Add solution #1473

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,318 LeetCode solutions in JavaScript
+# 1,319 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1126,6 +1126,7 @@
 1470|[Shuffle the Array](./solutions/1470-shuffle-the-array.js)|Easy|
 1471|[The k Strongest Values in an Array](./solutions/1471-the-k-strongest-values-in-an-array.js)|Medium|
 1472|[Design Browser History](./solutions/1472-design-browser-history.js)|Medium|
+1473|[Paint House III](./solutions/1473-paint-house-iii.js)|Hard|
 1475|[Final Prices With a Special Discount in a Shop](./solutions/1475-final-prices-with-a-special-discount-in-a-shop.js)|Easy|
 1480|[Running Sum of 1d Array](./solutions/1480-running-sum-of-1d-array.js)|Easy|
 1481|[Least Number of Unique Integers after K Removals](./solutions/1481-least-number-of-unique-integers-after-k-removals.js)|Medium|

solutions/1473-paint-house-iii.js

@@ -0,0 +1,60 @@
+/**
+ * 1473. Paint House III
+ * https://leetcode.com/problems/paint-house-iii/
+ * Difficulty: Hard
+ *
+ * There is a row of m houses in a small city, each house must be painted with one of the n
+ * colors (labeled from 1 to n), some houses that have been painted last summer should not
+ * be painted again.
+ *
+ * A neighborhood is a maximal group of continuous houses that are painted with the same color.
+ *
+ * For example: houses = [1,2,2,3,3,2,1,1] contains 5 neighborhoods [{1}, {2,2}, {3,3}, {2}, {1,1}].
+ *
+ * Given an array houses, an m x n matrix cost and an integer target where:
+ * - houses[i]: is the color of the house i, and 0 if the house is not painted yet.
+ * - cost[i][j]: is the cost of paint the house i with the color j + 1.
+ *
+ * Return the minimum cost of painting all the remaining houses in such a way that there are exactly
+ * target neighborhoods. If it is not possible, return -1.
+ */
+
+/**
+ * @param {number[]} houses
+ * @param {number[][]} cost
+ * @param {number} m
+ * @param {number} n
+ * @param {number} target
+ * @return {number}
+ */
+var minCost = function(houses, cost, m, n, target) {
+  const cache = new Map();
+  const result = findMinCost(0, 0, 0);
+  return result === Infinity ? -1 : result;
+
+  function findMinCost(index, prevColor, neighborhoods) {
+    if (index === m) return neighborhoods === target ? 0 : Infinity;
+    if (neighborhoods > target) return Infinity;
+
+    const key = `${index}:${prevColor}:${neighborhoods}`;
+    if (cache.has(key)) return cache.get(key);
+
+    let minCost = Infinity;
+    const currentColor = houses[index];
+
+    if (currentColor !== 0) {
+      const newNeighborhoods = prevColor === currentColor ? neighborhoods : neighborhoods + 1;
+      minCost = findMinCost(index + 1, currentColor, newNeighborhoods);
+    } else {
+      for (let color = 1; color <= n; color++) {
+        const newNeighborhoods = prevColor === color ? neighborhoods : neighborhoods + 1;
+        const currentCost = cost[index][color - 1]
+          + findMinCost(index + 1, color, newNeighborhoods);
+        minCost = Math.min(minCost, currentCost);
+      }
+    }
+
+    cache.set(key, minCost);
+    return minCost;
+  }
+};