Snapshottable List - Google Top Interview Questions
Problem Statement :
Implement a data structure with the following methods: SnapshottableList(int size) which instantiates the list with the given size. The initial value of each element is 0. void set(int idx, int val) which sets the value at index idx to val. int snapshot() which returns the number of times snapshot was called previously. int get(int idx, int snapshotId) which returns the value at index idx after we took snapshot snapshotId. You can assume that snapshotId is valid. Constraints 0 ≤ n ≤ 100,000 where n is the number of calls to set, snapshot and get 0 ≤ size ≤ 100,000 Example 1 Input methods = ["constructor", "set", "snapshot", "get", "set", "snapshot", "get"] arguments = [[2], [1, 1], [], [1, 0], [1, 100], [], [1, 1]]` Output [None, None, 0, 1, None, 1, 100]
Solution :
Solution in C++ :
class SnapshottableList {
vector<map<int, int>> v; // v[i] --> snap_id: val
int snaps = -1;
public:
SnapshottableList(int size) {
v.resize(size);
for (auto &x : v) {
x[-1] = 0;
}
}
void set(int idx, int val) {
v[idx][snaps] = val;
}
int snapshot() {
snaps++;
return snaps;
}
int get(int idx, int snapshotId) {
return (--(v[idx].lower_bound(snapshotId)))->second;
}
};
Solution in Java :
import java.util.*;
class SnapshottableList {
TreeMap<Integer, Integer>[] values;
int snapshotId = 0;
public SnapshottableList(int size) {
values = new TreeMap[size];
for (int i = 0; i < values.length; i++) {
values[i] = new TreeMap<>();
}
}
public void set(int idx, int val) {
values[idx].put(snapshotId, val);
}
public int snapshot() {
int ss = snapshotId;
snapshotId++;
return ss;
}
public int get(int idx, int snapshotId) {
Map.Entry<Integer, Integer> entry = values[idx].floorEntry(snapshotId);
return entry == null ? 0 : entry.getValue();
}
}
Solution in Python :
class SnapshottableList:
def __init__(self, size):
# for each index, store a list of (snapshot_id, val)
self.slist = [[] for _ in range(size)]
self.current_snapshot = 0
def set(self, idx, val):
l = self.slist[idx]
if l:
# if we already have a value for the current snapshot, overwrite it
if l[-1][0] == self.current_snapshot:
l[-1] = (self.current_snapshot, val)
return
# if the existing value is the same as the new value, no need to append a new value
if l[-1][1] == val:
return
l.append((self.current_snapshot, val))
def snapshot(self):
res = self.current_snapshot
self.current_snapshot += 1
return res
def get(self, idx, snapshotId):
l = self.slist[idx]
i = bisect_right(l, (snapshotId, float("inf")))
if i == 0:
# no snapshots <= snapshotId
return 0
return l[i - 1][1]
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