Add solution #778

Author: JoshCrozier

README.md

@@ -588,6 +588,7 @@
 773|[Sliding Puzzle](./0773-sliding-puzzle.js)|Hard|
 775|[Global and Local Inversions](./0775-global-and-local-inversions.js)|Medium|
 777|[Swap Adjacent in LR String](./0777-swap-adjacent-in-lr-string.js)|Medium|
+778|[Swim in Rising Water](./0778-swim-in-rising-water.js)|Hard|
 783|[Minimum Distance Between BST Nodes](./0783-minimum-distance-between-bst-nodes.js)|Easy|
 784|[Letter Case Permutation](./0784-letter-case-permutation.js)|Medium|
 790|[Domino and Tromino Tiling](./0790-domino-and-tromino-tiling.js)|Medium|

solutions/0778-swim-in-rising-water.js

@@ -0,0 +1,71 @@
+/**
+ * 778. Swim in Rising Water
+ * https://leetcode.com/problems/swim-in-rising-water/
+ * Difficulty: Hard
+ *
+ * You are given an n x n integer matrix grid where each value grid[i][j] represents the elevation
+ * at that point (i, j).
+ *
+ * The rain starts to fall. At time t, the depth of the water everywhere is t. You can swim from
+ * a square to another 4-directionally adjacent square if and only if the elevation of both squares
+ * individually are at most t. You can swim infinite distances in zero time. Of course, you must
+ * stay within the boundaries of the grid during your swim.
+ *
+ * Return the least time until you can reach the bottom right square (n - 1, n - 1) if you start
+ * at the top left square (0, 0).
+ */
+
+/**
+ * @param {number[][]} grid
+ * @return {number}
+ */
+var swimInWater = function(grid) {
+  const n = grid.length;
+  let left = grid[0][0];
+  let right = n * n - 1;
+
+  const directions = [[0, 1], [1, 0], [0, -1], [-1, 0]];
+
+  const canReachDestination = (time) => {
+    if (grid[0][0] > time) return false;
+
+    const visited = Array(n).fill().map(() => Array(n).fill(false));
+    const queue = [[0, 0]];
+    visited[0][0] = true;
+
+    while (queue.length > 0) {
+      const [row, col] = queue.shift();
+
+      if (row === n - 1 && col === n - 1) {
+        return true;
+      }
+
+      for (const [dr, dc] of directions) {
+        const newRow = row + dr;
+        const newCol = col + dc;
+
+        if (
+          newRow >= 0 && newRow < n && newCol >= 0 && newCol < n
+          && !visited[newRow][newCol] && grid[newRow][newCol] <= time
+        ) {
+          queue.push([newRow, newCol]);
+          visited[newRow][newCol] = true;
+        }
+      }
+    }
+
+    return false;
+  };
+
+  while (left < right) {
+    const mid = Math.floor((left + right) / 2);
+
+    if (canReachDestination(mid)) {
+      right = mid;
+    } else {
+      left = mid + 1;
+    }
+  }
+
+  return left;
+};