Add solution #879

Author: JoshCrozier

README.md

@@ -689,6 +689,7 @@
 876|[Middle of the Linked List](./0876-middle-of-the-linked-list.js)|Easy|
 877|[Stone Game](./0877-stone-game.js)|Medium|
 878|[Nth Magical Number](./0878-nth-magical-number.js)|Hard|
+879|[Profitable Schemes](./0879-profitable-schemes.js)|Hard|
 884|[Uncommon Words from Two Sentences](./0884-uncommon-words-from-two-sentences.js)|Easy|
 889|[Construct Binary Tree from Preorder and Postorder Traversal](./0889-construct-binary-tree-from-preorder-and-postorder-traversal.js)|Medium|
 890|[Find and Replace Pattern](./0890-find-and-replace-pattern.js)|Medium|

solutions/0879-profitable-schemes.js

@@ -0,0 +1,55 @@
+/**
+ * 879. Profitable Schemes
+ * https://leetcode.com/problems/profitable-schemes/
+ * Difficulty: Hard
+ *
+ * There is a group of n members, and a list of various crimes they could commit. The ith crime
+ * generates a profit[i] and requires group[i] members to participate in it. If a member
+ * participates in one crime, that member can't participate in another crime.
+ *
+ * Let's call a profitable scheme any subset of these crimes that generates at least minProfit
+ * profit, and the total number of members participating in that subset of crimes is at most n.
+ *
+ * Return the number of schemes that can be chosen. Since the answer may be very large, return
+ * it modulo 109 + 7.
+ */
+
+/**
+ * @param {number} n
+ * @param {number} minProfit
+ * @param {number[]} group
+ * @param {number[]} profit
+ * @return {number}
+ */
+var profitableSchemes = function(n, minProfit, group, profit) {
+  const MOD = 1e9 + 7;
+  const dp = Array.from({ length: group.length + 1 }, () =>
+    Array.from({ length: n + 1 }, () => new Array(minProfit + 1).fill(0))
+  );
+
+  dp[0][0][0] = 1;
+
+  for (let crime = 1; crime <= group.length; crime++) {
+    const membersNeeded = group[crime - 1];
+    const profitGained = profit[crime - 1];
+
+    for (let members = 0; members <= n; members++) {
+      for (let currentProfit = 0; currentProfit <= minProfit; currentProfit++) {
+        dp[crime][members][currentProfit] = dp[crime - 1][members][currentProfit];
+
+        if (members >= membersNeeded) {
+          const prevProfit = Math.max(0, currentProfit - profitGained);
+          dp[crime][members][currentProfit] = (dp[crime][members][currentProfit]
+            + dp[crime - 1][members - membersNeeded][prevProfit]) % MOD;
+        }
+      }
+    }
+  }
+
+  let result = 0;
+  for (let members = 0; members <= n; members++) {
+    result = (result + dp[group.length][members][minProfit]) % MOD;
+  }
+
+  return result;
+};