Leetcode #1146: Snapshot Array
In this guide, we solve Leetcode #1146 Snapshot Array in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Design, Array, Hash Table, Binary Search
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
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
Python Solution
class SnapshotArray:
def __init__(self, length: int):
self.arr = [[] for _ in range(length)]
self.i = 0
def set(self, index: int, val: int) -> None:
self.arr[index].append((self.i, val))
def snap(self) -> int:
self.i += 1
return self.i - 1
def get(self, index: int, snap_id: int) -> int:
i = bisect_left(self.arr[index], (snap_id, inf)) - 1
return 0 if i < 0 else self.arr[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)
Complexity
The time complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.