LeetCode Snapshot Array Solution

Implement a SnapshotArray that supports the following interface:

  • SnapshotArray(int length) initializes an array-like data structure with the given length. Initially, each element equals 0.
  • void set(index, val) sets the element at the given index to be equal to val.
  • int snap() takes a snapshot of the array and returns the snap_id: the total number of times we called snap() minus 1.
  • int get(index, snap_id) returns the value at the given index, at the time we took the snapshot with the given snap_id

Example 1:

Input: ["SnapshotArray","set","snap","set","get"]

[[3],[0,5],[],[0,6],[0,0]]

Output: [null,null,0,null,5]

Explanation:

1SnapshotArray snapshotArr = new SnapshotArray(3); // set the length to be 3
2
3snapshotArr.set(0,5);  // Set array[0] = 5
4
5snapshotArr.snap();  // Take a snapshot, return snap_id = 0
6
7snapshotArr.set(0,6);
8
9snapshotArr.get(0,0);  // Get the value of array[0] with snap_id = 0, return 5

Constraints:

  • 1 <= length <= 5 * 104
  • 0 <= index < length
  • 0 <= val <= 109
  • 0 <= snap_id < (the total number of times we call snap())
  • At most 5 * 104 calls will be made to set, snap, and get.

Problem Link: https://leetcode.com/problems/snapshot-array/




Solution

We wish to find the pos for the most recent value at the time we took the snapshot with the given snap_id, we are trying to find the rightmost index in history=histories[i] such that the snap_id at history[pos] is less or equal to the target snap_id (a[i][0] <= snap_id). This means that the feasible function is a[i][0] <= snap_id, whenever this is true, we must check the positions on its right to find the rightmost position that makes this condition hold.

Implementation

1class SnapshotArray(object):
2
3def __init__(self, n):
4    # set up histories so that each index has its own history
5    self.histories = [[[-1, 0]] for _ in range(n)]
6    self.snap_id = 0
7
8def set(self, index, val):
9    self.histories[index].append([self.snap_id, val])
10
11def snap(self):
12    self.snap_id += 1
13    return self.snap_id - 1
14
15def get(self, index, snap_id):
16    left, right, pos = 0, len(self.histories[index])-1, -1
17    while left <= right:
18        mid = (left+right) // 2
19        if self.histories[index][mid][0] <= snap_id:
20            left = mid + 1
21            pos = mid
22        else:
23            right = mid - 1
24    return self.histories[index][pos][1]