Tuesday, August 9, 2016

Google – Toggle Problem


Google – Toggle Problem
类似开关灯泡的题。要求实现
isToggled(long idx)
toggle(long start, long end)
这题难点在于有无限数量的灯泡,所以不能另开数据结构来存每个灯泡的状态。
[Solution]
[Solution #1]
Merge Interval
维护一个sorted list of interval表示已经toggled的灯泡。每次call toggle(start, end), 就生成一个新的Inveral [start, end], 扫描整个list,把和新interval重叠的部分删掉,把没有重叠的部分加进sorted list就可以了。
Query的时候就看idx是不是在list里面。
Pros: Query time O(k), where k is the number of toggled bubbles.
Cons: A little bit hard to implement
[Solution #2]
Bit Manipulation
因为这道题说了是无限量的灯泡,所以bit manipulation其实没法解。但是如果数量有限且在可接受范围之内的话,bit manipulation是个不错的解法。就有几个灯泡就开几个bit然后toggle就好了。时间空间复杂度都还可以。
[Solution #3]
Interval Tree
Interval Tree是一个BST, 每个node是记录一个range [low, high],BST结构根据low来维护。除了记录range,每个node还记录一个maxValue表示以该结点为root的整个subtree的最大high值。
Insert node的时候不需要像segment tree那样考虑partial overlap的情况,如果不是complete overlap, 都作为一个新的node加入到bst,沿路更新每个node的maxValue就好了。
那么这道题就是维护一个interval tree, 每次call toggle()就加入新node,query的时候就计算包含idx的node有多少个。
class Solution {
   
  ITNode root;
   
  public boolean isToggled(long idx) {
    int[] count = {0};
    dfs(root, count, idx);
    return count[0] % 2 != 0;
  }
   
  private void dfs(ITNode root, int[] count, long idx) {
    if (root == null) {
      return;
    }
     
    if (root.interval.low <= idx && idx <= root.interval.high) {
      count[0]++;
    }
    dfs(root.left, count, idx);
    dfs(root.right, count, idx);
  }
   
  public void toggle(long start, long end) {
    if (root == null) {
      root = new ITNode(new Interval(start, end));
    }
    else {
      insert(root, new Interval(start, end));
    }
  }
   
  private ITNode insert(ITNode root, Interval interval) {
    if (root == null) {
      return new ITNode(interval);
    }
    if (interval.low < root.interval.low) {
      root.left = insert(root.left, interval);
    }
    else {
      root.right = insert(root.right, interval);
    }
  }
   
  private class ITNode {
    Interval interval;
     
    ITNode left;
    ITNode right;
     
    ITNode(Interval interval) {
      this.interval = interval;
    }
  }
   
  private class Interval {
    long low;
    long high;
     
    Interval(long low, long high) {
      this.low = low;
      this.high = high;
    }
  }
   
}
https://instant.1point3acres.com/thread/130512
有一排数量无限的object,每个object有两个状态,可以用true和false来表示,object的状态是可切换的,初始情况下所有object的状态都是false。让你设计一个class,实现两个方法:isToggled(long idx), toggle(long start, long end)。这题没记错的话之前面经里也出现过。我最后给出的方法是维护一个list of intervals, 每次执行toggle时都要新加入一个新interval,然后把这个新interval merge进已有的interval list里面,这个解法的问题在于实现起来非常复杂,因为做merge操作时你需要反转所有和新interval重叠的interval的状态,最后才把新interval里不重叠的部分加进list里面。最后代码也没写完,

但是写到中间我突然想起来之前看到一个面经里的比较类似的题,可以只记录每个object的toggle操作的次数,最后根据操作次数的奇偶来判断object的状态,如此一来就不需要实际记录状态,只需要维护一个balanced interval tree,每次toggle就把新interval加进tree里,执行isToggled(long idx)时计算所有包含给定idx的interval的数量就能决定对应的object的状态了。跟面试官提了提这个做法,面试官表示这就是他的解法,但是他觉得我给的解法也不错,查询的复杂度要比他的方法低,就是难实现。

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