysis:

Time Complexity : O(n).
Only one traversal of the array is required. So the time complexity is O(n). 
Space Complexity: O(1).
As constant extra space is required.