Off-by-one on range boundaries
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.
Build confidence with an intuition-first walkthrough focused on core interview patterns fundamentals.
Table: Logins
+----------------+----------+ | Column Name | Type | +----------------+----------+ | user_id | int | | time_stamp | datetime | +----------------+----------+ (user_id, time_stamp) is the primary key (combination of columns with unique values) for this table. Each row contains information about the login time for the user with ID user_id.
Write a solution to report the latest login for all users in the year 2020. Do not include the users who did not login in 2020.
Return the result table in any order.
The result format is in the following example.
Example 1:
Input: Logins table: +---------+---------------------+ | user_id | time_stamp | +---------+---------------------+ | 6 | 2020-06-30 15:06:07 | | 6 | 2021-04-21 14:06:06 | | 6 | 2019-03-07 00:18:15 | | 8 | 2020-02-01 05:10:53 | | 8 | 2020-12-30 00:46:50 | | 2 | 2020-01-16 02:49:50 | | 2 | 2019-08-25 07:59:08 | | 14 | 2019-07-14 09:00:00 | | 14 | 2021-01-06 11:59:59 | +---------+---------------------+ Output: +---------+---------------------+ | user_id | last_stamp | +---------+---------------------+ | 6 | 2020-06-30 15:06:07 | | 8 | 2020-12-30 00:46:50 | | 2 | 2020-01-16 02:49:50 | +---------+---------------------+ Explanation: User 6 logged into their account 3 times but only once in 2020, so we include this login in the result table. User 8 logged into their account 2 times in 2020, once in February and once in December. We include only the latest one (December) in the result table. User 2 logged into their account 2 times but only once in 2020, so we include this login in the result table. User 14 did not login in 2020, so we do not include them in the result table.
Problem summary: Table: Logins +----------------+----------+ | Column Name | Type | +----------------+----------+ | user_id | int | | time_stamp | datetime | +----------------+----------+ (user_id, time_stamp) is the primary key (combination of columns with unique values) for this table. Each row contains information about the login time for the user with ID user_id. Write a solution to report the latest login for all users in the year 2020. Do not include the users who did not login in 2020. Return the result table in any order. The result format is in the following example.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: General problem-solving
{"headers": {"Logins": ["user_id", "time_stamp"]}, "rows": {"Logins": [[6, "2020-06-30 15:06:07"], [6, "2021-04-21 14:06:06"], [6, "2019-03-07 00:18:15"], [8, "2020-02-01 05:10:53"], [8, "2020-12-30 00:46:50"], [2, "2020-01-16 02:49:50"], [2, "2019-08-25 07:59:08"], [14, "2019-07-14 09:00:00"], [14, "2021-01-06 11:59:59"]]}}Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #1890: The Latest Login in 2020
// Auto-generated Java example from rust.
class Solution {
public void exampleSolution() {
}
}
// Reference (rust):
// // Accepted solution for LeetCode #1890: The Latest Login in 2020
// pub fn sql_example() -> &'static str {
// r#"
// -- Accepted solution for LeetCode #1890: The Latest Login in 2020
// # Write your MySQL query statement below
// SELECT user_id, MAX(time_stamp) AS last_stamp
// FROM Logins
// WHERE YEAR(time_stamp) = 2020
// GROUP BY 1;
// "#
// }
// Accepted solution for LeetCode #1890: The Latest Login in 2020
// Auto-generated Go example from rust.
func exampleSolution() {
}
// Reference (rust):
// // Accepted solution for LeetCode #1890: The Latest Login in 2020
// pub fn sql_example() -> &'static str {
// r#"
// -- Accepted solution for LeetCode #1890: The Latest Login in 2020
// # Write your MySQL query statement below
// SELECT user_id, MAX(time_stamp) AS last_stamp
// FROM Logins
// WHERE YEAR(time_stamp) = 2020
// GROUP BY 1;
// "#
// }
# Accepted solution for LeetCode #1890: The Latest Login in 2020
# Auto-generated Python example from rust.
def example_solution() -> None:
return
# Reference (rust):
# // Accepted solution for LeetCode #1890: The Latest Login in 2020
# pub fn sql_example() -> &'static str {
# r#"
# -- Accepted solution for LeetCode #1890: The Latest Login in 2020
# # Write your MySQL query statement below
# SELECT user_id, MAX(time_stamp) AS last_stamp
# FROM Logins
# WHERE YEAR(time_stamp) = 2020
# GROUP BY 1;
# "#
# }
// Accepted solution for LeetCode #1890: The Latest Login in 2020
pub fn sql_example() -> &'static str {
r#"
-- Accepted solution for LeetCode #1890: The Latest Login in 2020
# Write your MySQL query statement below
SELECT user_id, MAX(time_stamp) AS last_stamp
FROM Logins
WHERE YEAR(time_stamp) = 2020
GROUP BY 1;
"#
}
// Accepted solution for LeetCode #1890: The Latest Login in 2020
// Auto-generated TypeScript example from rust.
function exampleSolution(): void {
}
// Reference (rust):
// // Accepted solution for LeetCode #1890: The Latest Login in 2020
// pub fn sql_example() -> &'static str {
// r#"
// -- Accepted solution for LeetCode #1890: The Latest Login in 2020
// # Write your MySQL query statement below
// SELECT user_id, MAX(time_stamp) AS last_stamp
// FROM Logins
// WHERE YEAR(time_stamp) = 2020
// GROUP BY 1;
// "#
// }
Use this to step through a reusable interview workflow for this problem.
Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.
Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.
Review these before coding to avoid predictable interview regressions.
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.