diff --git a/src/solution/mod.rs b/src/solution/mod.rs
index 842f2d20..8354b983 100644
--- a/src/solution/mod.rs
+++ b/src/solution/mod.rs
@@ -1369,3 +1369,4 @@ mod s1813_sentence_similarity_iii;
 mod s1814_count_nice_pairs_in_an_array;
 mod s1815_maximum_number_of_groups_getting_fresh_donuts;
 mod s1816_truncate_sentence;
+mod s1817_finding_the_users_active_minutes;
diff --git a/src/solution/s1817_finding_the_users_active_minutes.rs b/src/solution/s1817_finding_the_users_active_minutes.rs
new file mode 100644
index 00000000..18def33a
--- /dev/null
+++ b/src/solution/s1817_finding_the_users_active_minutes.rs
@@ -0,0 +1,93 @@
+/**
+ * [1817] Finding the Users Active Minutes
+ *
+ * You are given the logs for users' actions on LeetCode, and an integer k. The logs are represented by a 2D integer array logs where each logs[i] = [IDi, timei] indicates that the user with IDi performed an action at the minute timei.
+ * Multiple users can perform actions simultaneously, and a single user can perform multiple actions in the same minute.
+ * The user active minutes (UAM) for a given user is defined as the number of unique minutes in which the user performed an action on LeetCode. A minute can only be counted once, even if multiple actions occur during it.
+ * You are to calculate a 1-indexed array answer of size k such that, for each j (1 <= j <= k), answer[j] is the number of users whose UAM equals j.
+ * Return the array answer as described above.
+ *  
+ * Example 1:
+ *
+ * Input: logs = [[0,5],[1,2],[0,2],[0,5],[1,3]], k = 5
+ * Output: [0,2,0,0,0]
+ * Explanation:
+ * The user with ID=0 performed actions at minutes 5, 2, and 5 again. Hence, they have a UAM of 2 (minute 5 is only counted once).
+ * The user with ID=1 performed actions at minutes 2 and 3. Hence, they have a UAM of 2.
+ * Since both users have a UAM of 2, answer[2] is 2, and the remaining answer[j] values are 0.
+ *
+ * Example 2:
+ *
+ * Input: logs = [[1,1],[2,2],[2,3]], k = 4
+ * Output: [1,1,0,0]
+ * Explanation:
+ * The user with ID=1 performed a single action at minute 1. Hence, they have a UAM of 1.
+ * The user with ID=2 performed actions at minutes 2 and 3. Hence, they have a UAM of 2.
+ * There is one user with a UAM of 1 and one with a UAM of 2.
+ * Hence, answer[1] = 1, answer[2] = 1, and the remaining values are 0.
+ *
+ *  
+ * Constraints:
+ *
+ * 	1 <= logs.length <= 10^4
+ * 	0 <= IDi <= 10^9
+ * 	1 <= timei <= 10^5
+ * 	k is in the range [The maximum UAM for a user, 10^5].
+ *
+ */
+pub struct Solution {}
+
+// problem: https://leetcode.com/problems/finding-the-users-active-minutes/
+// discuss: https://leetcode.com/problems/finding-the-users-active-minutes/discuss/?currentPage=1&orderBy=most_votes&query=
+
+// submission codes start here
+
+impl Solution {
+    pub fn finding_users_active_minutes(logs: Vec<Vec<i32>>, k: i32) -> Vec<i32> {
+        let mut result = vec![0; k as usize];
+        let mut hm: std::collections::HashMap<i32, Vec<i32>> = std::collections::HashMap::new();
+
+        for log in logs {
+            let user_id = log[0];
+            let minute = log[1];
+
+            hm.entry(user_id).or_default().push(minute)
+        }
+
+        for (_user_id, mut minutes) in hm {
+            minutes.sort();
+            minutes.dedup();
+
+            result[minutes.len() - 1] += 1;
+        }
+
+        result
+    }
+}
+
+// submission codes end
+
+#[cfg(test)]
+mod tests {
+    use super::*;
+
+    #[test]
+    fn test_1817_example_1() {
+        let logs = vec![vec![0, 5], vec![1, 2], vec![0, 2], vec![0, 5], vec![1, 3]];
+        let k = 5;
+
+        let result = vec![0, 2, 0, 0, 0];
+
+        assert_eq!(Solution::finding_users_active_minutes(logs, k), result);
+    }
+
+    #[test]
+    fn test_1817_example_2() {
+        let logs = vec![vec![1, 1], vec![2, 2], vec![2, 3]];
+        let k = 4;
+
+        let result = vec![1, 1, 0, 0];
+
+        assert_eq!(Solution::finding_users_active_minutes(logs, k), result);
+    }
+}