What is the time complexity of Snapshot Array?

The optimal solution has a time complexity of O(log(snap_id)).

What is the space complexity of Snapshot Array?

The space complexity is O(length * snap_id).

What data structures are used to solve Snapshot Array?

This problem uses Array, Hash Table, Binary Search, Design. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Snapshot Array?

Insight 1: The core idea is to not store the entire array for each snapshot, as this would lead to excessive memory usage. Instead, we only store the changes (index and value) that occur between snapshots. Insight 2: Using a list of (snap_id, value) pairs for each index allows efficient retrieval of values at specific snapshot IDs using binary search.

Which companies ask LeetCode 1146. Snapshot Array in interviews?

This problem is frequently asked at Coupang, Databricks, Nvidia, Rubrik, Snowflake and 2 more companies.

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Snapshot Array - LeetCode 1146 Solution

Snapshot Array - Complete Solution Guide

Snapshot Array is LeetCode problem 1146, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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 . 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 s

Detailed Explanation

The problem requires you to implement a `SnapshotArray` class that simulates taking snapshots of an array over time. The array is initialized with a given length and all elements are initially 0. The `set(index, val)` method allows you to update the value at a specific index. The `snap()` method takes a snapshot of the current array state and returns a unique `snap_id` (representing the snapshot number). Finally, the `get(index, snap_id)` method allows you to retrieve the value at a given index at a specific snapshot ID.

Solution Approach

The solution uses a list (or vector) of lists (or vectors). For each index in the array, we store a list of (snap_id, value) pairs. Whenever `set` is called, we append a new (snap_id, value) pair to the list at the given index, but only if the current snap_id is different from the last snap_id in the list. If it's the same snap_id, we overwrite the value. When `snap` is called, we simply increment a counter (`snap_id`) to keep track of the current snapshot number. When `get` is called, we perform a binary search on the list of snap IDs for the given index to find the most recent value set before or at the requested snap ID.

Step-by-Step Algorithm

Step 1: Initialize the `SnapshotArray` with a given length. For each index in the array, create an empty list (or vector) to store (snap_id, value) pairs. Initialize the `snap_id` to 0.
Step 2: When `set(index, val)` is called, retrieve the list of (snap_id, value) pairs for the given index.
Step 3: Check if the current `snap_id` is the same as the last snap_id in the list. If it is, update the last value in the list. Otherwise, append a new (snap_id, value) pair to the list.
Step 4: When `snap()` is called, increment the `snap_id` and return the previous `snap_id`.
Step 5: When `get(index, snap_id)` is called, retrieve the list of (snap_id, value) pairs for the given index.
Step 6: Perform a binary search on the list of snap IDs to find the largest snap ID that is less than or equal to the given `snap_id`.
Step 7: Return the value corresponding to the found snap ID.

Key Insights

Insight 1: The core idea is to not store the entire array for each snapshot, as this would lead to excessive memory usage. Instead, we only store the changes (index and value) that occur between snapshots.
Insight 2: Using a list of (snap_id, value) pairs for each index allows efficient retrieval of values at specific snapshot IDs using binary search.
Insight 3: The binary search is crucial to achieve the O(log(snap_id)) time complexity for the `get` operation. We must find the latest snapshot ID (<= the target snap_id) for which a value was set at a specific index.

Complexity Analysis

Time Complexity: O(log(snap_id))

Space Complexity: O(length * snap_id)

Topics

This problem involves: Array, Hash Table, Binary Search, Design.

Companies

Asked at: Coupang, Databricks, Nvidia, Rubrik, Snowflake, StackAdapt, Verkada.

Snapshot Array | LeetCode 1146

Medium

Array Hash Table Binary Search Design

Time: O(log(snap_id))Space: O(length * snap_id)

Solution Code

Available in:

import java.util.ArrayList;
import java.util.List;

class SnapshotArray {

    private List<List<Integer>> snap_ids;
    private List<List<Integer>> values;
    private int snap_id;

    public SnapshotArray(int length) {
        snap_ids = new ArrayList<>(length);
        values = new ArrayList<>(length);
        for (int i = 0; i < length; i++) {
            List<Integer> snapIdList = new ArrayList<>();
            snapIdList.add(0);
            snap_ids.add(snapIdList);

            List<Integer> valueList = new ArrayList<>();
            valueList.add(0);
            values.add(valueList);
        }
        snap_id = 0;
    }

    public void set(int index, int val) {
        List<Integer> snapIdList = snap_ids.get(index);
        List<Integer> valueList = values.get(index);

        if (snapIdList.get(snapIdList.size() - 1) == snap_id) {
            valueList.set(valueList.size() - 1, val);
        } else {
            snapIdList.add(snap_id);
            valueList.add(val);
        }
    }

    public int snap() {
        int current_snap_id = snap_id;
        snap_id++;
        return current_snap_id;
    }

    public int get(int index, int snap_id) {
        List<Integer> snapIdList = snap_ids.get(index);
        List<Integer> valueList = values.get(index);

        int i = binarySearchRight(snapIdList, snap_id);
        return valueList.get(i - 1);
    }

    private int binarySearchRight(List<Integer> arr, int target) {
        int low = 0;
        int high = arr.size();
        while (low < high) {
            int mid = low + (high - low) / 2;
            if (arr.get(mid) <= target) {
                low = mid + 1;
            } else {
                high = mid;
            }
        }
        return low;
    }
}

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Problem Info

Description

Problem 1146: Snapshot Array - Detailed explanation and solution approaches.

Topics

Array Hash Table Binary Search Design

Companies

CoupangDatabricksNvidiaRubrikSnowflakeStackAdaptVerkada

Stats

Difficulty:Medium

Languages:4

Companies:7

Problem #:1146

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

7 steps

1
Step 1: Initialize the `SnapshotArray` with a given length. For each index in the array, create an empty list (or vector) to store (snap_id, value) pairs. Initialize the `snap_id` to 0.
2
Step 2: When `set(index, val)` is called, retrieve the list of (snap_id, value) pairs for the given index.
3
Step 3: Check if the current `snap_id` is the same as the last snap_id in the list. If it is, update the last value in the list. Otherwise, append a new (snap_id, value) pair to the list.
4
Step 4: When `snap()` is called, increment the `snap_id` and return the previous `snap_id`.
5
Step 5: When `get(index, snap_id)` is called, retrieve the list of (snap_id, value) pairs for the given index.
6
Step 6: Perform a binary search on the list of snap IDs to find the largest snap ID that is less than or equal to the given `snap_id`.
7
Step 7: Return the value corresponding to the found snap ID.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(log(snap_id))

🚀 Logarithmic - Very efficient! Halves the problem each step.

💾

Space Complexity

O(length * snap_id)

📝

Why?

The `get` operation uses binary search on a list of snapshot IDs, so its time complexity is O(log n) where n is the number of snapshots. The set and snap operations usually take constant time to execute. The space complexity arises from storing a log of changes for each index in the array. In the worst case, every set operation will create a new entry if called with unique snap_ids, and we will have a unique snap id for each snapshot.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Using a Map for each index

Instead of a list of pairs, each index could store a map (TreeMap in Java) where the key is the snap_id and the value is the value at that snapshot. Then, get() would use `floorKey()` (or similar functions) to find the largest snap_id <= the given snap_id.

Time: O(log(snap_id)) for `get` (due to map lookup). O(1) for `set` and `snap`.Space: O(length * snap_id)

Trade-offs: The TreeMap approach might be slightly easier to implement and understand, but it has the same time and space complexity as the list with binary search approach. TreeMap lookups may have slightly different constant factor compared to binary search over an ArrayList depending on language implementations.

✨

Key Insights

1Insight 1: The core idea is to not store the entire array for each snapshot, as this would lead to excessive memory usage. Instead, we only store the changes (index and value) that occur between snapshots.
2Insight 2: Using a list of (snap_id, value) pairs for each index allows efficient retrieval of values at specific snapshot IDs using binary search.
3Insight 3: The binary search is crucial to achieve the O(log(snap_id)) time complexity for the `get` operation. We must find the latest snapshot ID (<= the target snap_id) for which a value was set at a specific index.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

Edge Cases

Edge case 1: Initial value of elements. The problem specifies that the array is initialized with all elements set to 0. The provided solutions initialize the data structure accordingly.
Edge case 2: `snap_id` is 0 in `get`. The binary search correctly handles the case where we're trying to get the initial snapshot value.
Edge case 3: `snap_id` is greater than the maximum snap_id recorded for a specific index. The binary search ensures that we find the closest valid snap_id and return the corresponding value.
Edge case 4: No `set` operation has been performed for an index before a `get` call with a non-zero snap_id. The initial value (0) should be returned in this case.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Storing the entire array for each snapshot. This approach leads to extremely high memory usage and violates the constraints.

✗Common mistake 2: Not using binary search for the `get` operation. A linear search would result in O(snap_id) time complexity, which is too slow.

✗Common mistake 3: Failing to handle the edge case where no `set` operation has been performed for a specific index before a `get` call.

✗Common mistake 4: Incorrect binary search implementation, leading to wrong results or infinite loops.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Version control systems like Git. Snapshots of files are taken, and you can retrieve the state of a file at any given snapshot (commit).

🏦Real-world scenario 2: Database systems. Databases often use snapshot isolation to provide consistent views of the data for long-running transactions. A transaction sees the data as it was at a particular point in time (snapshot).

🎮Real-world scenario 3: Game development. Saving game states at different checkpoints allows players to revert to previous states.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: August 6, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 13

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

What are the trade-offs of using a hash table?

Hash tables provide O(1) average-case lookup, insertion, and deletion, making them ideal for frequency counting and duplicate detection. However, they use O(n) extra space and may degrade to O(n) time in worst-case scenarios with hash collisions. Always consider if the space trade-off is acceptable.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Snapshot Array"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether hash table techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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