diff --git a/cpp/2206/220621.cpp b/cpp/2206/220621.cpp
new file mode 100644
index 0000000..2182f9e
--- /dev/null
+++ b/cpp/2206/220621.cpp
@@ -0,0 +1,41 @@
+#include <vector>
+#include <queue>
+#include <iostream>
+
+/**
+ * 1642. Furthest Building You Can Reach
+ * You are given an integer array heights representing the heights of buildings, some bricks, and some ladders.
+ * You start your journey from building 0 and move to the next building by possibly using bricks or ladders.
+ * While moving from building i to building i+1 (0-indexed),
+ * If the current building's height is greater than or equal to the next building's height, you do not need a ladder or bricks.
+ * If the current building's height is less than the next building's height, you can either use one ladder or (h[i+1] - h[i]) bricks.
+ * Return the furthest building index (0-indexed) you can reach if you use the given ladders and bricks optimally.
+ */
+
+class Solution {
+public:
+	static int furthestBuilding(const std::vector<int>& h, int bricks, int ladders) {
+		const int n = h.size();
+		std::priority_queue<int, std::vector<int>, std::greater<>> q;
+		int sum = 0;
+		for (int i = 1; i < n; ++i) {
+			if (h[i] - h[i - 1] <= 0)
+				continue;
+			q.push(h[i] - h[i - 1]);
+			if (q.size() > ladders) {
+				sum += q.top();
+				q.pop();
+			}
+			if (sum > bricks)
+				return i - 1;
+		}
+		return n - 1;
+	}
+};
+
+int main() {
+	std::cout << Solution::furthestBuilding({4,2,7,6,9,14,12}, 5, 1) << "\n";
+	std::cout << Solution::furthestBuilding({4,12,2,7,3,18,20,3,19}, 10, 2) << "\n";
+	std::cout << Solution::furthestBuilding({14,3,19,3}, 17, 0) << "\n";
+	return 0;
+}
diff --git a/cpp/2206/CMakeLists.txt b/cpp/2206/CMakeLists.txt
index 5906648..9321f92 100644
--- a/cpp/2206/CMakeLists.txt
+++ b/cpp/2206/CMakeLists.txt
@@ -3,4 +3,4 @@ PROJECT(2206)
 
 SET(CMAKE_CXX_STANDARD 23)
 
-ADD_EXECUTABLE(2206 220619.cpp)
+ADD_EXECUTABLE(2206 220621.cpp)