Add solution #2327

JoshCrozier · JoshCrozier · commit 217e147183af · 2025-07-29T23:45:18.000-05:00
diff --git a/README.md b/README.md
@@ -2112,6 +2112,7 @@
 2323|[Find Minimum Time to Finish All Jobs II](./solutions/2323-find-minimum-time-to-finish-all-jobs-ii.js)|Medium|
 2325|[Decode the Message](./solutions/2325-decode-the-message.js)|Easy|
 2326|[Spiral Matrix IV](./solutions/2326-spiral-matrix-iv.js)|Medium|
+2327|[Number of People Aware of a Secret](./solutions/2327-number-of-people-aware-of-a-secret.js)|Medium|
 2328|[Number of Increasing Paths in a Grid](./solutions/2328-number-of-increasing-paths-in-a-grid.js)|Hard|
 2330|[Valid Palindrome IV](./solutions/2330-valid-palindrome-iv.js)|Medium|
 2331|[Evaluate Boolean Binary Tree](./solutions/2331-evaluate-boolean-binary-tree.js)|Easy|
diff --git a/solutions/2327-number-of-people-aware-of-a-secret.js b/solutions/2327-number-of-people-aware-of-a-secret.js
@@ -0,0 +1,43 @@
+/**
+ * 2327. Number of People Aware of a Secret
+ * https://leetcode.com/problems/number-of-people-aware-of-a-secret/
+ * Difficulty: Medium
+ *
+ * On day 1, one person discovers a secret.
+ *
+ * You are given an integer delay, which means that each person will share the secret with a new
+ * person every day, starting from delay days after discovering the secret. You are also given
+ * an integer forget, which means that each person will forget the secret forget days after
+ * discovering it. A person cannot share the secret on the same day they forgot it, or on any
+ * day afterwards.
+ *
+ * Given an integer n, return the number of people who know the secret at the end of day n.
+ * Since the answer may be very large, return it modulo 109 + 7.
+ */
+
+/**
+ * @param {number} n
+ * @param {number} delay
+ * @param {number} forget
+ * @return {number}
+ */
+var peopleAwareOfSecret = function(n, delay, forget) {
+  const MOD = 1e9 + 7;
+  const dp = new Array(n + 1).fill(0);
+  dp[1] = 1;
+
+  for (let day = 1; day <= n; day++) {
+    if (dp[day] === 0) continue;
+
+    for (let shareDay = day + delay; shareDay < day + forget && shareDay <= n; shareDay++) {
+      dp[shareDay] = (dp[shareDay] + dp[day]) % MOD;
+    }
+  }
+
+  let result = 0;
+  for (let day = Math.max(1, n - forget + 1); day <= n; day++) {
+    result = (result + dp[day]) % MOD;
+  }
+
+  return result;
+};
