[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);
});

 

posted @   Zhentiw  阅读(60)  评论(0编辑  收藏  举报
相关博文:
阅读排行:
· 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
历史上的今天:
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
点击右上角即可分享
微信分享提示