Snapshot - Google


https://docs.google.com/document/d/1qxA2wps0IhVRWULulQ55W4SGPMu2AE5MkBB37h8Dr58


实现三个functions, get, set, take snapshots。其实就是一个长度为N的Array。Set可以设置Index i的值,每次take snapshot, version + 1,并且记录下当前version下 Array里面的值。然后get方法可以得到某一个Version下,每一个Index的Array的值。就是非常Naive的方法,在Chromebook上写完了。写完之后有一个变量名Typo被指出。口头跑了Test case。Follow up 时空复杂度,并且要节省空间。
举例
初始 100001
take a snapshot 返回sid 1
改变成 100201.
take a snapshot 返回sid 2
这时lookup get(3,1)( get(index, snapshotID))应该返回0而不是2,这是version control的原理可是关键在怎么存最省空间

可以参考视频压缩,每隔若干帧保存一份完整的图像,期间只保存delta。重建图像的时候,从最近的full snapshot开始,然后apply 每个version的delta

类似LC 981? - Google Snap (高频)


第二轮是面经高频题,Google Snapshot
就是一个array,然后这个array有不同的版本数,然后有不同的函数,比如set(int index, int val),get(int index, int version), snapshot()。比如说我先
set(1,1),set(0,0),set(2,2),这样我有一个[0,1,2],然后snapshot()之后系统里会记住version 1的array是[0,1,2],然后set(1,0),array就变成[0,0,2],这个时候如果get(1,1)得到的是1,但get(1,2)得到的是0。
思路:
  1. 模拟过程,用map存下snapshot即可,主要要和面试官交流得到各种异常的处理方式
包括数组长度是否固定,idx是否一定合法,version不存在是返回最近的还是抛出异常等等

2. hash map嵌套treemap,牺牲了一些时间,但是如果一些idx在某些version一直没用过可以节省空间


code:
Provider: null
class Snapshot {
int version = 1;
int length;
Map<Integer, TreeMap<Integer, Integer>> map = new HashMap<>();
public Snapshot(int length) {
this.length = length;
for(int i = 0 ; i < length ; i++) {
map.put(i, new TreeMap<>());
}
}
public void set(int idx, int val) {
// sanity check and exception process
map.get(idx).put(version, val);
}
public int get(int idx, int version) {
// sanity check and exception process
Integer key = map.get(idx).floorKey(version);
if(key == null) return Integer.MIN_VALUE;
return map.get(idx).get(key);
}
public void snapshot() {
version++;
}
}

假如我在[0,0,2]再叫snapshot(), 那[0,1,2] 的reference 就會不見了?
这里用了treeMap。题意中是否要求如果输入了不存在的版本号,那么就返回这个版本号的lower_bound?如果没有这个要求那么treeMap是能用hashMap取代的

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