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+# Leetcode Triangle
+
+#### 2022-07-20 22:59
+
+> ##### Algorithms:
+> #algorithm #dynamic_programming 
+> ##### Difficulty:
+> #coding_problem #difficulty-medium
+> ##### Additional tags:
+> #leetcode 
+> ##### Revisions:
+> N/A
+
+##### Related topics:
+```expander
+tag:#dynamic_programming
+```
+
+
+##### Links:
+- [Link to problem](https://leetcode.com/problems/triangle/)
+___
+### Problem
+
+Given a `triangle` array, return _the minimum path sum from top to bottom_.
+
+For each step, you may move to an adjacent number of the row below. More formally, if you are on index `i` on the current row, you may move to either index `i` or index `i + 1` on the next row.
+
+#### Examples
+
+**Example 1:**
+
+```
+**Input:** triangle = [[2],[3,4],[6,5,7],[4,1,8,3]]
+**Output:** 11
+**Explanation:** The triangle looks like:
+   2
+  3 4
+ 6 5 7
+4 1 8 3
+The minimum path sum from top to bottom is 2 + 3 + 5 + 1 = 11 (underlined above).
+```
+
+**Example 2:**
+
+```
+**Input:** triangle = [[-10]]
+**Output:** -10
+```
+
+#### Constraints
+
+-   `1 <= triangle.length <= 200`
+-   `triangle[0].length == 1`
+-   `triangle[i].length == triangle[i - 1].length + 1`
+-   `-104 <= triangle[i][j] <= 104`
+
+### Thoughts
+
+> [!summary]
+> This is a #dynamic_programming problem.
+
+Same as in [[Leetcode House-Robber]], there are four stages to optimization:
+
+#### Stage 1: ordinary recursion
+
+#### Stage 2: recursion with caching
+
+### Solution