Add solution #799

Author: JoshCrozier

README.md

@@ -609,6 +609,7 @@
 796|[Rotate String](./0796-rotate-string.js)|Easy|
 797|[All Paths From Source to Target](./0797-all-paths-from-source-to-target.js)|Medium|
 798|[Smallest Rotation with Highest Score](./0798-smallest-rotation-with-highest-score.js)|Hard|
+799|[Champagne Tower](./0799-champagne-tower.js)|Medium|
 802|[Find Eventual Safe States](./0802-find-eventual-safe-states.js)|Medium|
 804|[Unique Morse Code Words](./0804-unique-morse-code-words.js)|Easy|
 819|[Most Common Word](./0819-most-common-word.js)|Easy|

solutions/0799-champagne-tower.js

@@ -0,0 +1,44 @@
+/**
+ * 799. Champagne Tower
+ * https://leetcode.com/problems/champagne-tower/
+ * Difficulty: Medium
+ *
+ * We stack glasses in a pyramid, where the first row has 1 glass, the second row has 2 glasses, and
+ * so on until the 100th row. Each glass holds one cup of champagne.
+ *
+ * Then, some champagne is poured into the first glass at the top. When the topmost glass is full,
+ * any excess liquid poured will fall equally to the glass immediately to the left and right of it.
+ * When those glasses become full, any excess champagne will fall equally to the left and right of
+ * those glasses, and so on. (A glass at the bottom row has its excess champagne fall on the
+ * floor.)
+ *
+ * For example, after one cup of champagne is poured, the top most glass is full. After two cups
+ * of champagne are poured, the two glasses on the second row are half full. After three cups of
+ * champagne are poured, those two cups become full - there are 3 full glasses total now. After
+ * four cups of champagne are poured, the third row has the middle glass half full, and the two
+ * outside glasses are a quarter full, as pictured below.
+ */
+
+/**
+ * @param {number} poured
+ * @param {number} queryRow
+ * @param {number} queryGlass
+ * @return {number}
+ */
+var champagneTower = function(poured, queryRow, queryGlass) {
+  const tower = new Array(101).fill().map(() => new Array(101).fill(0));
+  tower[0][0] = poured;
+
+  for (let row = 0; row <= queryRow; row++) {
+    for (let glass = 0; glass <= row; glass++) {
+      const excess = (tower[row][glass] - 1) / 2;
+
+      if (excess > 0) {
+        tower[row + 1][glass] += excess;
+        tower[row + 1][glass + 1] += excess;
+      }
+    }
+  }
+
+  return Math.min(1, tower[queryRow][queryGlass]);
+};