LeetCode #1604 — MEDIUM

Alert Using Same Key-Card Three or More Times in a One Hour Period

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

LeetCode company workers use key-cards to unlock office doors. Each time a worker uses their key-card, the security system saves the worker's name and the time when it was used. The system emits an alert if any worker uses the key-card three or more times in a one-hour period.

You are given a list of strings keyName and keyTime where [keyName[i], keyTime[i]] corresponds to a person's name and the time when their key-card was used in a single day.

Access times are given in the 24-hour time format "HH:MM", such as "23:51" and "09:49".

Return a list of unique worker names who received an alert for frequent keycard use. Sort the names in ascending order alphabetically.

Notice that "10:00" - "11:00" is considered to be within a one-hour period, while "22:51" - "23:52" is not considered to be within a one-hour period.

Example 1:

Input: keyName = ["daniel","daniel","daniel","luis","luis","luis","luis"], keyTime = ["10:00","10:40","11:00","09:00","11:00","13:00","15:00"]
Output: ["daniel"]
Explanation: "daniel" used the keycard 3 times in a one-hour period ("10:00","10:40", "11:00").

Example 2:

Input: keyName = ["alice","alice","alice","bob","bob","bob","bob"], keyTime = ["12:01","12:00","18:00","21:00","21:20","21:30","23:00"]
Output: ["bob"]
Explanation: "bob" used the keycard 3 times in a one-hour period ("21:00","21:20", "21:30").

Constraints:

1 <= keyName.length, keyTime.length <= 10⁵
keyName.length == keyTime.length
keyTime[i] is in the format "HH:MM".
[keyName[i], keyTime[i]] is unique.
1 <= keyName[i].length <= 10
keyName[i] contains only lowercase English letters.

Step 01

Brute Force Baseline

Problem summary: LeetCode company workers use key-cards to unlock office doors. Each time a worker uses their key-card, the security system saves the worker's name and the time when it was used. The system emits an alert if any worker uses the key-card three or more times in a one-hour period. You are given a list of strings keyName and keyTime where [keyName[i], keyTime[i]] corresponds to a person's name and the time when their key-card was used in a single day. Access times are given in the 24-hour time format "HH:MM", such as "23:51" and "09:49". Return a list of unique worker names who received an alert for frequent keycard use. Sort the names in ascending order alphabetically. Notice that "10:00" - "11:00" is considered to be within a one-hour period, while "22:51" - "23:52" is not considered to be within a one-hour period.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map

Example 1

["daniel","daniel","daniel","luis","luis","luis","luis"]
["10:00","10:40","11:00","09:00","11:00","13:00","15:00"]

Example 2

["alice","alice","alice","bob","bob","bob","bob"]
["12:01","12:00","18:00","21:00","21:20","21:30","23:00"]

Step 02

Core Insight

What unlocks the optimal approach

Group the times by the name of the card user, then sort each group

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1604: Alert Using Same Key-Card Three or More Times in a One Hour Period
class Solution {
    public List<String> alertNames(String[] keyName, String[] keyTime) {
        Map<String, List<Integer>> d = new HashMap<>();
        for (int i = 0; i < keyName.length; ++i) {
            String name = keyName[i];
            String time = keyTime[i];
            int t
                = Integer.parseInt(time.substring(0, 2)) * 60 + Integer.parseInt(time.substring(3));
            d.computeIfAbsent(name, k -> new ArrayList<>()).add(t);
        }
        List<String> ans = new ArrayList<>();
        for (var e : d.entrySet()) {
            var ts = e.getValue();
            int n = ts.size();
            if (n > 2) {
                Collections.sort(ts);
                for (int i = 0; i < n - 2; ++i) {
                    if (ts.get(i + 2) - ts.get(i) <= 60) {
                        ans.add(e.getKey());
                        break;
                    }
                }
            }
        }
        Collections.sort(ans);
        return ans;
    }
}

// Accepted solution for LeetCode #1604: Alert Using Same Key-Card Three or More Times in a One Hour Period
func alertNames(keyName []string, keyTime []string) (ans []string) {
	d := map[string][]int{}
	for i, name := range keyName {
		var a, b int
		fmt.Sscanf(keyTime[i], "%d:%d", &a, &b)
		t := a*60 + b
		d[name] = append(d[name], t)
	}
	for name, ts := range d {
		n := len(ts)
		if n > 2 {
			sort.Ints(ts)
			for i := 0; i < n-2; i++ {
				if ts[i+2]-ts[i] <= 60 {
					ans = append(ans, name)
					break
				}
			}
		}
	}
	sort.Strings(ans)
	return
}

# Accepted solution for LeetCode #1604: Alert Using Same Key-Card Three or More Times in a One Hour Period
class Solution:
    def alertNames(self, keyName: List[str], keyTime: List[str]) -> List[str]:
        d = defaultdict(list)
        for name, t in zip(keyName, keyTime):
            t = int(t[:2]) * 60 + int(t[3:])
            d[name].append(t)
        ans = []
        for name, ts in d.items():
            if (n := len(ts)) > 2:
                ts.sort()
                for i in range(n - 2):
                    if ts[i + 2] - ts[i] <= 60:
                        ans.append(name)
                        break
        ans.sort()
        return ans

// Accepted solution for LeetCode #1604: Alert Using Same Key-Card Three or More Times in a One Hour Period
struct Solution;
use std::collections::BTreeSet;
use std::collections::HashMap;
use std::collections::HashSet;

impl Solution {
    fn alert_names(key_name: Vec<String>, key_time: Vec<String>) -> Vec<String> {
        let n = key_name.len();
        let mut alerted: HashSet<String> = HashSet::new();
        let mut queues: HashMap<String, BTreeSet<i32>> = HashMap::new();
        for i in 0..n {
            if alerted.contains(&key_name[i]) {
                continue;
            }
            let times = queues.entry(key_name[i].to_string()).or_default();
            let t = Self::parse_time(&key_time[i]);
            times.insert(t);
            let sorted_times: Vec<i32> = times.iter().copied().collect();
            if sorted_times.windows(3).any(|w| w[0] + 60 >= w[2]) {
                alerted.insert(key_name[i].to_string());
            }
        }
        let mut res: Vec<String> = alerted.into_iter().collect();
        res.sort_unstable();
        res
    }
    fn parse_time(s: &str) -> i32 {
        s[..2].parse::<i32>().unwrap() * 60 + s[3..].parse::<i32>().unwrap()
    }
}

#[test]
fn test() {
    let key_name = vec_string!["daniel", "daniel", "daniel", "luis", "luis", "luis", "luis"];
    let key_time = vec_string!["10:00", "10:40", "11:00", "09:00", "11:00", "13:00", "15:00"];
    let res = vec_string!["daniel"];
    assert_eq!(Solution::alert_names(key_name, key_time), res);
    let key_name = vec_string!["alice", "alice", "alice", "bob", "bob", "bob", "bob"];
    let key_time = vec_string!["12:01", "12:00", "18:00", "21:00", "21:20", "21:30", "23:00"];
    let res = vec_string!["bob"];
    assert_eq!(Solution::alert_names(key_name, key_time), res);
    let key_name = vec_string!["john", "john", "john"];
    let key_time = vec_string!["23:58", "23:59", "00:01"];
    let res = vec_string![];
    assert_eq!(Solution::alert_names(key_name, key_time), res);
    let key_name = vec_string!["leslie", "leslie", "leslie", "clare", "clare", "clare", "clare"];
    let key_time = vec_string!["13:00", "13:20", "14:00", "18:00", "18:51", "19:30", "19:49"];
    let res = vec_string!["clare", "leslie"];
    assert_eq!(Solution::alert_names(key_name, key_time), res);
}

// Accepted solution for LeetCode #1604: Alert Using Same Key-Card Three or More Times in a One Hour Period
function alertNames(keyName: string[], keyTime: string[]): string[] {
    const d: { [name: string]: number[] } = {};
    for (let i = 0; i < keyName.length; ++i) {
        const name = keyName[i];
        const t = keyTime[i];
        const minutes = +t.slice(0, 2) * 60 + +t.slice(3);
        if (d[name] === undefined) {
            d[name] = [];
        }
        d[name].push(minutes);
    }
    const ans: string[] = [];
    for (const name in d) {
        if (d.hasOwnProperty(name)) {
            const ts = d[name];
            if (ts.length > 2) {
                ts.sort((a, b) => a - b);
                for (let i = 0; i < ts.length - 2; ++i) {
                    if (ts[i + 2] - ts[i] <= 60) {
                        ans.push(name);
                        break;
                    }
                }
            }
        }
    }
    ans.sort();
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × log n)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

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.

OPTIMIZED

O(n) time

O(1) space

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.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

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.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.