Snapshot Array - Leetcode Solution

The Snapshot Array problem on LeetCode requires us to implement a data structure that supports two operations:

1. update(index, val) - Update the element at the given index to be val.
2. snap() - Take a snapshot of the array at its current state.

When we take a snapshot, we want to be able to efficiently store the current state of the array in memory so that we can access it later. One important consideration is that we want the snapshots to be able to capture changes made to the array after previous snapshots were taken.

To solve this problem, we can use an array of arrays, where each sub-array represents the state of the overall array at a particular snapshot. We can keep track of the current snapshot index and update the appropriate sub-array whenever we perform an update operation.

Here’s a step-by-step guide to implementing this solution:

1. Initialize our snapshot array with a single sub-array containing all 0’s.
2. Implement an update(index, val) function that updates the element at the given index to be val. We can simply update the value in the current sub-array at the given index.
3. Implement a snap() function that takes a snapshot of the current state of the array. We can create a new sub-array with the current state of the array by making a copy of the current sub-array and appending it to the snapshot array. We also need to increment the snapshot index to indicate that a new snapshot has been taken.
4. To retrieve the value of an element at a certain snapshot, we need to traverse the snapshot array backwards and find the closest snapshot that was taken before or at the desired index. We can start at the most recent snapshot and work backwards until we find the first snapshot with an index less than or equal to the desired index. We can then return the value of the element in that snapshot at the desired index.

Here is the code implementation for the above solution:

class SnapshotArray:
    def __init__(self, length: int):
        self.snapshots = [[0] * length]
        self.snapshot_index = 0
    
    def update(self, index: int, val: int) -> None:
        self.snapshots[self.snapshot_index][index] = val
    
    def snap(self) -> int:
        self.snapshots.append(self.snapshots[self.snapshot_index][:])
        self.snapshot_index += 1
        return self.snapshot_index - 1
    
    def get(self, index: int, snap_id: int) -> int:
        snapshot = self.snapshots[snap_id]
        for i in range(snap_id, -1, -1):
            if self.snapshots[i][index] != snapshot[index]:
                snapshot = self.snapshots[i]
                break
        return snapshot[index]

In this implementation, we use the self.snapshots list to store all the snapshots taken so far. We initialize it with a single sub-array containing all 0's. We also initialize the self.snapshot_index variable to 0, indicating that no snapshots have been taken yet.

The update() function simply updates the value in the current sub-array at the given index.

The snap() function takes a snapshot of the current state of the array and stores it as a new sub-array in self.snapshots. We first make a copy of the current sub-array by using the slice notation [:]. We then append the copy to self.snapshots and increment self.snapshot_index to indicate that a new snapshot has been taken. Finally, we return the index of the new snapshot, which is equal to self.snapshot_index - 1.

The get() function returns the value of the element at the given index in the given snapshot. We start by retrieving the snapshot array using the given snap_id. We then traverse self.snapshots backwards from the given snapshot index to 0 and find the closest snapshot that was taken before or at the desired index. We do this by checking if the value of the element at the given index has changed in each sub-array. If it has, we update our snapshot variable to the current snapshot and break out of the loop. Finally, we return the value of the element at the desired index in the snapshot array.

With this implementation, all operations take constant time except for the get() function, which takes O(n) time in the worst case, where n is the number of snapshots taken. However, since the number of snapshots is likely to be much smaller than the length of the array, this should not be a significant issue in practice.

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