Kadane's Algorithm | Maximum Subarray Sum

We will look at kadane's algorithm and its problem-solving properties in order to solve the "Maximum Subarray Sum" problem in this post. We'll go over the algorithm and python code for it, as well as an example and its related output. Finally, we will look at the algorithm's temporal complexity as well as its real-world application. So, let's get this party started!

What is Kadane’s Algorithm?

In every given array of numbers, Kadane's procedure is used to find the "maximum subarray sum." Some of you may believe it will be the sum of all array elements. There will be negative elements in the array, which will reduce the array's total sum.

Within a one-dimensional numeric array, a method is used to locate the contiguous subarray with the largest sum. Most CS students nowadays begin learning difficult concepts without first mastering the fundamentals. To put it another way, you can start learning segment trees without knowing Kadane's algorithms. It's a dynamic programming algorithm that iterates.

It is fairly typical in iterative DP algorithms to reduce one dimension of the DP matrix along the principal axis of the algorithm's advancement in order to improve performance. The traditional ‘longest common subsequence' technique, for example, is normally stated with a 2D matrix, but if the procedure proceeds from left to right, only two columns are required.

The full DP array vanishes with Kadane's approach, which is a similar optimization applied to a 1D problem. For whatever reason, the DP code in your inquiry includes a 2D matrix. I'm not sure why – it doesn't make sense.

Explanation:

The Kadane algorithm's basic idea is to search the array for all positive contiguous segments (max ending here is utilized for this). Also, among all positive segments, maintain track of the maximum total contiguous segment (max so far is utilized for this). We compare each positive-sum to the max so far and update max so far if it is more than max so far.

Kadane's algorithm works by keeping track of a subarray's start point and deciding whether or not to look at the next element in the array.

• Append that element to the subarray to extend it, or

• After that element, discard the subarray and start a new one.

It shouldn't be too difficult to identify the best subarray if you maintain track of the current subarray's start point (which is initiated before the first element of the array and resets every time the total dips below zero). You either update the maximum subarray discovered every time you update the maximum subarray found.

Have just appended a new element to the existing maximum subarray (so just append to the best item you've found so far), or Have discovered a new subarray that starts at a different location than the previous best subarray, so replace the old max with this new one.

You can do this by keeping track of not just the maximum subarray thus far, but also it's beginning location. If the max subarray starts at the same location as your current array, you're in case (1); otherwise, you're in case (2).

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