From 9c7bbe71d3b6b5201c51ff83d9a010f78e181086 Mon Sep 17 00:00:00 2001
From: aiueoriku <112373068+aiueoriku@users.noreply.github.com>
Date: Fri, 23 May 2025 19:19:04 +0900
Subject: [PATCH] Create Subarray Sum Equals K.md

Problem:
https://leetcode.com/problems/subarray-sum-equals-k/description/
---
 .../Subarray Sum Equals K.md                  | 40 +++++++++++++++++++
 1 file changed, 40 insertions(+)
 create mode 100644 Subarray Sum Equals K/Subarray Sum Equals K.md

diff --git a/Subarray Sum Equals K/Subarray Sum Equals K.md b/Subarray Sum Equals K/Subarray Sum Equals K.md
new file mode 100644
index 0000000..c232170
--- /dev/null
+++ b/Subarray Sum Equals K/Subarray Sum Equals K.md	
@@ -0,0 +1,40 @@
+# Step1
+何も見ないで解く
+```python
+class Solution:
+    def subarraySum(self, nums: List[int], k: int) -> int:
+        num_subarrays = 0
+        for steps in range(len(nums)):
+            for i in range(len(nums)-steps):
+                print(f"nums[i:i+steps+1]:{nums[i:i+steps+1]}")
+                if sum(nums[i:i+steps+1]) == k:
+                    num_subarrays += 1
+        return num_subarrays
+```
+まずは総当たりでSubarrayを求めることを試みた．テストケースは突破したが，Submitの際，Output Limit Exceededとなってしまう．
+
+# Step2
+他の人の解答を見る．
+https://github.com/rinost081/LeetCode/pull/15/
+
+- 累積和がキーワードぽい．今の総和-累積和をすることで1回のfor文で事足りる．
+- 累積和を保持するためにHashmapを使うのが良さそう
+
+# Step3
+他の人の解答を参考に解き直し
+```python
+class Solution:
+    def subarraySum(self, nums: List[int], k: int) -> int:
+        subarray_count = 0
+        current_sum = 0
+        prefix_sum = defaultdict(int)
+        prefix_sum[0] = 1
+        for num in nums:
+            current_sum += num
+            complement = current_sum - k
+            if complement in prefix_sum:
+                subarray_count += prefix_sum[complement]
+            prefix_sum[current_sum] += 1
+        return subarray_count
+```
+prefix_sum[0]=1を書かないと，例えばnums=[1,2,3], k=1で最初の要素がヒットするときに対応出来ないというポイントがある．(prefix_sum[complement]=0となってしまう)