Two pointers and sliding window

The cable car crosses the river in 10-seat cabins; the queue marshal fills them and the controller watches the line.

At the cable car	In the code
the marshal lines the waiting groups up by size, smallest first, and offers one cabin to the smallest and the largest group together	pair from both ends of a sorted list (step 2)
too few riders: the next bigger small group steps forward; too many: the largest group waits and the next smaller big group is tried	move the end that can close the gap (step 3)
the "riders on the line" board: the controller adds each cabin that boards and subtracts each that unloads, never recounting the whole line	keep the summary in step (step 5)
the controller's one-minute frame over the departure log: each new departure enters at the bottom, departures a minute or more older slide off the top	a window over time (step 10)

In real terms: two pointers replace a nested loop with one pass by never letting an index move back.

Breaks down when: a cabin never carries fewer than zero riders, but a list of net changes (riders boarding minus riders unloading) has negative values, and then a window's goal no longer changes one way (step 9).

Contrasting case: three groups in the queue. The marshal tries every pairing by eye; the pattern pays as the queue grows.

4.1 ExamHall EXH-4930: tab-switch bursts, a filter per switch

ExamHall flags a candidate when 5 or more tab switches fall inside any 60 seconds. Its first check filtered the whole log once per switch:

const flagged = (times) => times.some((t) => times.filter((u) => u >= t && u - t < 60000).length >= 5);

On long logs with no burst, the proctor queue fell behind (EXH-4930). MR !5230 walks one window over the sorted times instead (step 10). Both were counted on logs with one switch every 15 s: neither finds a burst, so both read to the end, the worst case:

4.1 EXH-4930 filter per switch, 500 switches, one every 15 s: 250,500 time reads · no burst
4.1 MR !5230 one window, same log: 1,992 time reads · no burst
4.1 EXH-4930 filter per switch, 1,000 switches, one every 15 s: 1,001,000 time reads · no burst
4.1 MR !5230 one window, same log: 3,992 time reads · no burst
4.1 doubling the switches: filter ×4.00 · one window ×2.00; a log with 5 switches inside 40 s: both flag it

The filter reads the log once per switch, so doubling the log quadruples the reads; the window's indices only move forward, so its reads double (steps 1, 8).

4.2 Chat: the busiest five minutes

A chat app shows each room's busiest five consecutive minutes. Re-adding every five-minute window reads a minute up to five times; a running sum adds the new minute and drops the one five back (step 6):

4.2 60 minutes: re-adding each window 280 reads · running sum 115 reads · both pick minutes 25–29 (62 messages)
4.2 120 minutes: re-adding each window 580 reads · running sum 235 reads · both pick minutes 25–29 (62 messages)

Both pick the same minutes; re-adding reads most minutes five times, the running sum at most twice.