diff --git a/solutions/silver/ccc-Firehose.mdx b/solutions/silver/ccc-Firehose.mdx
index 0954c00909..9fee7c6bbc 100644
--- a/solutions/silver/ccc-Firehose.mdx
+++ b/solutions/silver/ccc-Firehose.mdx
@@ -16,8 +16,8 @@ We use a greedy algorithm to find $\texttt{possible}(i)$. Firstly, we want to so
 **Time Complexity**: $\mathcal O(h^2\times log(\texttt{max}(H_i)))$
 
 <LanguageSection>
-
 <CPPSection>
+
 ```cpp
 #include <bits/stdc++.h>
 using namespace std;
@@ -71,7 +71,7 @@ int main() {
 	// Binary search for the lowest hose length
 	int l = 0;
 	int r = STREET_SIZE;
-	int ans = 0;
+	int ans = -1;
 	while (l <= r) {
 		int m = (l + r) / 2;
 		if (possible(m)) {
@@ -82,9 +82,143 @@ int main() {
 		}
 	}
 
-	cout << ans << "\n";
+	cout << ans << endl;
 }
 ```
+
 </CPPSection>
+<JavaSection>
+
+```java
+import java.io.*;
+import java.util.*;
+
+public class Firehose {
+	private static final int MAX_H = 1000;
+	private static final int STREET_SIZE = 1000000;
+
+	private static boolean possible(int length, int numHydrants, int[] houses) {
+		for (int i = 0; i < houses.length; i++) {
+			int needed = 0;  // Number of fire hydrants that are needed.
+			// The house that we need to connect a fire hydrant to.
+			int start = houses[i];
+			for (int j = 1; j < houses.length; j++) {
+				// Address of house at index j.
+				int end = houses[(i + j) % houses.length];
+				/*
+				 * If the distance between the start and end houses is greater than
+				 * two times the length of the hose, a single fire hydrant will not
+				 * be enough to cover both houses.
+				 */
+				int dist = (end - start + STREET_SIZE) % STREET_SIZE;
+				if (dist > 2 * length) {
+					start = end;
+					needed++;
+				}
+			}
+
+			// Increment needed as we need another hydrant to reach the last houses.
+			needed++;
+
+			if (needed <= numHydrants) { return true; }
+		}
+		return false;
+	}
+
+	public static void main(String[] args) {
+		Kattio io = new Kattio();
+
+		int numHouses = io.nextInt();
+		int[] houses = new int[numHouses];
+
+		for (int i = 0; i < numHouses; i++) { houses[i] = io.nextInt(); }
+
+		int numHydrants = io.nextInt();
+
+		Arrays.sort(houses);
+
+		// Binary search for the lowest hose length
+		int left = 0;
+		int right = STREET_SIZE;
+		int ans = -1;
+		while (left <= right) {
+			int mid = (left + right) / 2;
+			if (possible(mid, numHydrants, houses)) {
+				right = mid - 1;
+				ans = mid;
+			} else {
+				left = mid + 1;
+			}
+		}
+
+		System.out.println(ans);
+		io.close();
+	}
+
+	// CodeSnip{Kattio}
+}
+```
+
+</JavaSection>
+<PySection>
+
+```py
+MAX_H = 1000
+STREET_SIZE = 1000000
+
+
+def possible(length: int, num_hydrants: int, houses: list) -> bool:
+	for i in range(len(houses)):
+		# Number of fire hydrants that are needed.
+		needed = 0
+		# The house that we need to connect a fire hydrant to.
+		start = houses[i]
+		for j in range(1, len(houses)):
+			# Address of house at index j.
+			end = houses[(i + j) % len(houses)]
+
+			"""
+			If the distance between the start and end houses is greater than
+			two times the length of the hose, a single fire hydrant will not
+			be enough to cover both houses.
+			"""
+			dist = (end - start + STREET_SIZE) % STREET_SIZE
+			if dist > 2 * length:
+				start = end
+				needed += 1
+
+		# Increment needed as we need another hydrant to reach the last houses.
+		needed += 1
+
+		if needed <= num_hydrants:
+			return True
+
+	return False
+
+
+num_houses = int(input())
+houses = []
+for _ in range(num_houses):
+	houses.append(int(input()))
+
+num_hydrants = int(input())
+
+houses.sort()
+
+# Binary search for the lowest hose length
+l = 0
+r = STREET_SIZE
+ans = -1
+while l <= r:
+	m = (l + r) // 2
+	if possible(m, num_hydrants, houses):
+		r = m - 1
+		ans = m
+	else:
+		l = m + 1
+
+print(ans)
+```
 
+</PySection>
 </LanguageSection>