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Okay, let's dive into finding the maximum length subarray with a given sum. This is a classic problem with several approaches. I'll present a detailed tutorial covering multiple methods, focusing on the most efficient one (using a hash map) along with explanations, code examples (in Python), and considerations for different scenarios.

*Understanding the Problem*

The problem is defined as follows:

Given an array (or list) of integers (`nums`) and a target sum (`k`), find the maximum length subarray within `nums` that sums to `k`. If no such subarray exists, return 0.

*Example:*

`nums = [1, -1, 5, -2, 3]`
`k = 3`

The maximum length subarray that sums to 3 is `[5, -2]`, which has a length of 2.

*Approaches*

We'll cover these approaches:

1. *Brute Force:* (Simple but inefficient)
2. *Using Prefix Sum and Two Loops:* (Slightly better than brute force, but still not optimal)
3. *Prefix Sum and Hash Map (Optimal):* (Most efficient and widely used)

*1. Brute Force Approach*

The brute force approach is the most straightforward: Check every possible subarray and see if its sum equals `k`. Keep track of the maximum length found so far.

*Algorithm:*

1. Iterate through all possible starting indices `i` from 0 to `len(nums) - 1`.
2. For each starting index `i`, iterate through all possible ending indices `j` from `i` to `len(nums) - 1`.
3. Calculate the sum of the subarray `nums[i:j+1]`.
4. If the sum equals `k`, update the maximum length if the current subarray's length is greater.

*Python Code:*



*Time Complexity:* O(n^3) (Nested loops and `sum()` within the loops)
*Space Complexity:* O(1) (Constant extra space)

*Why it's bad:* The triple nested loops make it incredibly slow, especially for larger arrays. The `sum()` function is also a hidden loop within the inner loops, adding to the complexity.

*2. Using Prefix Sum and Two Loops*

We can improve upon the brute force approach by pre ...

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