Monday, August 8, 2016

Google – Rain Drop Problem


Google – Rain Drop Problem
有一个1m长的side walk, 假设是0到1。雨滴的宽度为0.01m,每次下雨的位置随机,雨滴和雨滴之间可能no overlap, complete overlap和partial overlap。问多少雨滴才能覆盖整个side walk。
  1. 模拟整个下雨过程,返回总共需要的雨滴数
  2. 算期望值。
[Analysis]
(1) 首先这道题我在模拟的过程中碰到一个问题,因为double精度的关系,在[0.0, 1.0]之间相当于有"无数"个点,导致在模拟过程中下了几十万滴雨都没有把整个side walk覆盖。因为想象一下,比如现在下了2滴雨, range分别为
[0.2000000000000001, 0.2100000000000001]
[0.2200000000000001, 0.2300000000000001]
但是其实这两滴雨之间还有"无数"个点没有被覆盖。
所以我觉得这道题应该要保留两位小数,这样像上面那两滴雨已经算把0.20 ~ 023这段完全覆盖了。
(2) 另一个需要保留两位小数的理由就是Java的double type真是能虐死人。
0.09 + 0.01 = 0.09999999999999999
(3) 还有一点需要注意的是,这道题random雨滴的范围不应该是[0.0, 1.0],而应该是[-0.01, 1.0],因为雨滴宽度的关系,如果random范围是[0.0, 1.0]的话,那么0.0被覆盖的概率和其他点被覆盖的概率是不一样的,只有当random到0.0的时候才会被覆盖,但是比如1.0这个点,random到0.99和1.0都能被覆盖。
所以我们需要等概率的来random[-0.01, 1.0]的之间的数并保留两位小数。方法有点类似于rand2 to rand6,保证概率的一致性。
1.01 * Math.random() – 0.01

[Solution]
模拟的过程有两种解法
__[Solution #1]__
merge interval + rand2 to rand6 + binary search
思路就是维护一个list of interval, 每次random一个新的雨滴,通过binary search来找插入位置。然后merge.
注意这道题的merge并不像一般的merge interval一样,因为这道题确定每个新的interval range都是0.01, 所以找到插入点后只需要merge前后两个就好了。
但是就算这样,其实这道题merge interval并不好做,因为比如[0.20, 0.21]和[0.22, 0.23]这两个interval就要merge了,而不是非要curr.start = prev.end才能和前面merge。另外binary search也是难写的一比。
后面有merge interval的code,但是真不知道对不对,binary search也偷懒了直接linear search了。
__[Solution #2]__
后来发现如果保留两位小数的话,其实0.0到1.0之间就是100个点。我们完全可以避免double带来的一堆麻烦。也可以不用merge来merge去的。
直接开个size为100的array作为bucket,每次random一个[-1, 100]的数,那么雨滴范围就是[start, start + 1]。看start和start + 1的bucket是否为空,是的话,bucket置1,bucketCnt++。 直到bucketCnt为100为止。
[期望]
网上看来的:
一共100个buckets, 假设现在剩余x个buckets为空,那么
雨滴落在这x个buckets里的概率为x / 100,这样剩余的buckets就为x – 1
雨滴没有落在这x个buckets里的概率为(100 – x) / 100,剩余的buckets仍然为x
那么
f(x) = f(x – 1) * (x / 100) + f(x) * (100 – x) / 100 + 1
100 * f(x) = x * f(x – 1) + (100 – x) * f(x) + 100
f(x) = f(x – 1) + 100 / x
f(x) = 100 * (1 + 1/2 + 1/3 + 1/4 + … + 1/x),后面那一串称为调和级数。
那么对于这道题来说,期望就是f(100) = 518.737752

这道题如果真被问到,模拟过程的算法肯定是重点。
期望的话可以说个大概的数字。能分析出概率情况写出上面第一个式子已经非常不错了。调和级数不调和级数这种东西,面试官是不会期待有人能答出来的。
class Solution {
   
  int N = 100;
   
  public int rainDrop() {
    Random rand = new Random();
    boolean[] buckets = new boolean[N];
    int cnt = 0;
    int bucketCount = 0;
    while (true) {
      int randStart = rand.nextInt(N) - 1;
      int randEnd = randStart + 1;
      cnt++;
      if (randStart >= 0 && !buckets[randStart]) {
        buckets[randStart] = true;
        bucketCount++;
      }
      else if (!buckets[randEnd]) {
        buckets[randEnd] = true;
        bucketCount++;
      }
       
      if (bucketCount == N) {
        break;
      }
    }
     
    return cnt;
  }
}
 
class Solution2 {
  public int rainDrop() {
    List<Interval> intervals = new ArrayList<>();
    int cnt = 0;
     
    while (true) {
       
      double randStart = 1.01 * Math.random() - 0.01;
       
      randStart = round(randStart);
      double randEnd = randStart + 0.01;
      randEnd = round(randEnd);
 
      Interval newInterval =  new Interval(randStart, randEnd);
      int index = binarySearch(intervals, newInterval);
      mergeIntervals(intervals, index, newInterval);
 
      if (intervals.size() == 1 && intervals.get(0).start <= 0.0 && intervals.get(0).end >= 1.0) {
        break;
      }
      else {
        cnt++;
      }
    }
    return cnt;
  }
 
  private int binarySearch(List<Interval> intervals, Interval newInterval) {
    for (int i = 0; i < intervals.size(); i++) {
      if (newInterval.start <= intervals.get(i).start) {
        return i;
      }
    }
    return intervals.size();
  }
 
  private void mergeIntervals(List<Interval> intervals, int index, Interval newInterval) {
    Interval curr = newInterval;
    if (intervals.isEmpty()) {
      intervals.add(newInterval);
    }
    else if (index == 0) {
      if (newInterval.start == intervals.get(0).start) {
        return;
      }
      else if (round(newInterval.end + 0.01) >= intervals.get(0).start) {
        intervals.get(0).start = Math.min(intervals.get(0).start, newInterval.start);
      }
      else {
        intervals.add(0, newInterval);
      }
    }
    else if (index == intervals.size()) {
      Interval last = intervals.get(intervals.size() - 1);
      if (round(last.end + 0.01) >= newInterval.start) {
        last.end = Math.max(last.end, newInterval.end);
      }
      else {
        intervals.add(newInterval);
      }
    }
    else {
      Interval prev = intervals.get(index - 1);
      Interval next = intervals.get(index);
 
      if (round(prev.end + 0.01) < newInterval.start && round(newInterval.end + 0.01) < next.start) {
        intervals.add(index, newInterval);
      }
      else {
        if (round(prev.end + 0.01) >= newInterval.start && round(newInterval.end + 0.01) >= next
            .start) {
          prev.end = next.end;
          intervals.remove(next);
        }
        else if (round(prev.end + 0.01) >= newInterval.start) {
          prev.end = Math.max(prev.end, newInterval.end);
          if (round(prev.end + 0.01) >= next.start) {
            prev.end = next.end;
            intervals.remove(next);
          }
        }
        else if (round(newInterval.end + 0.01) >= next.start) {
          next.start = Math.min(newInterval.start, next.start);
          if (round(prev.end + 0.01) >= next.start) {
            prev.end = next.end;
            intervals.remove(next);
          }
        }
        else {
          intervals.add(index, newInterval);
        }
      }
    }
  }
 
  private double round(double d) {
    return (double) Math.round(d * 100) / 100;
  }
 
  private class Interval {
    double start;
    double end;
 
    Interval(double start, double end) {
      this.start = start;
      this.end = end;
    }
  }
}
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