How do you solve Snapshot Array on LeetCode?

Snapshot Array (LeetCode #1146) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n * s) where s is the number of snapshots space complexity. Key insight: Using a list of dictionaries allows for efficient storage of values with their corresponding snap_ids.

What is the brute force solution for Snapshot Array?

The brute force approach for Snapshot Array is Brute Force, with O(n²) time complexity and O(n * s) where s is the number of snapshots space complexity. The brute force approach involves storing the entire array state for each snapshot. Each time we take a snapshot, we create a copy of the current array. This is simple but inefficient in terms of space and time. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Snapshot Array?

Snapshot Array uses the following concepts: Array, Hash Table, Binary Search, Design. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Snapshot Array?

Think about how you can optimize space and time complexity. Explain your thought process clearly, especially when transitioning from brute force to optimal solutions. Practice dry runs of your code to ensure you understand how it works.

What are common mistakes when solving Snapshot Array?

Failing to handle cases where a snap_id has no corresponding value set. Not considering the need to find the largest snap_id less than the requested one.

#1146

Snapshot Array

Medium

ArrayHash TableBinary SearchDesignHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(n * s) where s is the number of snapshots	O(n * s) where s is the number of snapshots

💡

Intuition

Time O(n)Space O(n * s) where s is the number of snapshots

The optimal approach uses a list of dictionaries to store the values at each index along with their corresponding snap_ids. This allows for efficient retrieval without needing to copy the entire array for each snapshot.

⚙️

Algorithm

4 steps

1Step 1: Initialize an array of dictionaries to store values and their corresponding snap_ids.
2Step 2: For the 'set' operation, update the value at the specified index with the current snap_id.
3Step 3: For the 'snap' operation, increment the snap_id counter and return the current snap_id.
4Step 4: For the 'get' operation, retrieve the value from the dictionary for the specified index, using the largest snap_id that is less than or equal to the requested snap_id.

solution.py20 lines

1class SnapshotArray:
2    def __init__(self, length: int):
3        self.data = [{} for _ in range(length)]
4        self.snap_id = 0
5
6    def set(self, index: int, val: int) -> None:
7        self.data[index][self.snap_id] = val
8
9    def snap(self) -> int:
10        self.snap_id += 1
11        return self.snap_id - 1
12
13    def get(self, index: int, snap_id: int) -> int:
14        if snap_id in self.data[index]:
15            return self.data[index][snap_id]
16        # Find the largest snap_id less than the requested one
17        for i in range(snap_id, -1, -1):
18            if i in self.data[index]:
19                return self.data[index][i]
20        return 0

ℹ

Complexity note: The time complexity for 'set' and 'snap' operations is O(1), while 'get' can take O(n) in the worst case if we have to search through all previous snapshots. However, this is efficient given the constraints.

1Using a list of dictionaries allows for efficient storage of values with their corresponding snap_ids.
2The optimal solution avoids unnecessary copying of the entire array, saving both time and space.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.