Add solution #417

Author: JoshCrozier

README.md

@@ -333,6 +333,7 @@
 414|[Third Maximum Number](./0414-third-maximum-number.js)|Easy|
 415|[Add Strings](./0415-add-strings.js)|Easy|
 416|[Partition Equal Subset Sum](./0416-partition-equal-subset-sum.js)|Medium|
+417|[Pacific Atlantic Water Flow](./0417-pacific-atlantic-water-flow.js)|Medium|
 419|[Battleships in a Board](./0419-battleships-in-a-board.js)|Medium|
 434|[Number of Segments in a String](./0434-number-of-segments-in-a-string.js)|Easy|
 435|[Non-overlapping Intervals](./0435-non-overlapping-intervals.js)|Medium|

solutions/0417-pacific-atlantic-water-flow.js

@@ -0,0 +1,63 @@
+/**
+ * 417. Pacific Atlantic Water Flow
+ * https://leetcode.com/problems/pacific-atlantic-water-flow/
+ * Difficulty: Medium
+ *
+ * There is an m x n rectangular island that borders both the Pacific Ocean and Atlantic Ocean.
+ * The Pacific Ocean touches the island's left and top edges, and the Atlantic Ocean touches
+ * the island's right and bottom edges.
+ *
+ * The island is partitioned into a grid of square cells. You are given an m x n integer
+ * matrix heights where heights[r][c] represents the height above sea level of the cell
+ * at coordinate (r, c).
+ *
+ * The island receives a lot of rain, and the rain water can flow to neighboring cells directly
+ * north, south, east, and west if the neighboring cell's height is less than or equal to the
+ * current cell's height. Water can flow from any cell adjacent to an ocean into the ocean.
+ *
+ * Return a 2D list of grid coordinates result where result[i] = [ri, ci] denotes that rain water
+ * can flow from cell (ri, ci) to both the Pacific and Atlantic oceans.
+ */
+
+/**
+ * @param {number[][]} heights
+ * @return {number[][]}
+ */
+var pacificAtlantic = function(heights) {
+  const result = [];
+  const pacific = new Set();
+  const atlantic = new Set();
+
+  for (let i = 0; i < heights.length; i++) {
+    dfs(i, 0, pacific);
+    dfs(i, heights[0].length - 1, atlantic);
+  }
+  for (let j = 0; j < heights[0].length; j++) {
+    dfs(0, j, pacific);
+    dfs(heights.length - 1, j, atlantic);
+  }
+
+  for (let i = 0; i < heights.length; i++) {
+    for (let j = 0; j < heights[0].length; j++) {
+      const key = `${i},${j}`;
+      if (pacific.has(key) && atlantic.has(key)) {
+        result.push([i, j]);
+      }
+    }
+  }
+
+  return result;
+
+  function dfs(r, c, ocean) {
+    const key = `${r},${c}`;
+    if (ocean.has(key)) return;
+    ocean.add(key);
+
+    [[r - 1, c], [r + 1, c], [r, c - 1], [r, c + 1]].forEach(([nr, nc]) => {
+      if (nr >= 0 && nr < heights.length && nc >= 0 && nc < heights[0].length
+        && heights[nr][nc] >= heights[r][c]) {
+        dfs(nr, nc, ocean);
+      }
+    });
+  }
+};