LeetCode 1146. Snapshot Array

Description

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

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 <= 50000
  • At most 50000 calls will be made to setsnap, and get.
  • 0 <= index < length
  • 0 <= snap_id < (the total number of times we call snap())
  • 0 <= val <= 10^9

Explanation

Use a hashmap to store the current array arrangements and use a list to store snap shots.

Python Solution

class SnapshotArray:

    def __init__(self, length: int):        
        self.snap_shots = []
        self.current = {}
        
    def set(self, index: int, val: int) -> None:
        self.current[index] = val        

    def snap(self) -> int:
        self.snap_shots.append(dict(self.current))
        
        return len(self.snap_shots) - 1

    def get(self, index: int, snap_id: int) -> int:
        snap_shot = self.snap_shots[snap_id]
        
        if index in snap_shot:
            return snap_shot[index]
        
        return 0


# 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)
  • Time Complexity: O(1).
  • Space Complexity: O(N).

where Q is the length of queries and N is the length of colors.

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