fixed some things, time for a revamp

.gitignore

@@ -1,4 +1,5 @@
-.vscode/settings.json
-.obsidian/*
-.obsidian/
-
+.vscode/settings.json
+.obsidian/*
+.obsidian/
+template/
+

docs/LeetCode 75/Day 12 Container With Most Water.md

@@ -0,0 +1,43 @@
+# 11. Container With Most Water
+
+Tags: #array #twopointers #greedy #medium 
+
+### Problem Statement
+You are given an integer array `height` of length `n`. There are `n` vertical lines drawn such that the two endpoints of the `ith` line are `(i, 0)` and `(i, height[i])`.
+
+Find two lines that together with the x-axis form a container, such that the container contains the most water.
+
+Return _the maximum amount of water a container can store_.
+
+**Notice** that you may not slant the container.
+
+![LeetCodeExample](https://s3-lc-upload.s3.amazonaws.com/uploads/2018/07/17/question_11.jpg)
+
+
+```Example
+Input: height = [1,8,6,2,5,4,8,3,7]
+Output: 49
+Explanation: The above vertical lines are represented by array [1,8,6,2,5,4,8,3,7]. In this case, the max area of water (blue section) the container can contain is 49.
+```
+
+### My Approach
+```Python
+class Solution:
+    def maxArea(self, height: List[int]) -> int:
+        i = 0
+        j = len(height)-1
+        temp = 0
+        primary = 0
+        while(i<j):
+            temp = min(height[i], height[j])*(j-i)
+            if temp > primary:
+                primary = temp
+            if height[j]>height[i]:
+                i+=1
+            else:
+                j-=1
+        return primary
+```
+
+
+### Personal Thoughts

docs/LeetCode 75/Day 13 Max Number of K-Sum Pairs.md

@@ -0,0 +1,43 @@
+# 1679. Max Number of K-Sum Pairs
+
+Tags: #array #twopointers #sorting #hashtable #medium 
+
+### Problem Statement
+You are given an integer array `nums` and an integer `k`.
+
+In one operation, you can pick two numbers from the array whose sum equals `k` and remove them from the array.
+
+Return _the maximum number of operations you can perform on the array_.
+
+```Example
+Input: nums = [1,2,3,4], k = 5
+Output: 2
+Explanation: Starting with nums = [1,2,3,4]:
+- Remove numbers 1 and 4, then nums = [2,3]
+- Remove numbers 2 and 3, then nums = []
+There are no more pairs that sum up to 5, hence a total of 2 operations.
+```
+
+### My Approach
+```Python
+class Solution:
+    def maxOperations(self, nums: List[int], k: int) -> int:
+        i = 0
+        j = len(nums)-1
+        count = 0
+        nums.sort()
+
+        while(i<j):
+            sum =nums[i]+nums[j] 
+            if  sum == k:
+                count +=1
+                i+=1
+                j-=1
+            elif sum < k:
+                i+=1
+            else:
+                j-=1
+        return count
+```
+
+### Personal Thoughts

docs/index.md

@@ -21,3 +21,15 @@ These are my notes for the ["Ace Coding Interview with 75 Qs"](https://leetcode.
 - [[Day 6 Reverse Words in a String]]
 
 - [[Day 7 Product of Array Except Self]]
+
+- [[Day 8 Increasing Triplet Subsequence]]
+
+- [[Day 9 String Compression]]
+
+- [[Day 10 Move Zeroes]]
+
+- [[Day 11 Is Subsequence]]
+
+- [[Day 12 Container With Most Water]]
+
+- [[Day 13 Max Number of K-Sum Pairs]]