Merge pull request #4881 from freakin23/master

minor nitpicks

content/5_Plat/Range_Sweep.mdx

@@ -32,6 +32,10 @@ upper $x$ coordinate of the segment.
 
 Our answer would be the sum of all the queries.
 
+## Implementation
+
+**Time Complexity:** $\mathcal{O}(N\log N)$
+
 <LanguageSection>
 <CPPSection>
 

content/5_Plat/Sweep_Line.mdx

@@ -38,7 +38,7 @@ number line?) never explicitly mentioned in module
 
 ### Solution 1
 
-**Time Complexity:** $\mathcal{O}(n\log n)$
+**Time Complexity:** $\mathcal{O}(N\log N)$
 
 We will use a divide and conquer algorithm. First, sort the points by x-coordinate. Now, let $P$ be the subarray of points in the current step. Then, partition $P$ into two groups $L$ and $R$ representing the left and right halves of $P$. Let $\delta_l$ and $\delta_r$ be the answer of $L$ and $R$ respectively, and define $\delta$ as $\min(\delta_l, \delta_r)$.
 
@@ -143,7 +143,7 @@ int main() {
 
 ### Solution 2
 
-**Time Complexity:** $\mathcal{O}(n\log n)$
+**Time Complexity:** $\mathcal{O}(N\log N)$
 
 Extending Solution 1, we can use a set instead of divide and conquer. Once again, we define $\delta$ as the shortest distance between two points so far. After sorting the points by x-coordinate, we iterate through them while maintaining a running window containing the y-coordinates of all points in $[x-\delta,x]$.
 

content/5_Plat/Sweep_Line.problems.json

@@ -165,18 +165,6 @@
         "url": "https://hsin.hr/ceoi2006/tasks/solutions.pdf"
       }
     },
-    {
-      "uniqueId": "cses-1740",
-      "name": "Intersection Points",
-      "url": "https://cses.fi/problemset/task/1740",
-      "source": "CSES",
-      "difficulty": "Easy",
-      "isStarred": false,
-      "tags": ["1DRQ", "Sweep Line"],
-      "solutionMetadata": {
-        "kind": "none"
-      }
-    },
     {
       "uniqueId": "poi-18-stone",
       "name": "2018 - Stone",