Add solution #1659

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,456 LeetCode solutions in JavaScript
+# 1,457 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1279,6 +1279,7 @@
 1656|[Design an Ordered Stream](./solutions/1656-design-an-ordered-stream.js)|Easy|
 1657|[Determine if Two Strings Are Close](./solutions/1657-determine-if-two-strings-are-close.js)|Medium|
 1658|[Minimum Operations to Reduce X to Zero](./solutions/1658-minimum-operations-to-reduce-x-to-zero.js)|Medium|
+1659|[Maximize Grid Happiness](./solutions/1659-maximize-grid-happiness.js)|Hard|
 1668|[Maximum Repeating Substring](./solutions/1668-maximum-repeating-substring.js)|Easy|
 1669|[Merge In Between Linked Lists](./solutions/1669-merge-in-between-linked-lists.js)|Medium|
 1672|[Richest Customer Wealth](./solutions/1672-richest-customer-wealth.js)|Easy|

solutions/1659-maximize-grid-happiness.js

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+/**
+ * 1659. Maximize Grid Happiness
+ * https://leetcode.com/problems/maximize-grid-happiness/
+ * Difficulty: Hard
+ *
+ * You are given four integers, m, n, introvertsCount, and extrovertsCount. You have an m x n grid,
+ * and there are two types of people: introverts and extroverts. There are introvertsCount
+ * introverts and extrovertsCount extroverts.
+ *
+ * You should decide how many people you want to live in the grid and assign each of them one grid
+ * cell. Note that you do not have to have all the people living in the grid.
+ *
+ * The happiness of each person is calculated as follows:
+ * - Introverts start with 120 happiness and lose 30 happiness for each neighbor (introvert or
+ *   extrovert).
+ * - Extroverts start with 40 happiness and gain 20 happiness for each neighbor (introvert or
+ *   extrovert).
+ *
+ * Neighbors live in the directly adjacent cells north, east, south, and west of a person's cell.
+ *
+ * The grid happiness is the sum of each person's happiness. Return the maximum possible grid
+ * happiness.
+ */
+
+/**
+ * @param {number} m
+ * @param {number} n
+ * @param {number} introvertsCount
+ * @param {number} extrovertsCount
+ * @return {number}
+ */
+var getMaxGridHappiness = function(m, n, introvertsCount, extrovertsCount) {
+  const memo = new Map();
+  const maxState = Math.pow(3, n);
+
+  return calculateHappiness(0, 0, introvertsCount, extrovertsCount, 0);
+
+  function calculateHappiness(row, col, introverts, extroverts, prevState) {
+    if (row === m || (introverts === 0 && extroverts === 0)) return 0;
+    if (col === n) return calculateHappiness(row + 1, 0, introverts, extroverts, prevState);
+
+    const key = `${row},${col},${introverts},${extroverts},${prevState}`;
+    if (memo.has(key)) return memo.get(key);
+
+    const nextState = (prevState * 3) % maxState;
+    let maxHappiness = calculateHappiness(row, col + 1, introverts, extroverts, nextState);
+
+    if (introverts > 0) {
+      let happiness = 120;
+      const leftNeighbor = col > 0 ? prevState % 3 : 0;
+      const upNeighbor = row > 0 ? Math.floor(prevState / Math.pow(3, n - 1)) % 3 : 0;
+
+      if (leftNeighbor) happiness -= 30;
+      if (upNeighbor) happiness -= 30;
+
+      let neighborHappiness = 0;
+      if (leftNeighbor === 1) neighborHappiness -= 30;
+      if (leftNeighbor === 2) neighborHappiness += 20;
+      if (upNeighbor === 1) neighborHappiness -= 30;
+      if (upNeighbor === 2) neighborHappiness += 20;
+
+      maxHappiness = Math.max(
+        maxHappiness,
+        happiness + neighborHappiness
+        + calculateHappiness(row, col + 1, introverts - 1, extroverts, nextState + 1)
+      );
+    }
+
+    if (extroverts > 0) {
+      let happiness = 40;
+      const leftNeighbor = col > 0 ? prevState % 3 : 0;
+      const upNeighbor = row > 0 ? Math.floor(prevState / Math.pow(3, n - 1)) % 3 : 0;
+
+      if (leftNeighbor) happiness += 20;
+      if (upNeighbor) happiness += 20;
+
+      let neighborHappiness = 0;
+      if (leftNeighbor === 1) neighborHappiness -= 30;
+      if (leftNeighbor === 2) neighborHappiness += 20;
+      if (upNeighbor === 1) neighborHappiness -= 30;
+      if (upNeighbor === 2) neighborHappiness += 20;
+
+      maxHappiness = Math.max(
+        maxHappiness,
+        happiness + neighborHappiness
+        + calculateHappiness(row, col + 1, introverts, extroverts - 1, nextState + 2)
+      );
+    }
+
+    memo.set(key, maxHappiness);
+    return maxHappiness;
+  }
+};