Add solution #1595

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,408 LeetCode solutions in JavaScript
+# 1,409 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1232,6 +1232,7 @@
 1592|[Rearrange Spaces Between Words](./solutions/1592-rearrange-spaces-between-words.js)|Easy|
 1593|[Split a String Into the Max Number of Unique Substrings](./solutions/1593-split-a-string-into-the-max-number-of-unique-substrings.js)|Medium|
 1594|[Maximum Non Negative Product in a Matrix](./solutions/1594-maximum-non-negative-product-in-a-matrix.js)|Medium|
+1595|[Minimum Cost to Connect Two Groups of Points](./solutions/1595-minimum-cost-to-connect-two-groups-of-points.js)|Hard|
 1598|[Crawler Log Folder](./solutions/1598-crawler-log-folder.js)|Easy|
 1657|[Determine if Two Strings Are Close](./solutions/1657-determine-if-two-strings-are-close.js)|Medium|
 1668|[Maximum Repeating Substring](./solutions/1668-maximum-repeating-substring.js)|Easy|

solutions/1595-minimum-cost-to-connect-two-groups-of-points.js

@@ -0,0 +1,63 @@
+/**
+ * 1595. Minimum Cost to Connect Two Groups of Points
+ * https://leetcode.com/problems/minimum-cost-to-connect-two-groups-of-points/
+ * Difficulty: Hard
+ *
+ * You are given two groups of points where the first group has size1 points, the second group
+ * has size2 points, and size1 >= size2.
+ *
+ * The cost of the connection between any two points are given in an size1 x size2 matrix where
+ * cost[i][j] is the cost of connecting point i of the first group and point j of the second group.
+ * The groups are connected if each point in both groups is connected to one or more points in the
+ * opposite group. In other words, each point in the first group must be connected to at least one
+ * point in the second group, and each point in the second group must be connected to at least one
+ * point in the first group.
+ *
+ * Return the minimum cost it takes to connect the two groups.
+ */
+
+/**
+* @param {number[][]} cost
+* @return {number}
+*/
+var connectTwoGroups = function(cost) {
+  const size1 = cost.length;
+  const size2 = cost[0].length;
+
+  const minCostGroup2 = new Array(size2).fill(Infinity);
+  for (let j = 0; j < size2; j++) {
+    for (let i = 0; i < size1; i++) {
+      minCostGroup2[j] = Math.min(minCostGroup2[j], cost[i][j]);
+    }
+  }
+
+  const memo = new Array(size1).fill(0).map(() => new Array(1 << size2).fill(-1));
+
+  return dfs(0, 0);
+
+  function dfs(i, mask) {
+    if (i === size1) {
+      let remainingCost = 0;
+      for (let j = 0; j < size2; j++) {
+        if ((mask & (1 << j)) === 0) {
+          remainingCost += minCostGroup2[j];
+        }
+      }
+      return remainingCost;
+    }
+
+    if (memo[i][mask] !== -1) return memo[i][mask];
+
+    let minCost = Infinity;
+
+    for (let j = 0; j < size2; j++) {
+      minCost = Math.min(
+        minCost,
+        cost[i][j] + dfs(i + 1, mask | (1 << j))
+      );
+    }
+
+    memo[i][mask] = minCost;
+    return minCost;
+  }
+};