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 :



title-img




                        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]
                    


View More Similar Problems

Delete a Node

Delete the node at a given position in a linked list and return a reference to the head node. The head is at position 0. The list may be empty after you delete the node. In that case, return a null value. Example: list=0->1->2->3 position=2 After removing the node at position 2, list'= 0->1->-3. Function Description: Complete the deleteNode function in the editor below. deleteNo

View Solution →

Print in Reverse

Given a pointer to the head of a singly-linked list, print each data value from the reversed list. If the given list is empty, do not print anything. Example head* refers to the linked list with data values 1->2->3->Null Print the following: 3 2 1 Function Description: Complete the reversePrint function in the editor below. reversePrint has the following parameters: Sing

View Solution →

Reverse a linked list

Given the pointer to the head node of a linked list, change the next pointers of the nodes so that their order is reversed. The head pointer given may be null meaning that the initial list is empty. Example: head references the list 1->2->3->Null. Manipulate the next pointers of each node in place and return head, now referencing the head of the list 3->2->1->Null. Function Descriptio

View Solution →

Compare two linked lists

You’re given the pointer to the head nodes of two linked lists. Compare the data in the nodes of the linked lists to check if they are equal. If all data attributes are equal and the lists are the same length, return 1. Otherwise, return 0. Example: list1=1->2->3->Null list2=1->2->3->4->Null The two lists have equal data attributes for the first 3 nodes. list2 is longer, though, so the lis

View Solution →

Merge two sorted linked lists

This challenge is part of a tutorial track by MyCodeSchool Given pointers to the heads of two sorted linked lists, merge them into a single, sorted linked list. Either head pointer may be null meaning that the corresponding list is empty. Example headA refers to 1 -> 3 -> 7 -> NULL headB refers to 1 -> 2 -> NULL The new list is 1 -> 1 -> 2 -> 3 -> 7 -> NULL. Function Description C

View Solution →

Get Node Value

This challenge is part of a tutorial track by MyCodeSchool Given a pointer to the head of a linked list and a specific position, determine the data value at that position. Count backwards from the tail node. The tail is at postion 0, its parent is at 1 and so on. Example head refers to 3 -> 2 -> 1 -> 0 -> NULL positionFromTail = 2 Each of the data values matches its distance from the t

View Solution →