CodingTour —
ARTS #141

Algorithm

本周选择的算法题是：Pairs of Songs With Total Durations Divisible by 60。

规则

You are given a list of songs where the ith song has a duration of time[i] seconds.

Return the number of pairs of songs for which their total duration in seconds is divisible by 60. Formally, we want the number of indices i, j such that i < j with (time[i] + time[j]) % 60 == 0.

Example 1:

Input: time = [30,20,150,100,40]
Output: 3
Explanation: Three pairs have a total duration divisible by 60:
(time[0] = 30, time[2] = 150): total duration 180
(time[1] = 20, time[3] = 100): total duration 120
(time[1] = 20, time[4] = 40): total duration 60

Example 2:

Input: time = [60,60,60]
Output: 3
Explanation: All three pairs have a total duration of 120, which is divisible by 60.

Constraints:

  • 1 <= time.length <= 6 * 104
  • 1 <= time[i] <= 500

Solution

class Solution:
    def numPairsDivisibleBy60(self, time: List[int]) -> int:
        ans = 0
        # counter = defaultdict(int)
        counter = [0] * 60
        for item in time:
            diff = -item % 60
            ans += counter[diff]
            counter[item % 60] += 1
        return ans

Review

No Title

Erik Engheim 的这篇文章和保罗·格雷厄姆在《黑客与画家》里的观点类似，语言是生产工具，但更重要的是，它是生产力工具，这意味着商业竞争的护城河已经延伸到了编程语言的选择上，当你选择比对手生产力更高的编程语言，你就比对手有更大的先发优势。

既然可以使用任何语言，那就不得不思考到底使用哪一种语言。如果你的公司对这种选择的自由视而不见，而竞争对手看到了，那么你就有被击败的危险。语言也能成为商业、技术的杠杆。

Tip

学习了 Kafka 时间轮(TimingWheel)算法，真是一个精妙的算法，三层时间轮用60个数组元素(wheelSize=20)就可以承载[0-7999]个定时任务！

Share

最近在做业务跨端的规划，动作拆解为：

  1. 确定具体的页面后，先用原生技术优化到足够好 ⬅️ 价值体现：让用户得到最实在的体验收益
  2. 再用 Kotlin Native + 原生 UI 和纯原生做性能对比 ⬅️ 验证价值：不重写 UI 代码的情况下，KN 方案落地成本低，作为兜低方案，可以统一移动端一半左右的代码量
  3. 再用 Kotlin Native + Flutter UI 和前述方案做对比 ⬅️ 验证价值：业务代码不重写的情况下，验证 UI 性能、体验是否达到预期
  4. 最后用 Dart + Flutter UI 做跨端技术栈的最终价值验证