leetcode_1146
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
Solutions
binary search
class SnapshotArray {
public:
vector<vector<pair<int, int>>> m;
int snapid = 0;
SnapshotArray(int length) : m(length, {{0, 0}}) {
}
void set(int index, int val) {
auto & v = m[index];
if (v.back().first != snapid)
v.emplace_back(snapid, val);
else
v.back().second = val;
}
int snap() {
return snapid++;
}
int get(int index, int snap_id) {
auto & vp = m[index];
pair<int, int> target = {snap_id, INT_MAX};
auto find = lower_bound(vp.begin(), vp.end(), target);
if (find != vp.end() && find->first == snap_id)
return find->second;
else
return prev(find)->second;
}
};
/**
* 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);
*/
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