What is the main approach for Minimum Time Difference?

Convert all time strings to minutes, sort the array, then check consecutive differences including the circular midnight case.

Can timePoints contain duplicates?

Yes, duplicates are allowed and can result in a minimum difference of zero.

Why do we need to consider circular differences?

Because the clock wraps from 23:59 to 00:00, ignoring this can miss the true minimal difference.

What is the optimal time complexity?

Using bucket sort or counting sort, the problem can be solved in O(N) time, otherwise standard sorting gives O(N log N).

Which pattern does this problem follow?

This is an Array plus Math pattern where array manipulation and numeric calculations are combined for efficient solution.

#539

Medium

auto_awesome数组·数学

LeetCode 题解工作台

最小时间差

给定一个 24 小时制（小时:分钟 "HH:MM" ）的时间列表，找出列表中任意两个时间的最小时间差并以分钟数表示。示例 1：输入： timePoints = ["23:59","00:00"] 输出： 1 示例 2：输入： timePoints = ["00:00","23:59","00:…

数组数学字符串排序

题目描述

给定一个 24 小时制（小时:分钟 "HH:MM"）的时间列表，找出列表中任意两个时间的最小时间差并以分钟数表示。

示例 1：

输入：timePoints = ["23:59","00:00"]
输出：1

示例 2：

输入：timePoints = ["00:00","23:59","00:00"]
输出：0

提示：

2 <= timePoints.length <= 2 * 10⁴
timePoints[i] 格式为 "HH:MM"

lightbulb

解题思路

方法一：排序

我们注意到，时间点最多只有 $24 \times 60 = 1440$ 个，因此，当 $timePoints$ 长度超过 $1440$ ，说明有重复的时间点，提前返回 $0$ 。

接下来，我们首先遍历时间列表，将其转换为“分钟制”列表 $nums$ ，比如，对于时间点 13:14，将其转换为 $13 \times 60 + 14$ 。

接着将“分钟制”列表按升序排列，然后将此列表的最小时间 $nums[0]$ 加上 $1440$ 追加至列表尾部，用于处理最大值、最小值的差值这种特殊情况。

最后遍历“分钟制”列表，找出相邻两个时间的最小值即可。

时间复杂度 $O(n \times \log n)$ ，空间复杂度 $O(n)$ 。其中 $n$ 为时间点个数。

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class Solution:
    def findMinDifference(self, timePoints: List[str]) -> int:
        if len(timePoints) > 1440:
            return 0
        nums = sorted(int(x[:2]) * 60 + int(x[3:]) for x in timePoints)
        nums.append(nums[0] + 1440)
        return min(b - a for a, b in pairwise(nums))

speed

复杂度分析

指标	值
时间	complexity is O(N) for converting and sorting if bucket counting is used, otherwise O(N log N) with standard sort. Space complexity is O(1) aside from the array of minutes since in-place computations are possible.
空间	O(1)

psychology

面试官常问的追问

外企场景

question_mark
Asks about handling duplicates and zero-minute differences.
question_mark
Checks awareness of circular time calculation and modulo logic.
question_mark
Tests for edge cases like adjacent midnight values.

warning

常见陷阱

外企场景

error
Forgetting the wrap-around from 23:59 to 00:00 and missing minimal differences.
error
Incorrectly comparing string times instead of numeric minutes.
error
Not handling duplicate times which can produce zero difference.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Compute maximum time difference instead of minimum.
arrow_right_alt
Allow times with seconds and compute minimal difference including seconds.
arrow_right_alt
Find k smallest differences among all time points.

help

常见问题

外企场景

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