remove sol 3 yeah

solutions/silver/cf-702C.mdx

@@ -7,7 +7,7 @@ author: Nathan Wang, Benjamin Qi, Maggie Liu, Brad Ma, George Pong
 
 [Official Editorial](https://codeforces.com/blog/entry/46324?locale=en)
 
-## Solution 1 - Binary Search on a Sorted Array
+## Solution 1 - Binary Search
 
 **Time Complexity:** $\mathcal{O}(N \log N)$
 
@@ -142,7 +142,7 @@ public class CellularNetwork {
 ```py
 def first_at_least(value: int) -> int:
 	"""
-	returns the first index in the array that is >= value, or len(arr)
+	:return the first index in the array that is >= value, or len(arr)
 	if no such index exists
 	"""
 	lo = 0
@@ -339,111 +339,3 @@ public class CellularNetwork2 {
 
 </JavaSection>
 </LanguageSection>
-
-## Solution 3 - Using a Set
-
-We use the set $\texttt{towers}$ to store the coordinates of the cellular
-towers. For each city, we want to find the closest tower and calculate the
-distance. The minimal $r$ will be the maximum of all these distances. To find
-the closest tower to a certain city, we check the towers to the left and right
-and determine which one is closer.
-
-For each city, we use
-[`lower_bound`](http://www.cplusplus.com/reference/algorithm/lower_bound/) to
-find the closest tower to the right of the city. If this tower is found,
-calculate the distance between this tower and the city. If the tower is not at
-the beginning of the set, the previous tower in the set will be the tower to the
-left of the city. Define $\texttt{dist}$ to be the minimum of the distance to
-the tower on the right and the distance to the tower on the left. Then, set $r$
-to be the maximum of itself and $\texttt{dist}$.
-
-**Time Complexity:** $\mathcal{O}(N \log M)$
-
-<LanguageSection>
-<CPPSection>
-
-```cpp
-#include <algorithm>
-#include <iostream>
-#include <set>
-using namespace std;
-
-int main() {
-	int n, m;
-	cin >> n >> m;
-	int cities[n];
-	set<int> towers;
-	int tower;
-	for (int i = 0; i < n; i++) { cin >> cities[i]; }
-	for (int i = 0; i < m; i++) {
-		cin >> tower;
-		towers.insert(tower);
-	}
-	int r = 0;
-	for (int i = 0; i < n; i++) {
-		int dist = 2e9 + 1;
-		// find closest tower to the right of the city
-		auto closest_tower = towers.lower_bound(cities[i]);
-		if (closest_tower != towers.end()) {
-			// if a tower is found, update the distance
-			dist = *closest_tower - cities[i];
-		}
-		// find closest tower to the left of the city
-		if (closest_tower != towers.begin()) {
-			closest_tower--;
-			// update dist with the minimum of the distances
-			dist = min(dist, cities[i] - *closest_tower);
-		}
-		r = max(r, dist);
-	}
-	cout << r << endl;
-	return 0;
-}
-```
-
-</CPPSection>
-<JavaSection>
-
-```java
-import java.io.*;
-import java.util.*;
-
-public class CellularNetwork {
-	// CodeSnip{Kattio}
-	public static void main(String[] args) {
-		Kattio io = new Kattio();
-
-		int n = io.nextInt();
-		int m = io.nextInt();
-		int[] cities = new int[n];
-		TreeSet<Integer> towers = new TreeSet<>();
-		for (int i = 0; i < n; i++) { cities[i] = io.nextInt(); }
-
-		for (int i = 0; i < m; i++) {
-			int tower = io.nextInt();
-			towers.add(tower);
-		}
-
-		int r = 0;
-		for (int i = 0; i < n; i++) {
-			int distance = Integer.MAX_VALUE;
-			/*
-			 * .lower() and .higher() methods return a value of type
-			 * Integer if the value exists and null if it doesn't
-			 */
-			Integer closestTower = towers.lower(cities[i]);
-			Integer farthestTower = towers.higher(cities[i]);
-			if (closestTower != null) { distance = cities[i] - closestTower; }
-			if (farthestTower != null) {
-				distance = Math.min(distance, farthestTower - cities[i]);
-			}
-			r = Math.max(r, distance);
-		}
-		io.println(r);
-		io.close();
-	}
-}
-```
-
-</JavaSection>
-</LanguageSection>