1146. Snapshot Array

Problem Link

Description

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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: 
SnapshotArray snapshotArr = new SnapshotArray(3); // set the length to be 3
snapshotArr.set(0,5);  // Set array[0] = 5
snapshotArr.snap();  // Take a snapshot, return snap_id = 0
snapshotArr.set(0,6);
snapshotArr.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.

Solution

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Python3

class SnapshotArray:
 
    def __init__(self, length: int):
        self.N = length
        self.A = [[[-1, 0]] for _ in range(self.N)]
        self.curr = 0
 
    def set(self, index: int, val: int) -> None:
        if self.A[index][-1][0] != self.curr:
            self.A[index].append([self.curr, val])
        else:
            self.A[index][-1][1] = val
 
    def snap(self) -> int:
        self.curr += 1
 
        return self.curr - 1
 
    def get(self, index: int, snap_id: int) -> int:
        i = bisect_right(self.A[index], [snap_id + 1, ]) - 1
 
        return self.A[index][i][1]
 
 
# Your SnapshotArray object will be instantiated and called as such:
# obj = SnapshotArray(length)
# obj.set(index,val)
# param_2 = obj.snap()
# param_3 = obj.get(index,snap_id)