# Maximum Size Subarray Sum Equals k

Given an arraynumsand a target valuek, find the maximum length of a subarray that sums tok. If there isn't one, return 0 instead.

**Note:**\
The sum of the entirenumsarray is guaranteed to fit within the 32-bit signed integer range.

**Example 1:**

```
Input: 
nums
 = 
[1, -1, 5, -2, 3]
, 
k
 = 
3
Output: 
4 

Explanation: 
The subarray 
[1, -1, 5, -2]
 sums to 3 and is the longest.
```

**Example 2:**

```
Input: 
nums
 = 
[-2, -1, 2, 1]
, 
k
 = 
1
Output: 
2 

Explanation: 
The subarray 
[-1, 2]
 sums to 1 and is the longest.
```

**Follow Up:**\
Can you do it in O(n) time?

分析

不用subsum数组，直接cursum+map，一遍loop,第一次遇到sum加入，后来的就算subsum == k。

注意这里subsum不含头，所以res=j-i就行，不用+1

```
class Solution:
    def maxSubArrayLen(self, nums: List[int], k: int) -> int:
        sm = 0 
        mm = {}
        res = 0
        for i,v in enumerate(nums):
            sm += v
            if sm == k:
                res = i+1
            elif sm - k in mm:
                res = max(res,i - mm[sm - k]) #subsum不包含起始的Index，所以这里不用加1
            if sm not in mm: mm[sm] = i
        return res
```


---

# Agent Instructions: Querying This Documentation

If you need additional information that is not directly available in this page, you can query the documentation dynamically by asking a question.

Perform an HTTP GET request on the current page URL with the `ask` query parameter:

```
GET https://nataliekung.gitbook.io/ladder_code/l7shu-zu/maximum-size-subarray-sum-equals-k.md?ask=<question>
```

The question should be specific, self-contained, and written in natural language.
The response will contain a direct answer to the question and relevant excerpts and sources from the documentation.

Use this mechanism when the answer is not explicitly present in the current page, you need clarification or additional context, or you want to retrieve related documentation sections.