[Algorithm] Two crystal balls problem
You're given two identical crystal balls and a 100-story building. The balls are incredibly tough, but there exists some floor in the building, above which the balls will break when dropped, and below which they will not. You don't know what this floor is, but you want to find out with the fewest possible drops. How do you do it?
/**
* Noramlly we would consider find middle point of the array
* But if assume the first ball break, then we need to walk through
* half of the array to find the breaking point.
*
* There fore the T: O(N/2) = O(N).
*
* We want better than that.
*
* So what we can do is we take step of sqrt(N) each time for the first ball
* If it breaks, then we can walk through the sqrt(N) array to find the breaking point.
*
* O(sqrt(N)) < O(N)
*/
export default function two_crystal_balls(breaks: boolean[]): number {
const jumpAmount = Math.floor(Math.sqrt(breaks.length))
// using first ball to find the breaking point range
let i = 0
while (jumpAmount * i < breaks.length) {
if (breaks[jumpAmount * i]) {
break
}
i++
}
// using second ball to find actual breaking point
let j = jumpAmount * (i - 1)
for (let k =0; k < jumpAmount; k++) {
if (breaks[j + k]) {
return j + k
}
}
return -1
}
Test:
test("two crystal balls", function () {
let idx = Math.floor(Math.random() * 10000);
const data = new Array(10000).fill(false);
for (let i = idx; i < 10000; ++i) {
data[i] = true;
}
expect(two_crystal_balls(data)).toEqual(idx);
expect(two_crystal_balls(new Array(821).fill(false))).toEqual(-1);
});
【推荐】国内首个AI IDE,深度理解中文开发场景,立即下载体验Trae
【推荐】编程新体验,更懂你的AI,立即体验豆包MarsCode编程助手
【推荐】抖音旗下AI助手豆包,你的智能百科全书,全免费不限次数
【推荐】轻量又高性能的 SSH 工具 IShell:AI 加持,快人一步
· 阿里最新开源QwQ-32B,效果媲美deepseek-r1满血版,部署成本又又又降低了!
· Manus重磅发布:全球首款通用AI代理技术深度解析与实战指南
· 开源Multi-agent AI智能体框架aevatar.ai,欢迎大家贡献代码
· 被坑几百块钱后,我竟然真的恢复了删除的微信聊天记录!
· AI技术革命,工作效率10个最佳AI工具
2022-07-08 [Typescript] Difference between return a variable or Object directly
2018-07-08 [RxJS 6] The Catch and Rethrow RxJs Error Handling Strategy and the finalize Operator
2018-07-08 [RxJS 6] The Retry RxJs Error Handling Strategy