Analyze User Website Visit Pattern

An app's analytics log has one row per screen opening. Row i says that viewer viewer[i] opened screen screen[i] at clock reading moment[i]. The rows arrive in no particular order.

Put each viewer's own rows into increasing order of clock reading; when two of that viewer's rows share a reading, the one listed earlier in the input goes first. That gives every viewer a single ordered walk through the app.

A trail is a sequence of exactly three screen names. A viewer follows a trail when three rows can be picked out of their walk, at strictly increasing places in it, whose screen names spell the trail in order. The three places need not be next to each other, and the same screen name may appear more than once in a trail if the viewer opened that screen more than once. A viewer either follows a trail or does not; the number of different ways they could pick the rows makes no difference.

Return the trail followed by the greatest number of viewers, as a list of its three screen names. When several trails are followed by that many viewers, return the alphabetically smallest of them, comparing the three names in order.

Examples

Constraints

• 3 <= viewer.length <= 50
• 1 <= viewer[i].length <= 10
• moment.length == viewer.length
• 1 <= moment[i] <= 10^9
• screen.length == viewer.length
• 1 <= screen[i].length <= 10
• viewer[i] and screen[i] consist of lowercase English letters.
• At least one viewer has three or more rows in the log.
• All the triples [viewer[i], moment[i], screen[i]] are distinct.
• Two rows of the same viewer may share a clock reading; the earlier row in the input then comes first.

The signature

The editor starts with this. It matches the parent problem's shape, so a solution that works here works on a standard judge unchanged.

def busiest_trail(viewer: list[str], moment: list[int], screen: list[str]) -> list[str]:

public List<String> busiestTrail(String[] viewer, int[] moment, String[] screen)