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Formatted question description: https://leetcode.ca/all/1146.html

# 1146. Snapshot Array (Medium)

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 <= 50000
• At most 50000 calls will be made to set, snap, and get.
• 0 <= index < length
• 0 <= snap_id < (the total number of times we call snap())
• 0 <= val <= 10^9

Related Topics:
Array

## Solution 1. Brute Force

// OJ: https://leetcode.com/problems/snapshot-array/
// Time:
//      SnapshotArray, set, snap: O(1)
//      get: O(S) where S is the number of snapshots taken.
// Space: O(NS)
class SnapshotArray {
vector<unordered_map<int, int>> snapshot;
public:
SnapshotArray(int N) : snapshot(1) {}

void set(int index, int val) {
snapshot.back()[index] = val;
}

int snap() {
snapshot.emplace_back();
return snapshot.size() - 2;
}

int get(int index, int snap_id) {
for (int i = snap_id; i >= 0; --i) {
if (snapshot[i].count(index) == 0) continue;
return snapshot[i][index];
}
return 0;
}
};


## Solution 2. Brute Force

// OJ: https://leetcode.com/problems/snapshot-array/
// Time:
//      SnapshotArray: O(N)
//      set: O(1)
//      snap: O(N)
//      get: O(logS) where S is the number of snapshots taken
class SnapshotArray {
vector<vector<int>> snapIds; // snapIds[i] is a list of snapshot IDs related to the i-th element in array.
vector<unordered_map<int, int>> snapshot; // snapshot[i] contains all the values changed between snapshot i-1 and i
public:
SnapshotArray(int N) : snapshot(1), snapIds(N) {}

void set(int index, int val) {
snapshot.back()[index] = val;
}

int snap() {
int id = snapshot.size() - 1;
for (auto &[num, val] : snapshot.back()) snapIds[num].push_back(id);
snapshot.emplace_back();
return id;
}

int get(int index, int snap_id) {
auto &snaps = snapIds[index];
int i = upper_bound(begin(snaps), end(snaps), snap_id) - begin(snaps) - 1;
return i == -1 ? 0 : snapshot[snaps[i]][index];
}
};


The brute force solutions log all the snapshots in a single list.

To improve it, we make each A[i] have its own lists tracking all the snapshots related to it and the corresponding values.

// OJ: https://leetcode.com/problems/snapshot-array/
// Time:
//      SnapshotArray: O(N)
//      set: O(1)
//      snap: O(1)
//      get: O(logS)
// Space: O(NS)
class SnapshotArray {
vector<vector<int>> ids, vals; // If A[i] changed in between snapshot j-1 and j, we append j to ids[i], and append the changed value to vals[i].
int id = 0;
public:
SnapshotArray(int N) : ids(N, {-1}), vals(N, {0}) {}
void set(int i, int val) {
if (ids[i].back() != id) {
ids[i].push_back(id);
vals[i].push_back(val);
} else vals[i].back() = val;
}
int snap() {
return id++;
}
int get(int i, int snap_id) {
return vals[i][upper_bound(begin(ids[i]), end(ids[i]), snap_id) - begin(ids[i]) - 1];
}
};