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OJ notes/pages/Leetcode Subarray-Sum-Equals-K.md

@@ -16,7 +16,7 @@
 >
 > ##### Additional tags:
 >
-> #leetcode 
+> #leetcode #CS_list_need_practicing 
 >
 > ##### Revisions:
 >
@@ -25,6 +25,7 @@
 ##### Links:
 
 - [Link to problem](https://leetcode.com/problems/subarray-sum-equals-k/)
+- [Explanation](https://leetcode.com/problems/subarray-sum-equals-k/discuss/102111/Python-Simple-with-Explanation/694092)
 
 ---
 
@@ -64,10 +65,50 @@ A subarray is a contiguous **non-empty** sequence of elements within an array.
 > Tried using sliding window, but it doesn't work because of 
 > negative numbers.
 
+This one is harder to understand, making it a medium 
+difficulty problem.
 
+The intuition is using prefix to count the subarray sum, and
+use hashmaps to count the **number of subarray sums**. This 
+takes care of **negative numbers** which is not solvable 
+using sliding windows.
+
+We step one cell at a time, and try to ask ourselves:
+`Can we hop back K to land on a cell?`
+
+And we answer this question by using hash maps and recording
+the data on the fly.
 
 ### Solution
 
+```cpp
+class Solution {
+public:
+  int subarraySum(vector<int> &nums, int k) {
+    unordered_map<int, int> umap;
+    // sum == k is a solution, should be counted
+    umap[0] = 1;
+
+    int sum = 0;
+    int count = 0;
+
+    for (int i : nums) {
+      // update the prefix sum
+      sum += i;
+
+      if (umap.find(sum - k) != umap.end()) {
+        // we can land on a cell, then add the counter.
+        count += umap[sum - k];
+      }
+
+      // record this step.
+      umap[sum]++;
+    }
+
+    return count;
+  }
+};
+```
 
 TLE, brute force
 ```cpp