diff --git a/pacific-atlantic-water-flow/jamiebase.py b/pacific-atlantic-water-flow/jamiebase.py
new file mode 100644
index 0000000000..60df92c66e
--- /dev/null
+++ b/pacific-atlantic-water-flow/jamiebase.py
@@ -0,0 +1,56 @@
+"""
+# Approach
+태평양 가장자리에서 출발해서, 물이 거꾸로 올라올 수 있는 칸들을 찾고 대서양도 똑같이 찾은 뒤 둘의 교집합을 구한다.
+
+# Complexity
+- Time complexity: O(rows * cols)
+- Space complexity: O(rows * cols)
+"""
+
+from collections import deque
+
+
+class Solution:
+    def pacificAtlantic(self, heights: list[list[int]]) -> list[list[int]]:
+        rows, cols = len(heights), len(heights[0])
+        directions = [(0, 1), (0, -1), (1, 0), (-1, 0)]
+
+        def bfs(starts):
+            visited = set(starts)
+            q = deque(starts)
+
+            while q:
+                r, c = q.popleft()
+
+                for dr, dc in directions:
+                    nr, nc = r + dr, c + dc
+
+                    if not (0 <= nr < rows and 0 <= nc < cols):
+                        continue
+                    if (nr, nc) in visited:
+                        continue
+
+                    # 역방향: 현재보다 같거나 높은 곳으로 이동 가능
+                    if heights[nr][nc] < heights[r][c]:
+                        continue
+
+                    visited.add((nr, nc))
+                    q.append((nr, nc))
+
+            return visited
+
+        pacific_starts = []
+        atlantic_starts = []
+
+        for r in range(rows):
+            pacific_starts.append((r, 0))
+            atlantic_starts.append((r, cols - 1))
+
+        for c in range(cols):
+            pacific_starts.append((0, c))
+            atlantic_starts.append((rows - 1, c))
+
+        pacific = bfs(pacific_starts)
+        atlantic = bfs(atlantic_starts)
+
+        return [[r, c] for r, c in pacific & atlantic]