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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 givenindex
to be equal toval
.int snap()
takes a snapshot of the array and returns thesnap_id
: the total number of times we calledsnap()
minus1
.int get(index, snap_id)
returns the value at the givenindex
, at the time we took the snapshot with the givensnap_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 toset
,snap
, andget
. 0 <= index < length
0 <= snap_id <
(the total number of times we callsnap()
)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];
}
};
Solution 3. Binary Search
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];
}
};