Given an array of **non-negative** integers and a number `target`

, find a **continous subarray** whose sum is equal to **target**.

Return the start and end indices denoting this subarray.

If there are multiple solutions, you can return any of them.

**Example:**

Input:nums=`[1, 1,`

,5, 2, 1, 3, 10, 2, 1]k= 21Output:[2, 6]Explanation:5 + 2 + 1 + 3 + 10 = 21

1. The input array contains only **non-negative** integers.

2. Your algorithm should run in **O(n)** time and use **O(1)** extra space.

The core challenge of this problem is to find a continuous subarray within a given array of non-negative integers that sums up to a specified target value. This problem is significant in various applications such as financial analysis, where one might need to find a period with a specific sum of transactions.

Potential pitfalls include misunderstanding the requirement for a continuous subarray and not optimizing the solution to meet the O(n) time complexity constraint.

To solve this problem, we can use the sliding window technique, which is efficient for problems involving subarrays or substrings. The idea is to maintain a window that expands and contracts to find the target sum.

Here’s a step-by-step approach:

- Initialize two pointers,
`start`

and`end`

, both set to the beginning of the array. - Maintain a variable
`currentSum`

to store the sum of the current window. - Expand the window by moving the
`end`

pointer and adding the value at`end`

to`currentSum`

. - If
`currentSum`

exceeds the target, contract the window by moving the`start`

pointer and subtracting the value at`start`

from`currentSum`

. - Continue this process until you find a window where
`currentSum`

equals the target.

Here is a detailed breakdown of the algorithm:

- Initialize
`start`

to 0 and`currentSum`

to 0. - Iterate through the array with
`end`

from 0 to the length of the array. - Add the value at
`end`

to`currentSum`

. - While
`currentSum`

is greater than the target, subtract the value at`start`

from`currentSum`

and increment`start`

. - If
`currentSum`

equals the target, return the indices`[start, end]`

.

```
public class SubarraySum {
public static int[] findSubarrayWithGivenSum(int[] nums, int target) {
int start = 0;
int currentSum = 0;
for (int end = 0; end < nums.length; end++) {
// Add the current element to the currentSum
currentSum += nums[end];
// While currentSum is greater than target, subtract the element at start
while (currentSum > target) {
currentSum -= nums[start];
start++;
}
// Check if we have found the target sum
if (currentSum == target) {
return new int[]{start, end};
}
}
// If no subarray is found, return an empty array
return new int[]{};
}
public static void main(String[] args) {
int[] nums = {1, 1, 5, 2, 1, 3, 10, 2, 1};
int target = 21;
int[] result = findSubarrayWithGivenSum(nums, target);
if (result.length == 2) {
System.out.println("Subarray found from index " + result[0] + " to " + result[1]);
} else {
System.out.println("No subarray found with the given sum.");
}
}
}
```

The time complexity of this approach is O(n) because each element is processed at most twice, once by the `end`

pointer and once by the `start`

pointer. The space complexity is O(1) as we are using only a few extra variables.

Consider the following edge cases:

- An empty array: The function should return an empty array.
- An array where no subarray sums to the target: The function should return an empty array.
- An array where the entire array sums to the target: The function should return the indices of the entire array.

To test the solution comprehensively, consider the following test cases:

- Simple cases with small arrays.
- Cases where the subarray is at the beginning, middle, and end of the array.
- Edge cases such as an empty array or an array with no valid subarray.

When approaching such problems, it is crucial to:

- Understand the problem requirements and constraints thoroughly.
- Think about different approaches and their trade-offs.
- Start with a simple solution and then optimize it.
- Practice similar problems to improve problem-solving skills.

In this blog post, we discussed how to find a continuous subarray with a given sum in an array of non-negative integers. We explored the sliding window technique, provided a detailed algorithm, and implemented the solution in Java. Understanding and solving such problems is essential for improving algorithmic thinking and coding skills.

For further reading and practice, consider the following resources: