Given an input array that may contain both positive and negative integers, find the sum of **continuous subarray** of numbers which has the largest sum.

**Example:**

Input:nums=`[-2, -5, 6, -2, -3, 1, 5, -6]`

Output:7Explanation:sum([6, -2, -3, 1, 5]) = 7

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

The core challenge of this problem is to find the maximum sum of a contiguous subarray within a one-dimensional numeric array. This problem is significant in various fields such as finance (for finding maximum profit), computer science (for optimization problems), and more. A common pitfall is to consider non-contiguous subarrays or to use a brute-force approach that is not efficient.

To solve this problem, we can use Kadane's Algorithm, which is an efficient way to find the maximum sum subarray in linear time.

A naive solution would involve checking all possible subarrays and calculating their sums, which would result in a time complexity of O(n^2). This is not optimal for large arrays.

Kadane's Algorithm improves upon the naive solution by maintaining a running sum of the maximum subarray ending at the current position. It uses two variables: `maxCurrent`

and `maxGlobal`

. The algorithm iterates through the array, updating these variables based on the current element and the running sum.

- Initialize
`maxCurrent`

and`maxGlobal`

to the first element of the array. - Iterate through the array starting from the second element.
- For each element, update
`maxCurrent`

to be the maximum of the current element and the sum of`maxCurrent`

and the current element. - If
`maxCurrent`

is greater than`maxGlobal`

, update`maxGlobal`

. - Return
`maxGlobal`

as the result.

```
public class MaximumSumSubarray {
public static int maxSubArray(int[] nums) {
// Initialize maxCurrent and maxGlobal with the first element of the array
int maxCurrent = nums[0];
int maxGlobal = nums[0];
// Iterate through the array starting from the second element
for (int i = 1; i < nums.length; i++) {
// Update maxCurrent to be the maximum of the current element and the sum of maxCurrent and the current element
maxCurrent = Math.max(nums[i], maxCurrent + nums[i]);
// Update maxGlobal if maxCurrent is greater
if (maxCurrent > maxGlobal) {
maxGlobal = maxCurrent;
}
}
// Return the maximum sum of the subarray
return maxGlobal;
}
public static void main(String[] args) {
int[] nums = {-2, -5, 6, -2, -3, 1, 5, -6};
System.out.println("Maximum Sum Subarray: " + maxSubArray(nums)); // Output: 7
}
}
```

The time complexity of Kadane's Algorithm is O(n) because it involves a single pass through the array. The space complexity is O(1) as it uses a constant amount of extra space.

Potential edge cases include:

- All negative numbers: The algorithm should return the maximum single element.
- Single element array: The algorithm should return that element.
- Mixed positive and negative numbers: The algorithm should correctly identify the subarray with the maximum sum.

Examples:

Input: [-1, -2, -3] Output: -1 Input: [1] Output: 1 Input: [1, 2, 3, -2, 5] Output: 9

To test the solution comprehensively, consider a variety of test cases:

- Simple cases with small arrays.
- Arrays with all negative numbers.
- Arrays with a mix of positive and negative numbers.
- Large arrays to test performance.

Using a testing framework like JUnit can help automate and manage these tests effectively.

When approaching such problems:

- Understand the problem requirements and constraints thoroughly.
- Start with a brute-force solution to understand the basic approach.
- Look for patterns and optimizations to improve efficiency.
- Practice similar problems to build intuition and problem-solving skills.

Understanding and implementing Kadane's Algorithm is crucial for solving maximum sum subarray problems efficiently. This problem is a great example of how dynamic programming can optimize solutions. Practice and familiarity with such algorithms will enhance your problem-solving skills.

For further reading and practice: