From fdb29cf0ca8cdb532887965e0824b8a36869f04b Mon Sep 17 00:00:00 2001
From: juan <juancloudcomputing@protonmail.com>
Date: Tue, 26 Jul 2022 08:56:08 +0800
Subject: [PATCH] vault backup: 2022-07-26 08:56:08

---
 .../Leetcode Largest-Perimeter-Triangle.md    | 52 +++++++++++++++++--
 OJ notes/pages/Untitled.md                    |  0
 2 files changed, 48 insertions(+), 4 deletions(-)
 create mode 100644 OJ notes/pages/Untitled.md

diff --git a/OJ notes/pages/Leetcode Largest-Perimeter-Triangle.md b/OJ notes/pages/Leetcode Largest-Perimeter-Triangle.md
index 410e3ee..a57d05f 100644
--- a/OJ notes/pages/Leetcode Largest-Perimeter-Triangle.md	
+++ b/OJ notes/pages/Leetcode Largest-Perimeter-Triangle.md	
@@ -3,7 +3,7 @@
 #### 2022-07-26 08:39
 
 > ##### Algorithms:
-> #algorithm #greedy
+> #algorithm #greedy #math
 > ##### Difficulty:
 > #coding_problem #difficulty-easy
 > ##### Additional tags:
@@ -13,23 +13,67 @@
 
 ##### Related topics:
 ```expander
-tag:#<INSERT_TAG_HERE>
+tag:#math
 ```
  
  
 
 ##### Links:
-- [Link to problem]()
+- [Link to problem](https://leetcode.com/problems/largest-perimeter-triangle/)
 ___
 ### Problem
 
+Given an integer array `nums`, return _the largest perimeter of a triangle with a non-zero area, formed from three of these lengths_. If it is impossible to form any triangle of a non-zero area, return `0`.
+
 #### Examples
 
+**Example 1:**
+
+**Input:** nums = [2,1,2]
+**Output:** 5
+
+**Example 2:**
+
+**Input:** nums = [1,2,1]
+**Output:** 0
+
 #### Constraints
 
+-   `3 <= nums.length <= 104`
+-   `1 <= nums[i] <= 106`
+
 ### Thoughts
 
 > [!summary]
-> This is a #template_remove_me
+> This is a #math problem.
+
+After sorting, the answer must be consecutive 3 elements.
+
+#### Why?
+
+assume l, m , r.
+
+if nums[l] + nums[m] < nums[r], because after sorting,
+the order is ascending, anything before l and m will not suffice.
 
 ### Solution
+
+```cpp
+class Solution {
+public:
+  int largestPerimeter(vector<int> &nums) {
+    sort(nums.begin(), nums.end());
+    int r = nums.size() - 1;
+
+    // change l first until its boundary
+
+    for (int i = r; i > 1; i--) {
+      if (nums[i] < nums[i - 1] + nums[i - 2]) {
+        return nums[i] + nums[i - 1] + nums[i - 2];
+      }
+    }
+
+    return 0;
+  }
+};
+```
\ No newline at end of file
diff --git a/OJ notes/pages/Untitled.md b/OJ notes/pages/Untitled.md
new file mode 100644
index 0000000..e69de29