LeetCode #1146 — MEDIUM

Snapshot Array

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

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 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 <= 5 * 10⁴
0 <= index < length
0 <= val <= 10⁹
0 <= snap_id < (the total number of times we call snap())
At most 5 * 10⁴ calls will be made to set, snap, and get.

Step 01

Brute Force Baseline

Problem summary: 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

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map · Binary Search · Design

Example 1

["SnapshotArray","set","snap","set","get"]
[[3],[0,5],[],[0,6],[0,0]]

Step 02

Core Insight

What unlocks the optimal approach

Use a list of lists, adding both the element and the snap_id to each index.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1146: Snapshot Array
class SnapshotArray {
    private List<int[]>[] arr;
    private int idx;

    public SnapshotArray(int length) {
        arr = new List[length];
        Arrays.setAll(arr, k -> new ArrayList<>());
    }

    public void set(int index, int val) {
        arr[index].add(new int[] {idx, val});
    }

    public int snap() {
        return idx++;
    }

    public int get(int index, int snap_id) {
        int l = 0, r = arr[index].size();
        while (l < r) {
            int mid = (l + r) >> 1;
            if (arr[index].get(mid)[0] > snap_id) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        --l;
        return l < 0 ? 0 : arr[index].get(l)[1];
    }
}

/**
 * Your SnapshotArray object will be instantiated and called as such:
 * SnapshotArray obj = new SnapshotArray(length);
 * obj.set(index,val);
 * int param_2 = obj.snap();
 * int param_3 = obj.get(index,snap_id);
 */

// Accepted solution for LeetCode #1146: Snapshot Array
type SnapshotArray struct {
	arr [][][2]int
	i   int
}

func Constructor(length int) SnapshotArray {
	return SnapshotArray{make([][][2]int, length), 0}
}

func (this *SnapshotArray) Set(index int, val int) {
	this.arr[index] = append(this.arr[index], [2]int{this.i, val})
}

func (this *SnapshotArray) Snap() int {
	this.i++
	return this.i - 1
}

func (this *SnapshotArray) Get(index int, snap_id int) int {
	i := sort.Search(len(this.arr[index]), func(i int) bool { return this.arr[index][i][0] > snap_id }) - 1
	if i < 0 {
		return 0
	}
	return this.arr[index][i][1]
}

/**
 * Your SnapshotArray object will be instantiated and called as such:
 * obj := Constructor(length);
 * obj.Set(index,val);
 * param_2 := obj.Snap();
 * param_3 := obj.Get(index,snap_id);
 */

# Accepted solution for LeetCode #1146: Snapshot Array
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)

// Accepted solution for LeetCode #1146: Snapshot Array
type Pair = (usize, i32);

struct SnapshotArray {
    size: usize,
    data: Vec<Vec<Pair>>,
    snap_id: usize,
}

impl SnapshotArray {
    fn new(length: i32) -> Self {
        let size = length as usize;
        let data = vec![vec![(0, 0)]; size];
        let snap_id = 0;
        SnapshotArray {
            size,
            data,
            snap_id,
        }
    }

    fn set(&mut self, index: i32, val: i32) {
        let i = index as usize;
        if let Some(last) = self.data[i].pop() {
            if last.0 != self.snap_id {
                self.data[i].push(last);
            }
        }
        self.data[i].push((self.snap_id, val));
    }

    fn snap(&mut self) -> i32 {
        self.snap_id += 1;
        (self.snap_id - 1) as i32
    }

    fn get(&self, index: i32, snap_id: i32) -> i32 {
        let i = index as usize;
        let snap_id = snap_id as usize;
        match self.data[i].binary_search_by_key(&snap_id, |p| p.0) {
            Ok(j) => self.data[i][j].1,
            Err(j) => self.data[i][j - 1].1,
        }
    }
}

#[test]
fn test() {
    let mut obj = SnapshotArray::new(3);
    obj.set(0, 5);
    assert_eq!(obj.snap(), 0);
    obj.set(0, 6);
    assert_eq!(obj.get(0, 0), 5);
}

// Accepted solution for LeetCode #1146: Snapshot Array
class SnapshotArray {
    private arr: [number, number][][];
    private i: number = 0;
    constructor(length: number) {
        this.arr = Array.from({ length }, () => []);
    }

    set(index: number, val: number): void {
        this.arr[index].push([this.i, val]);
    }

    snap(): number {
        return this.i++;
    }

    get(index: number, snap_id: number): number {
        let [l, r] = [0, this.arr[index].length];
        while (l < r) {
            const mid = (l + r) >> 1;
            if (this.arr[index][mid][0] > snap_id) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        --l;
        return l < 0 ? 0 : this.arr[index][l][1];
    }
}

/**
 * Your SnapshotArray object will be instantiated and called as such:
 * var obj = new SnapshotArray(length)
 * obj.set(index,val)
 * var param_2 = obj.snap()
 * var param_3 = obj.get(index,snap_id)
 */

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(log n)

Space

O(1)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.