0610EasyArrayCountingPrefix Sum

Busiest Lease Year

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Trains the technique from

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

A market hall keeps one record per shop lease. Entry logs[i] = [start_i, end_i] means that lease occupied a unit from the start of year start_i and was handed back at the start of year end_i, so it counts as active in every year from start_i up to and including end_i - 1, and not in end_i itself.

A year's occupancy is the number of leases active in that year. Return the earliest year whose occupancy is as large as any year's occupancy. If several years share the largest occupancy, return the smallest of them.

Examples

Example 1

Input: logs = [[1972, 1986], [1968, 1978], [1980, 1992]]
Output: 1972

In 1972 the first lease has just begun and the second is still running, so the occupancy is 2. No year here reaches 3, and 1972 is the first year that reaches 2.

Example 2

Input: logs = [[1970, 1980], [1990, 2000]]
Output: 1970

The two leases never overlap, so no year has occupancy above 1. Both 1970 and 1990 have occupancy 1, and the smaller of the tied years is returned.

Example 3

Input: logs = [[2049, 2050]]
Output: 2049

The single lease is active in 2049 only, because a lease does not count in its end year.

Constraints

1 <= logs.length <= 100
logs[i].length == 2
1950 <= start_i < end_i <= 2050

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def maximum_population(logs: list[list[int]]) -> int:

Java

public int maximumPopulation(int[][] logs)

The intended approach

Time O(m + span) where m = len(logs) and span is the fixed 101-year window.Space O(span) for the array of yearly changes; span is a constant 102 here.

Only the years where the occupancy changes matter, so instead of counting leases year by year the code records a change of +1 at each start year and a change of -1 at each end year in a small array indexed by year minus 1950. Sweeping that array from 1950 upwards while adding the changes gives the occupancy of each year in turn. The sweep keeps the year of the first strictly larger running total, which is the earliest year holding the largest occupancy.

Pythonentry point: maximum_population

def maximum_population(logs):
    # O(m + span) time, O(span) space, where m = len(logs) and span is the
    # fixed 101-year window 1950..2050. One pass records the change in the
    # active count at each year boundary, a second pass adds those changes up
    # and keeps the first year that reaches the highest running total.
    span = 102
    delta = [0] * span
    for start, end in logs:
        delta[start - 1950] += 1
        delta[end - 1950] -= 1
    best_count = -1
    best_year = 1950
    running = 0
    for i in range(101):
        running += delta[i]
        if running > best_count:
            best_count = running
            best_year = 1950 + i
    return best_year

The obvious version

Walk every year from 1950 to 2050. For each one, scan the whole log list and count the leases whose start is at or before that year and whose end is strictly after it. Keep the first year whose count beats every count seen so far.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_maximum_population(logs):
    best_count = -1
    best_year = 1950
    for year in range(1950, 2051):
        count = 0
        for start, end in logs:
            if start <= year < end:
                count += 1
        if count > best_count:
            best_count = count
            best_year = year
    return best_year

How this problem was checked

Reference solution executed against 11 test cases — 11 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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