Add solution #2017

Author: JoshCrozier

README.md

@@ -417,6 +417,7 @@
 2000|[Reverse Prefix of Word](./2000-reverse-prefix-of-word.js)|Easy|
 2011|[Final Value of Variable After Performing Operations](./2011-final-value-of-variable-after-performing-operations.js)|Easy|
 2016|[Maximum Difference Between Increasing Elements](./2016-maximum-difference-between-increasing-elements.js)|Easy|
+2017|[Grid Game](./2017-grid-game.js)|Medium|
 2027|[Minimum Moves to Convert String](./2027-minimum-moves-to-convert-string.js)|Easy|
 2037|[Minimum Number of Moves to Seat Everyone](./2037-minimum-number-of-moves-to-seat-everyone.js)|Easy|
 2047|[Number of Valid Words in a Sentence](./2047-number-of-valid-words-in-a-sentence.js)|Easy|

solutions/2017-grid-game.js

@@ -0,0 +1,38 @@
+/**
+ * 2017. Grid Game
+ * https://leetcode.com/problems/grid-game/
+ * Difficulty: Medium
+ *
+ * You are given a 0-indexed 2D array grid of size 2 x n, where grid[r][c] represents the number
+ * of points at position (r, c) on the matrix. Two robots are playing a game on this matrix.
+ *
+ * Both robots initially start at (0, 0) and want to reach (1, n-1). Each robot may only move
+ * to the right ((r, c) to (r, c + 1)) or down ((r, c) to (r + 1, c)).
+ *
+ * At the start of the game, the first robot moves from (0, 0) to (1, n-1), collecting all the
+ * points from the cells on its path. For all cells (r, c) traversed on the path, grid[r][c] is
+ * set to 0. Then, the second robot moves from (0, 0) to (1, n-1), collecting the points on its
+ * path. Note that their paths may intersect with one another.
+ *
+ * The first robot wants to minimize the number of points collected by the second robot. In
+ * contrast, the second robot wants to maximize the number of points it collects. If both
+ * robots play optimally, return the number of points collected by the second robot.
+ */
+
+/**
+ * @param {number[][]} grid
+ * @return {number}
+ */
+var gridGame = function(grid) {
+  let result = Infinity;
+  let sum1 = grid[0].reduce((a, b) => a + b, 0);
+  let sum2 = 0;
+
+  for (let i = 0; i < grid[0].length; i++) {
+    sum1 -= grid[0][i];
+    result = Math.min(result, Math.max(sum1, sum2));
+    sum2 += grid[1][i];
+  }
+
+  return result;
+};