vault backup: 2022-09-04 13:41:08

.obsidian/plugins/obsidian-style-settings/data.json

@@ -8,5 +8,6 @@
   "obsidian-prism-theme@@pt-hide-view-header": false,
   "obsidian-prism-theme@@accent-style-dt": "pt-accent-style-borderandfilled-dt",
   "obsidian-prism-theme@@accent-style-lt": "pt-accent-style-borderandfilled-lt",
-  "obsidian-prism-theme@@background-type-lt": "pt-background-default-lt"
+  "obsidian-prism-theme@@background-type-lt": "pt-background-default-lt",
+  "obsidian-prism-theme@@light-mark-highlight-style": "pt-highlight-style-filled-lt"
 }

OJ notes/pages/Leetcode Non-Overlapping-Intervals.md

@@ -4,7 +4,7 @@
 
 > ##### Algorithms:
 >
-> #algorithm
+> #algorithm #greedy 
 >
 > ##### Data structures:
 >
@@ -12,7 +12,7 @@
 >
 > ##### Difficulty:
 >
-> #coding_problem #difficulty-
+> #coding_problem #difficulty-medium 
 >
 > ##### Additional tags:
 >
@@ -61,7 +61,14 @@ Given an array of intervals `intervals` where `intervals[i] = [starti, endi]`, r
 ### Thoughts
 
 > [!summary]
-> This is a #template_remove_me
+> This is a #greedy problem, similar to [[Leetcode Merge-Intervals]]
+
+#### Key concept:
+
+first sort the intervals.
+
+pick the intervals with smallest end, which will allow us to
+make more room for following ones.
 
 ### Solution