**Problem Statement**:

Given an array of integers, find the contiguous subarray (containing at least one number) which has the largest sum and return its sum.

**Example**:

```
Input: [-2, 1, -3, 4, -1, 2, 1, -5, 4]
Output: 6
```

**Explanation**: The contiguous subarray `[4, -1, 2, 1]`

has the largest `sum = 6`

.

**Understanding the problem**

This problem is all about finding the maximum sum of a continuous sequence of numbers within an array. All we need to do is look at each contiguous subarray and find the one with the largest sum. A contiguous subarray is just a subset of an array and must include at least one number. We just need to make sure that we return the maximum sum.

For example, given the array `[-2, 1, -3, 4, -1, 2, 1, -5, 4]`

, the maximum sum is 6 because the subarray [`4, -1, 2, 1]`

has the largest sum.

**Solving this problem**

Well, the first thing you should know is the concept of a subarray. A subarray is just a subset of an array and it must include at least one number. Then, you should also understand how to find the sum of a subarray. This can just be done by adding up all the numbers in the subarray. You should know how to compare different subarrays and find the one with the largest sum.

Once you understand the basics, solving this problem should be easier. You just have to make sure that you go through each subarray and compare the sums. Then, you can return the maximum sum.

**Subarrays in Arrays**

– Sum subarrays by adding their numbers.

– Maximum Subarray Sum finds max sum of subarrays.

– Subarrays must have consecutive elements.

A subarray is just a subset of an array. It must include at least one number and the elements must be consecutive within the same array. The subarrays can be of any size, as long as the elements are all next to each other. We will be using subarrays in the Maximum Subarray Sum problem to find the subarray with the largest sum.

**Example**: If we had the array `[2, 4, 6, 8]`

, then the subarrays could be `[2] [4], [6], [8], [2, 4], [4, 6], [6, 8], [2,4,6], [4, 6, 8]`

, and so on.

**Calculating Subarray Sums**

– Add up the numbers in a subarray to calculate its sum.

– Maximum Subarray Sum problem finds max sum of all subarrays.

– Sums need to be compared individually.

Finding the sum of a subarray is pretty straightforward. All you need to do is add up the numbers in the subarray. We will be using this principle to find the sum of each subarray and compare it to the other subarrays to find the one with the maximum sum.

**Example**: If we had the subarray `[3, 4, 7]`

, the sum would be `3 + 4 + 7 = 14`

.

**Calculating the Subarray with the Largest Sum**

– Calculate the sum of each subarray

– Compare the sums to all the other subarrays

– Return the subarray with the largest sum

Comparing different subarrays to find the one with the largest sum is simple. You need to calculate the sum of each subarray and compare it to all the other subarrays. The subarray with the largest sum will be the one you want to return.

**Example**: Given the array `[-2, 1, -3, 4, -1, 2, 1, -5, 4]`

, the maximum sum is 6 because the subarray `[4, -1, 2, 1]`

has the largest sum.

## Finding the Maximum Sum of a Subarray

We can use a dynamic programming approach to solve challenge. We can start by creating a list to store the maximum sum of all subarrays from the beginning of the array to the current position. We also need a variable to keep track of the maximum sum of all subarrays seen so far.

We then loop through the array, and for each element, we calculate the maximum sum of the subarray starting at the beginning, and add the current element to it. We then compare this maximum sum to the maximum sum of all subarrays seen so far and update the maximum sum if necessary. Once we have gone through the entire array, we will have the maximum sum of all subarrays, which we can then return.

Here is a solution in Python that uses dynamic programming to solve the maximum subarray sum problem:

```
def maxSubarraySum(arr):
# create a list to store the maximum sum of all subarrays from the
# beginning of the array to the current position
subarrays_sum = [0]
# variable to keep track of the maximum sum of all subarrays seen so far
max_sum = 0
# loop through the array and calculate the maximum sum of the subarray
# starting at the beginning, and add the
# current element to it
for num in arr:
subarrays_sum.append(max(subarrays_sum[-1] + num, num))
max_sum = max(max_sum, subarrays_sum[-1])
# return the maximum sum of all subarrays
return max_sum
# For example, given the array [-2, 1, -3, 4, -1, 2, 1, -5, 4], the maximum
# sum is 6 because the subarray [4, -1, 2,
# 1] has the largest sum.
print(maxSubarraySum([-2, 1, -3, 4, -1, 2, 1, -5, 4])) # 6
```

This code uses dynamic programming to solve the Maximum Subarray Sum problem. We start by creating a list to store the maximum sum of all subarrays from the beginning of the array to the current index. Then, we have a variable to keep track of the maximum sum of all subarrays seen so far.

We then loop through the array and for each element, we calculate the maximum sum of the subarray starting at the beginning and add the current element to it. We then compare this maximum sum to the maximum sum of all subarrays seen so far and update the maximum sum if necessary.

Once we have gone through the entire array, we will have the maximum sum of all subarrays, which we can then return.

**Time Complexity of Algorithm**

– O(n), where n is the length of the array

– Loop array once, calculate max sum with constant time

– Efficient and able to handle large inputs

The time complexity of this code is O(n), where n is the length of the array. This is because we only need to loop through the array once and for each element, the maximum sum of the subarray is calculated with a constant time operation. This makes the time complexity of the algorithm totally linear, making it really efficient and able to handle large input arrays without any problems.

**Space Complexity of Code**

– Code runs on O(n), same as linear time

– List stores max sum of subarrays from the beginning of the array

– List size same as array length, consistent space needs

In terms of space complexity, it runs on O(n), the same as linear time. We’re using a list to store the maximum sum of all subarrays from the beginning of the array, and the size of the list is the same as the array’s length. This means the amount of space needed remains consistent even for large input arrays.

**Final Thoughts**

– Efficient solution with time and space complexity of O(n)

– Think through a coding challenge

– Talk through the problem to ensure understanding

This is a great example of how powerful dynamic programming can be! It allows us to find the most efficient solution for a problem, as well as giving us the chance to think about it in-depth. Plus, having a solid understanding of the problem is key, so it’s always important to make sure you understand it before starting to code. Awesome work!