What is the time complexity of the maximum population year problem?

The time complexity can vary depending on the approach, but an optimal solution using prefix sums generally has a time complexity of O(k), where k is the range of years.

How does prefix sum help in solving this problem?

Prefix sum helps by efficiently calculating the population at each year without needing to count people repeatedly, leading to faster computations over large ranges of years.

What should I do if multiple years have the same maximum population?

If multiple years have the same population, return the earliest one. This can be handled by keeping track of both the population and the year in each iteration.

What is the role of counting in the maximum population year problem?

Counting helps determine how many people are alive during each year by tracking population changes at birth and death years, allowing for an efficient solution.

How does the maximum population year problem relate to other problems in array manipulation?

The problem uses techniques like array manipulation and counting, which are also commonly used in problems dealing with ranges, intervals, and aggregate calculations.

#1854

Easy

auto_awesome前缀和

LeetCode 题解工作台

人口最多的年份

给你一个二维整数数组 logs ，其中每个 logs[i] = [birth i , death i ] 表示第 i 个人的出生和死亡年份。年份 x 的人口定义为这一年期间活着的人的数目。第 i 个人被计入年份 x 的人口需要满足： x 在闭区间 [birth i , death i - 1]…

数组计数前缀和

题目描述

给你一个二维整数数组 logs ，其中每个 logs[i] = [birth_i, death_i] 表示第 i 个人的出生和死亡年份。

年份 x 的人口定义为这一年期间活着的人的数目。第 i 个人被计入年份 x 的人口需要满足：x 在闭区间 [birth_i, death_i - 1] 内。注意，人不应当计入他们死亡当年的人口中。

返回 人口最多 且最早的年份。

示例 1：

输入：logs = [[1993,1999],[2000,2010]]
输出：1993
解释：人口最多为 1 ，而 1993 是人口为 1 的最早年份。

示例 2：

输入：logs = [[1950,1961],[1960,1971],[1970,1981]]
输出：1960
解释： 
人口最多为 2 ，分别出现在 1960 和 1970 。
其中最早年份是 1960 。

提示：

1 <= logs.length <= 100
1950 <= birth_i < death_i <= 2050

lightbulb

解题思路

方法一：差分数组

我们注意到，年份的范围是 $[1950,..2050]$ ，因此我们可以将这些年份映射到一个长度为 $101$ 的数组 $d$ 中，数组的下标表示年份减去 $1950$ 的值。

接下来遍历 $logs$ ，对于每个人，我们将 $d[birth_i - 1950]$ 加 $1$ ，将 $d[death_i - 1950]$ 减 $1$ 。最后遍历数组 $d$ ，求出前缀和的最大值，即为人口最多的年份，再加上 $1950$ 即为答案。

时间复杂度 $O(n)$ ，空间复杂度 $O(C)$ 。其中 $n$ 为数组 $logs$ 的长度；而 $C$ 为年份的范围大小，即 $2050 - 1950 + 1 = 101$ 。

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class Solution:
    def maximumPopulation(self, logs: List[List[int]]) -> int:
        d = [0] * 101
        offset = 1950
        for a, b in logs:
            a, b = a - offset, b - offset
            d[a] += 1
            d[b] -= 1
        s = mx = j = 0
        for i, x in enumerate(d):
            s += x
            if mx < s:
                mx, j = s, i
        return j + offset

speed

复杂度分析

指标	值
时间	Depends on the final approach
空间	Depends on the final approach

psychology

面试官常问的追问

外企场景

question_mark
Check if the candidate optimizes the approach by using a prefix sum technique.
question_mark
Look for the candidate's ability to handle ties in population values.
question_mark
Test if the candidate can efficiently count population using a simple array.

warning

常见陷阱

外企场景

error
Not correctly handling the exclusive nature of death years.
error
Forgetting to return the earliest year in case of ties in population counts.
error
Using inefficient methods that do not optimize population tracking over a range of years.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Use a different method to track population changes, such as a hash map.
arrow_right_alt
Consider a problem where some birth years are earlier than others, requiring a more complex population calculation.
arrow_right_alt
Limit the input size and check how the candidate handles edge cases like maximum years in a small range.

help

常见问题

外企场景

继续练习

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